If A is a square matrix such that $$A^2 = A$$, what is the value of $$(I + A)^3 - 7A$$?

Correct Option: B. Solution: If $$A^2 = A$$, then $$(I + A)^3 - 7A = I - A$$.

Matrices

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Summary

Summary of Chapter 3: Matrices

Matrices are fundamental tools in mathematics, offering a compact and simplified method for solving systems of linear equations. Beyond their mathematical applications, matrices are utilized in various fields such as business, science, cryptography, genetics, economics, sociology, psychology, and industrial management. They are instrumental in operations like magnification, rotation, and reflection, and are used in electronic spreadsheets for tasks like budgeting and sales projection. This chapter introduces the fundamentals of matrix and matrix algebra, exploring their diverse applications and operations.

Learning Objectives

Understand the concept and definition of matrices.
Identify the various types of matrices, including symmetric and skew symmetric matrices.
Perform matrix operations such as addition, subtraction, and multiplication.
Apply the properties of matrix operations, including commutative, associative, and distributive laws.
Use matrices to solve systems of linear equations.
Explore the applications of matrices in different fields such as business, science, and cryptography.
Verify matrix identities using examples, such as $(AB)' = B'A'$ .
Calculate the inverse of a matrix and understand its uniqueness.
Express any square matrix as the sum of a symmetric and a skew symmetric matrix.
Prove properties of matrices using mathematical induction and other methods.

Detailed Notes

Matrices

Introduction

Matrices are essential in various branches of mathematics.
They simplify solving systems of linear equations and are used in electronic spreadsheets for business and science applications.
Matrices represent physical operations like magnification, rotation, and reflection, and are used in cryptography, genetics, economics, sociology, psychology, and industrial management.

Matrix Basics

A matrix is an ordered rectangular array of numbers or functions.
A matrix with $m$ rows and $n$ columns is of order $m \times n$ .
Types of matrices:
- Column Matrix: $[a_{ij}]_{m \times 1}$
- Row Matrix: $[a_{ij}]_{1 \times n}$
- Square Matrix: When $m = n$ .
- Diagonal Matrix: $a_{ij} = 0$ when $i \neq j$ .
- Scalar Matrix: $a_{ij} = 0$ when $i \neq j$ , $a_{ij} = k$ when $i = j$ .
- Identity Matrix: $a_{ij} = 1$ when $i = j$ , $a_{ij} = 0$ when $i \neq j$ .
- Zero Matrix: All elements are zero.

Matrix Operations

Addition: $A + B = B + A$
Multiplication: If $A = [a_{ij}]_{m \times n}$ and $B = [b_{jk}]_{n \times p}$ , then $AB = C = [c_{ik}]_{m \times p}$ .
Transpose: $A' = [a_{ji}]_{n \times m}$ $A^{'} = [a_{ji}]_{n \times m}$
- Properties:
  - $(A')' = A$
  - $(kA)' = kA'$
  - $(A + B)' = A' + B'$
  - $(AB)' = B'A'$

Special Matrices

Symmetric Matrix: $A' = A$
Skew Symmetric Matrix: $A' = -A$
Any square matrix can be represented as the sum of a symmetric and a skew symmetric matrix.

Invertible Matrices

A square matrix $A$ is invertible if there exists a matrix $B$ such that $AB = BA = I$ .
The inverse of a matrix, if it exists, is unique.
If $A$ and $B$ are invertible, then $(AB)^{-1} = B^{-1}A^{-1}$ .

Examples

Example 1: If $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$ , find $A'$ , $A^{-1}$ , and verify $AA^{-1} = I$ .
Example 2: Given matrices $A$ and $B$ , show $(A + B)' = A' + B'$ .

Applications

Used in solving linear equations, business applications, physical transformations, and various scientific fields.

Exercises

Prove properties of transpose and inverse matrices.
Solve matrix equations and verify results using matrix operations.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Matrix Multiplication Order: Remember that matrix multiplication is not commutative, i.e., $AB \neq BA$ . Always pay attention to the order of multiplication.
Inverse of a Matrix: A rectangular matrix does not possess an inverse. Only square matrices can have inverses, and the inverse is unique if it exists.
Symmetric and Skew Symmetric Matrices: For any square matrix $A$ , $A + A'$ is symmetric, and $A - A'$ is skew symmetric. Ensure you understand these properties to avoid confusion in problems.
Transpose Properties: The transpose of a product of matrices $(AB)' = B'A'$ . This property is crucial in simplifying expressions involving transposes.
Matrix Equations: When solving matrix equations, ensure the dimensions are compatible for operations like addition, subtraction, and multiplication.
Verification of Properties: Always verify properties such as $(A + B)' = A' + B'$ and $(kA)' = kA'$ with examples to solidify understanding.
Zero and Identity Matrices: Be cautious with zero and identity matrices in operations as they have unique properties that can simplify or complicate problems if not handled correctly.
Checking for Invertibility: A matrix is invertible if $AB = BA = I$ . Ensure to verify this condition when asked about invertibility.
Use of Induction: For proving results involving matrices, such as powers of matrices, mathematical induction can be a powerful tool. Ensure you are comfortable with this method.
Exam Practice: Practice problems involving matrix operations, properties, and proofs as they are common in exams and can often be tricky if not well understood.

Practice & Assessment

Multiple Choice Questions

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}

\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}

Correct Answer: A

Solution:

The inverse of the identity matrix

I

is itself,

I

. Therefore, the inverse of

A

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Correct Answer: C

Solution:

The determinant of a 2x2 matrix

\begin{bmatrix} a & b \\ c & d \end{bmatrix}

is calculated as

ad - bc

. For matrix A, the determinant is

3 \times 3 - 0 \times 0 = 9

I - A

Correct Answer: B

Solution:

A^2 = A

, then

(I + A)^3 - 7A = I - A

Correct Answer: A

Solution:

The eigenvalues of

A

are found by solving

\det(A - \lambda I) = 0

. For

A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}

, the eigenvalues are

3

and

1

₹10,000

₹12,000

₹15,000

₹18,000

Correct Answer: D

Solution:

Let the amount invested in the first bond be ₹x. Then, the amount in the second bond is ₹(30,000 - x). The total interest is given by the equation:

0.05x + 0.07(30,000 - x) = 1800

. Solving,

0.05x + 2100 - 0.07x = 1800 \Rightarrow -0.02x = -300 \Rightarrow x = 15,000

. Therefore, the amount invested in the first bond is ₹15,000.

-1

-2

Correct Answer: C

Solution:

The determinant of

A

is calculated as

(2 \times 5) - (3 \times 4) = 10 - 12 = -2

1 TT : 2 Tt : 1 tt

3 TT : 1 tt

1 Tt : 2 TT : 1 tt

2 TT : 1 Tt : 1 tt

Correct Answer: A

Solution:

The genotypic ratio for a monohybrid cross in the F2 generation is 1 homozygous dominant (TT) : 2 heterozygous (Tt) : 1 homozygous recessive (tt).

\begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}

\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

Correct Answer: A

Solution:

The product

AB = \begin{bmatrix} 1 \times 0 + 2 \times 1 & 1 \times 1 + 2 \times 0 \\ 3 \times 0 + 4 \times 1 & 3 \times 1 + 4 \times 0 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}

\begin{bmatrix} 3 & 2 \\ \sqrt{3} & -1 \end{bmatrix}

\begin{bmatrix} 3 & \sqrt{3} \\ -1 & 2 \end{bmatrix}

\begin{bmatrix} 3 & -1 \\ \sqrt{3} & 2 \end{bmatrix}

\begin{bmatrix} 3 & \sqrt{3} \\ 2 & -1 \end{bmatrix}

Correct Answer: A

Solution:

The transpose of a matrix is obtained by swapping its rows and columns. Thus, the transpose of

A = \begin{bmatrix} 3 & \sqrt{3} \\ 2 & -1 \end{bmatrix}

\begin{bmatrix} 3 & 2 \\ \sqrt{3} & -1 \end{bmatrix}

Both A and B

Neither A nor B

Correct Answer: B

Solution:

The matrix B is symmetric, and for a symmetric matrix, the transpose is equal to the matrix itself. Therefore, if B is also its own inverse, then the transpose is equal to the inverse.

₹3000

₹3500

₹4000

₹4500

Correct Answer: A

Solution:

The total cost is calculated as follows:

1000 \times 0.40 + 500 \times 1.00 + 5000 \times 0.50 = 400 + 500 + 2500 = 3400

paise, which is ₹34.00. Therefore, the total cost incurred is ₹34.00.

₹5,000

₹10,000

₹15,000

₹20,000

Correct Answer: A

Solution:

Calculate the total revenue from sales and subtract the total cost. Revenue: ₹(2.50 \times 10,000 + 1.50 \times 10,000 + 1.00 \times 10,000) = ₹50,000. Cost: ₹(2.00 \times 10,000 + 1.00 \times 10,000 + 0.50 \times 10,000) = ₹45,000. Gross profit = ₹50,000 - ₹45,000 = ₹5,000.

All eigenvalues are zero

The eigenvalues are 0, 23, and 40

The eigenvalues are 23 and 40

The eigenvalues are the roots of the characteristic polynomial

Correct Answer: D

Solution:

The equation

A^3 - 23A - 40I = O

implies that the eigenvalues of A satisfy the characteristic equation

\lambda^3 - 23\lambda - 40 = 0

It is a symmetric matrix.

It is a skew symmetric matrix.

It is a diagonal matrix.

It is an identity matrix.

Correct Answer: B

Solution:

If A and B are symmetric matrices, then AB - BA is a skew symmetric matrix.

Correct Answer: B

Solution:

The trace of a matrix is the sum of the elements on its main diagonal. For matrix A, the trace is

1 + 4 = 5

\begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}

\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Matrix A does not have an inverse.

Correct Answer: A

Solution:

The inverse of a 2x2 matrix

\begin{bmatrix} a & b \\ c & d \end{bmatrix}

\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

. For matrix A, the inverse is

\frac{1}{1\times4 - 2\times3}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}

Correct Answer: B

Solution:

First, calculate

A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 7 & 18 \\ 6 & 19 \end{bmatrix}

. Then,

2A = 2 \times \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ 2 & 8 \end{bmatrix}

. Now,

C = \begin{bmatrix} 7 & 18 \\ 6 & 19 \end{bmatrix} - \begin{bmatrix} 4 & 6 \\ 2 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 12 \\ 4 & 12 \end{bmatrix}

. The determinant of

C

4 \times 12 - 4 \times 12 = 0

Correct Answer: C

Solution:

The matrix

A - B = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} - \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 0 & 4 \end{bmatrix}

. The determinant is calculated as

2\times4 - 2\times0 = 8

\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\begin{bmatrix} -2 & -3 \\ -1 & -4 \end{bmatrix}

\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}

Correct Answer: A

Solution:

First, calculate

A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 7 & 18 \\ 6 & 19 \end{bmatrix}

. Then, compute

5A = 5 \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 10 & 15 \\ 5 & 20 \end{bmatrix}

and

6B = 6 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}

. Therefore,

C = A^2 - 5A + 6B = \begin{bmatrix} 7 & 18 \\ 6 & 19 \end{bmatrix} - \begin{bmatrix} 10 & 15 \\ 5 & 20 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

Correct Answer: A

Solution:

The product

AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1\times2 + 2\times1 & 1\times0 + 2\times2 \\ 3\times2 + 4\times1 & 3\times0 + 4\times2 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 10 & 8 \end{bmatrix}

. The trace of a matrix is the sum of its diagonal elements:

4 + 8 = 12

\begin{bmatrix} -3 & -4 \\ 7 & 8 \end{bmatrix}

\begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}

Correct Answer: A

Solution:

To find

X

, solve the equation

X = B \cdot A^{-1}

. Calculate the inverse of

A

and then multiply it by

B

\begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \end{bmatrix}

\begin{bmatrix} 69 & -6 & 23 \\ 63 & 46 & 69 \end{bmatrix}

\begin{bmatrix} 46 & 69 & 63 \\ -6 & 23 & 69 \end{bmatrix}

\begin{bmatrix} 23 & 69 & 46 \\ 69 & 23 & -6 \end{bmatrix}

Correct Answer: A

Solution:

From the given excerpt,

A^3 = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \end{bmatrix}

True, the equation holds.

False, the equation does not hold.

Cannot be determined from the given information.

None of above is valid answer

Correct Answer: A

Solution:

Calculate

A^3

, then subtract

23A

and

40I

. The resulting matrix should be the zero matrix, confirming the equation holds.

\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}

\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}

Correct Answer: A

Solution:

The product

AB = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

is calculated by multiplying corresponding elements and summing them.

₹62,000

₹61,000

₹60,000

₹59,000

Correct Answer: A

Solution:

The total revenue from Market I is calculated as follows:

10,000 \times 2.50 + 2,000 \times 1.50 + 18,000 \times 1.00 = 25,000 + 3,000 + 18,000 = 46,000

. Therefore, the total revenue is ₹62,000.

\begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}

\begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}

\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}

\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

Correct Answer: A

Solution:

To find the sum of two matrices, add the corresponding elements. So,

A + B = \begin{bmatrix} 1+5 & 2+6 \\ 3+7 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}

3 Tall: 1 Dwarf

1 Tall: 3 Dwarf

2 Tall: 2 Dwarf

4 Tall: 0 Dwarf

Correct Answer: A

Solution:

The phenotypic ratio in the F2 generation is 3 Tall: 1 Dwarf as per Mendel's principles.

\begin{bmatrix} \frac{1}{5} & 0 \\ 0 & \frac{1}{5} \end{bmatrix}

\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\begin{bmatrix} 0 & 5 \\ 5 & 0 \end{bmatrix}

Correct Answer: A

Solution:

The inverse of a diagonal matrix

\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}

\begin{bmatrix} \frac{1}{a} & 0 \\ 0 & \frac{1}{b} \end{bmatrix}

. Therefore, the inverse of

A = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}

\begin{bmatrix} \frac{1}{5} & 0 \\ 0 & \frac{1}{5} \end{bmatrix}

Both option_a and option_b

Correct Answer: D

Solution:

The characteristic equation

A^2 - 5A + 6I = 0

can be factored as

(A - 2I)(A - 3I) = 0

. Therefore, the eigenvalues of

A

are 2 and 3.

AB is always symmetric

AB is always skew-symmetric

AB is symmetric if and only if A and B commute

AB is neither symmetric nor skew-symmetric

Correct Answer: C

Solution:

For AB to be symmetric, it must hold that (AB)' = AB. Since A and B are symmetric, this implies AB = BA.

Correct Answer: A

Solution:

First, calculate

C = A + B = \begin{bmatrix} 1+2 & 2+0 \\ 3+1 & 4+2 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 4 & 6 \end{bmatrix}

. The determinant of

C

3 \times 6 - 2 \times 4 = 18 - 8 = 10

\begin{bmatrix} 58 & 64 \\ 139 & 154 \end{bmatrix}

\begin{bmatrix} 58 & 64 \\ 139 & 150 \end{bmatrix}

\begin{bmatrix} 58 & 62 \\ 139 & 154 \end{bmatrix}

\begin{bmatrix} 58 & 64 \\ 140 & 154 \end{bmatrix}

Correct Answer: A

Solution:

The product

AB = \begin{bmatrix} 58 & 64 \\ 139 & 154 \end{bmatrix}

is calculated by multiplying corresponding elements and summing them.

Correct Answer: B

Solution:

The trace of a matrix is the sum of its diagonal elements. So, the trace of

A

3 + 3 = 6

\begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix}

\begin{bmatrix} 3 & 2 \\ 5 & 4 \end{bmatrix}

\begin{bmatrix} 5 & 4 \\ 3 & 2 \end{bmatrix}

\begin{bmatrix} 4 & 2 \\ 5 & 3 \end{bmatrix}

Correct Answer: A

Solution:

The transpose of a matrix is obtained by swapping its rows and columns. Thus, the transpose of

A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}

\begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix}

\begin{bmatrix} 2 & 8 \\ 6 & 15 \end{bmatrix}

\begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix}

\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}

\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

Correct Answer: A

Solution:

The product of two matrices is obtained by multiplying rows by columns. Thus,

AB = \begin{bmatrix} 2\times1 + 0\times2 & 2\times4 + 0\times5 \\ 0\times1 + 3\times2 & 0\times4 + 3\times5 \end{bmatrix} = \begin{bmatrix} 2 & 8 \\ 6 & 15 \end{bmatrix}

Matrix

A

is invertible.

Matrix

A

is a zero matrix.

Matrix

A

is a scalar matrix.

Matrix

A

is a singular matrix.

Correct Answer: D

Solution:

Matrix

A

is singular because its determinant is zero:

\det(A) = 2(-6) - (-3)(4) = -12 + 12 = 0

₹ 1200

₹ 7000

₹ 7100

₹ 7200

Correct Answer: D

Solution:

The total cost for city Y is calculated by multiplying the matrix A by the second row of matrix B:

40 \times 3000 + 100 \times 1000 + 50 \times 10000 = 7200

paise, which is ₹ 7200.

₹50,000

₹45,000

₹47,000

₹48,000

Correct Answer: D

Solution:

The total revenue in Market I is calculated as follows: (10,000 * 2.50) + (2,000 * 1.50) + (18,000 * 1.00) = ₹48,000.

AB = BA

A

and

B

are both diagonal matrices

A

B

is a zero matrix

AB = 0

Correct Answer: A

Solution:

For the product

AB

to be symmetric, it must hold that

AB = BA

A^2 = I

, the identity matrix.

A^2 = -I

, the negative identity matrix.

A^2 = A

A^2 = -A

Correct Answer: A

Solution:

Calculating

A^2

A^2 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

A is a diagonal matrix

A is a zero matrix

A is a square matrix

None of these

Correct Answer: B

Solution:

A matrix that is both symmetric (

A = A^T

) and skew-symmetric (

A = -A^T

) must satisfy

A = -A

, which implies

2A = 0

. Hence,

A

must be a zero matrix.

100

Correct Answer: A

Solution:

The genotypic ratio of the F2 generation is 1 TT : 2 Tt : 1 tt. The number of homozygous recessive plants (tt) is 1/4 of the total, which is 1/4 of 100 = 25 plants.

₹ 10,000 in 5% bonds and ₹ 20,000 in 7% bonds

₹ 15,000 in each bond

₹ 20,000 in 5% bonds and ₹ 10,000 in 7% bonds

₹ 12,000 in 5% bonds and ₹ 18,000 in 7% bonds

Correct Answer: A

Solution:

Let x be the amount invested in 5% bonds and y in 7% bonds. We have x + y = ₹ 30,000 and 0.05x + 0.07y = ₹ 2000. Solving these equations gives x = ₹ 10,000 and y = ₹ 20,000.

1:2:1

3:1

1:3

2:1:1

Correct Answer: A

Solution:

In Mendel's monohybrid cross, the F2 generation shows a genotypic ratio of 1 homozygous dominant (TT) : 2 heterozygous (Tt) : 1 homozygous recessive (tt), which corresponds to the phenotypic ratio of 3 tall : 1 dwarf.

True or False

Correct Answer: False

Solution:

The product of two symmetric matrices

A

and

B

is symmetric if and only if

A

and

B

commute, i.e.,

AB = BA

. Otherwise,

AB

is not necessarily symmetric.

Correct Answer: True

Solution:

Given

A^2 = A

, we have

(I + A)^3 = I + 3A + 3A^2 + A^3 = I + 3A + 3A + A = I + 7A

. Therefore,

(I + A)^3 - 7A = I

Correct Answer: True

Solution:

Mendel's monohybrid cross experiment results in a phenotypic ratio of 3 tall plants to 1 dwarf plant in the F2 generation.

Correct Answer: False

Solution:

A rectangular matrix does not possess an inverse because for the inverse to exist, the matrix must be square.

Correct Answer: True

Solution:

A matrix that is both symmetric and skew-symmetric must have all its elements as zero, because the only way for

A = A'

and

A = -A'

to hold simultaneously is if all elements of

A

are zero.

Correct Answer: False

Solution:

The correct formula for the inverse of a product of two invertible matrices is

(AB)^{-1} = B^{-1}A^{-1}

, not

A^{-1}B^{-1}

Correct Answer: True

Solution:

The matrix

F(x)

represents a rotation matrix. The property

F(x)F(y) = F(x + y)

holds for rotation matrices, as it reflects the addition of angles.

Correct Answer: True

Solution:

Theorem 3 states that the inverse of a square matrix, if it exists, is unique. This is because if

B

and

C

are both inverses of

A

, then

B = C

Correct Answer: True

Solution:

A matrix

A

is symmetric if

A' = A

, meaning the transpose of

A

is equal to

A

itself.

Correct Answer: True

Solution:

A

is symmetric, then

B'AB

is symmetric. This is because the symmetry of

A

ensures that the product

B'AB

retains symmetry.

Correct Answer: True

Solution:

For any square matrix

A

, the transpose of the transpose of

A

A

itself, i.e.,

(A')' = A

. This is a fundamental property of matrix transposition.

Correct Answer: True

Solution:

The given matrix

A

is such that its transpose

A'

is equal to

A

, hence

(A')' = A

holds true.

Correct Answer: True

Solution:

The transpose of a sum of matrices is the sum of their transposes, i.e.,

(A + B)' = A' + B'

. This property holds for any matrices

A

and

B

Correct Answer: True

Solution:

For symmetric matrices

A

and

B

, the product

AB

is symmetric if

AB = BA

. Therefore,

AB - BA

is skew-symmetric because

(AB - BA)' = - (AB - BA)

Correct Answer: True

Solution:

The solution provided shows that for

A = \begin{bmatrix} 3 & -2 & 1 \\ 4 & 2 & 1 \\ 1 & 0 & 1 \end{bmatrix}

A^3 - 23A - 40I = O

is verified.

Correct Answer: True

Solution:

The equation

A^3 - 23A - 40I = O

is a matrix equation that

A

satisfies, meaning that when

A

is substituted into the equation, it results in the zero matrix.

Correct Answer: True

Solution:

A matrix

A

is called idempotent if

A^2 = A

. Therefore, by definition, if

A^2 = A

A

is idempotent.

Correct Answer: True

Solution:

For

AB

to be symmetric, it must hold that

(AB)' = AB

. Since

A

and

B

are symmetric,

A' = A

and

B' = B

. Therefore,

(AB)' = B'A' = BA

. Thus,

AB

is symmetric if and only if

AB = BA

Correct Answer: True

Solution:

By definition, a zero matrix is one where all elements are zero.

Descriptive Questions

Expected Answer:

The total revenue for each market calculated using matrix multiplication.

Detailed Solution:

Represent the sales quantities and prices as matrices and perform matrix multiplication to find the total revenue for each market.

Expected Answer:

The total cost in city X is ₹340,000, and in city Y is ₹720,000. Matrices simplify the calculation of total costs by organizing data and operations efficiently, especially when dealing with large datasets.

Detailed Solution:

Compute

AB = \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix} \begin{bmatrix} 1000 & 3000 \\ 500 & 1000 \\ 5000 & 10000 \end{bmatrix} = \begin{bmatrix} 40 \times 1000 + 100 \times 500 + 50 \times 5000 & 40 \times 3000 + 100 \times 1000 + 50 \times 10000 \end{bmatrix} = \begin{bmatrix} 340,000 & 720,000 \end{bmatrix}

Expected Answer:

The property

(AB)^{-1} = B^{-1}A^{-1}

is significant because it shows how the inverse of a product of matrices is related to the inverses of the individual matrices, which is useful in solving systems of linear equations.

Detailed Solution:

First, compute

AB = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}

. The inverse of

B

B^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

, and the inverse of

A

A^{-1} = \frac{1}{(5)(3) - (-1)(2)} \begin{bmatrix} 3 & 1 \\ -2 & 5 \end{bmatrix} = \frac{1}{17} \begin{bmatrix} 3 & 1 \\ -2 & 5 \end{bmatrix}

. Thus,

B^{-1}A^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \times \frac{1}{17} \begin{bmatrix} 3 & 1 \\ -2 & 5 \end{bmatrix} = \frac{1}{17} \begin{bmatrix} 3 & 1 \\ -2 & 5 \end{bmatrix}

. Therefore,

(AB)^{-1} = B^{-1}A^{-1}

, confirming the property.

Expected Answer:

The inverse of the product of two matrices is equal to the product of the inverses of the matrices in reverse order.

Detailed Solution:

To prove

(AB)^{-1} = B^{-1}A^{-1}

, we use the property of matrix inverses. Since

A

and

B

are invertible,

AB

is also invertible. By definition,

(AB)(AB)^{-1} = I

. Pre-multiply both sides by

A^{-1}

A^{-1}(AB)(AB)^{-1} = A^{-1}I

. Simplifying,

(A^{-1}A)B(AB)^{-1} = A^{-1}

, which gives

IB(AB)^{-1} = A^{-1}

, or

B(AB)^{-1} = A^{-1}

. Pre-multiply by

B^{-1}

B^{-1}B(AB)^{-1} = B^{-1}A^{-1}

, resulting in

(AB)^{-1} = B^{-1}A^{-1}

Expected Answer:

In Mendel's monohybrid cross experiment, the F2 generation shows a phenotypic ratio of 3 tall plants to 1 dwarf plant and a genotypic ratio of 1 homozygous tall (TT), 2 heterozygous tall (Tt), and 1 homozygous dwarf (tt).

Detailed Solution:

The F1 generation, obtained by crossing homozygous tall (TT) with homozygous dwarf (tt), results in all heterozygous tall (Tt) plants. Self-pollinating the F1 generation (Tt x Tt) produces the F2 generation. Using a Punnett square, the possible genotypes are TT, Tt, Tt, and tt, leading to a phenotypic ratio of 3 tall:1 dwarf and a genotypic ratio of 1 TT:2 Tt:1 tt.

Expected Answer:

Compute

AB = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}

and

BA = \begin{bmatrix} 23 & 34 \\ 31 & 46 \end{bmatrix}

. Matrix multiplication is not commutative, meaning

AB \neq BA

. This non-commutativity is crucial in many applications, such as transformations in graphics, where the order of operations affects the outcome.

Detailed Solution:

Calculate

AB = \begin{bmatrix} 1 \times 5 + 2 \times 7 & 1 \times 6 + 2 \times 8 \\ 3 \times 5 + 4 \times 7 & 3 \times 6 + 4 \times 8 \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}

. Similarly,

BA = \begin{bmatrix} 5 \times 1 + 6 \times 3 & 5 \times 2 + 6 \times 4 \\ 7 \times 1 + 8 \times 3 & 7 \times 2 + 8 \times 4 \end{bmatrix} = \begin{bmatrix} 23 & 34 \\ 31 & 46 \end{bmatrix}

Expected Answer:

The equation

A^3 - 23A - 40I = O

demonstrates that

A

satisfies its own characteristic equation, which is a consequence of the Cayley-Hamilton theorem.

Detailed Solution:

Compute

A^2 = A \times A = \begin{bmatrix} 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 14 & 6 & 15 \\ 14 & 6 & 15 \end{bmatrix}

. Then,

A^3 = A \times A^2 = \begin{bmatrix} 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix} \times \begin{bmatrix} 14 & 6 & 15 \\ 14 & 6 & 15 \end{bmatrix} = \begin{bmatrix} 63 & 46 & 69 \\ 92 & 46 & 63 \end{bmatrix}

. Now, calculate

23A = 23 \times \begin{bmatrix} 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 69 & -46 & 23 \\ 92 & 46 & 23 \end{bmatrix}

. The identity matrix

I

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}

, so

40I = \begin{bmatrix} 40 & 0 & 0 \\ 0 & 40 & 0 \end{bmatrix}

. Finally,

A^3 - 23A - 40I = \begin{bmatrix} 63 & 46 & 69 \\ 92 & 46 & 63 \end{bmatrix} - \begin{bmatrix} 69 & -46 & 23 \\ 92 & 46 & 23 \end{bmatrix} - \begin{bmatrix} 40 & 0 & 0 \\ 0 & 40 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O

. This confirms that

A^3 - 23A - 40I = O

Expected Answer:

The expected answer should involve setting up matrices for quantities and prices, performing matrix multiplication to find total costs, and discussing the efficiency of matrices in handling such data.

Detailed Solution:

Represent the quantities as a matrix

Q = \begin{bmatrix} 2 & 5 \\ 8 & 10 \end{bmatrix}

and prices as

P_1 = \begin{bmatrix} 5 \\ 50 \end{bmatrix}

and

P_2 = \begin{bmatrix} 4 \\ 40 \end{bmatrix}

. Calculate

Q \cdot P_1

and

Q \cdot P_2

to find total expenditures. Matrices streamline calculations by organizing data in a structured format, reducing computational errors.

Expected Answer:

To verify that

(A')' = A

, compute the transpose of

A

to obtain

A'

, and then compute the transpose of

A'

to check if it equals

A

Detailed Solution:

The transpose of

A

A' = \begin{bmatrix} 4 & 2 \\ 3 & 0 \\ \sqrt{3} & 2 \end{bmatrix}

. The transpose of

A'

(A')' = \begin{bmatrix} 4 & 3 & \sqrt{3} \\ 2 & 0 & 2 \end{bmatrix}

, which is equal to

A

. Thus,

(A')' = A

is verified.

Expected Answer:

To verify the identity, compute

A^3

23A

, and

40I

, and then check if their combination results in the zero matrix. Such verifications are important as they confirm theoretical properties and ensure the correctness of mathematical models in applications.

Detailed Solution:

Calculate

A^3

23A

, and

40I

. Then,

A^3 - 23A - 40I = O

. This confirms the identity, showing the consistency of matrix operations and their properties.

Expected Answer:

The transpose of a matrix sum is equal to the sum of the transposes of the matrices. Calculate

(A + B)'

and

A' + B'

to verify this property.

A + B = \begin{bmatrix} 4 & 3 \\ \sqrt{3} & 2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 5 & 5 \\ \sqrt{3} + 4 & 7 \end{bmatrix}.

Then,

(A + B)' = \begin{bmatrix} 5 & \sqrt{3} + 4 \\ 5 & 7 \end{bmatrix}.

Now,

A' = \begin{bmatrix} 4 & \sqrt{3} \\ 3 & 2 \end{bmatrix}

and

B' = \begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix}.

Therefore,

A' + B' = \begin{bmatrix} 5 & \sqrt{3} + 4 \\ 5 & 7 \end{bmatrix} = (A + B)'.

This property is significant as it shows that transposition distributes over addition, which is important in simplifying expressions and proofs in linear algebra.

Detailed Solution:

Calculate the transpose of the sum and the sum of the transposes:

A + B = \begin{bmatrix} 4 & 3 \\ \sqrt{3} & 2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 5 & 5 \\ \sqrt{3} + 4 & 7 \end{bmatrix}.

Then,

(A + B)' = \begin{bmatrix} 5 & \sqrt{3} + 4 \\ 5 & 7 \end{bmatrix}.

Now,

A' = \begin{bmatrix} 4 & \sqrt{3} \\ 3 & 2 \end{bmatrix}

and

B' = \begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix}.

Therefore,

A' + B' = \begin{bmatrix} 5 & \sqrt{3} + 4 \\ 5 & 7 \end{bmatrix} = (A + B)'.

This demonstrates that the transpose of a sum is the sum of the transposes.

Expected Answer:

The expected answer should show the multiplication of

F(x)

and

F(y)

, resulting in

F(x+y)

. This demonstrates that the matrices represent rotations, and their product corresponds to the composition of these rotations.

Detailed Solution:

Calculate

F(x)F(y)

and show it equals

\begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}

. This corresponds to a rotation matrix, showing that successive rotations are additive in angle, reflecting the composition of rotations.

Expected Answer:

For Market I, the revenue is

10000 \times 2.50 + 2000 \times 1.50 + 18000 \times 1.00 = ₹44500

. For Market II, the revenue is

6000 \times 2.50 + 20000 \times 1.50 + 8000 \times 1.00 = ₹47000

. Total revenue is ₹91500.

Detailed Solution:

Calculate the revenue for Market I:

10000 \times 2.50 + 2000 \times 1.50 + 18000 \times 1.00 = ₹25000 + ₹3000 + ₹18000 = ₹44500

. For Market II:

6000 \times 2.50 + 20000 \times 1.50 + 8000 \times 1.00 = ₹15000 + ₹30000 + ₹8000 = ₹47000

. Total revenue is ₹44500 + ₹47000 = ₹91500.

Expected Answer:

Calculate

A + B = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}

, then

(A + B)C = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}

. Separately,

AC = A

and

BC = B

, so

AC + BC = A + B = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}

. Therefore,

(A + B)C = AC + BC

Detailed Solution:

First, calculate

A + B = \begin{bmatrix} 1+5 & 2+6 \\ 3+7 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}

. Then,

(A + B)C = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}

. Compute

AC = A \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = A

and

BC = B \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = B

. Thus,

AC + BC = A + B = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}

. Therefore,

(A + B)C = AC + BC

is verified.

Expected Answer:

The expected answer should conclude the properties of a matrix that is both symmetric and skew-symmetric.

Detailed Solution:

If a matrix

A

is both symmetric and skew-symmetric, then

A^T = A

and

A^T = -A

. This implies that

A = -A

, which can only be true if all elements of

A

are zero. Therefore,

A

must be a zero matrix.

Expected Answer:

The expected answer should involve showing that for symmetric matrices, the transpose of

AB - BA

is the negative of itself, thus proving it is skew-symmetric. The implications include insights into the behavior of linear transformations and their compositions.

Detailed Solution:

For symmetric matrices,

A' = A

and

B' = B

. Calculate

(AB - BA)' = B'A' - A'B' = BA - AB = -(AB - BA)

, proving skew-symmetry. This property is important in understanding the anti-commutative nature of certain matrix operations.

Expected Answer:

The expression simplifies to the zero matrix, confirming the identity.

Detailed Solution:

Calculate

A^3

, then compute

23A

and

40I

. Subtract these from

A^3

and verify that the result is the zero matrix.

Expected Answer:

The expected answer is the total cost calculated using matrix multiplication.

Detailed Solution:

Multiply matrix

A

by matrix

B

to find the total cost in each city. The result will be a matrix indicating the total expenditure in cities X and Y.

Expected Answer:

The total cost in each city can be calculated by multiplying the matrices

A

and

B

. The resulting matrix is

C = A \cdot B = \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix} \cdot \begin{bmatrix} 1000 & 3000 \\ 500 & 1000 \\ 5000 & 10000 \end{bmatrix} = \begin{bmatrix} 40 \times 1000 + 100 \times 500 + 50 \times 5000 & 40 \times 3000 + 100 \times 1000 + 50 \times 10000 \end{bmatrix} = \begin{bmatrix} 340000 & 720000 \end{bmatrix}

. Thus, the total cost in city X is ₹3400 and in city Y is ₹7200. Using matrices simplifies the computation by organizing data efficiently and allowing for systematic calculations.

Detailed Solution:

Perform matrix multiplication:

C = A \cdot B = \begin{bmatrix} 40 \times 1000 + 100 \times 500 + 50 \times 5000 & 40 \times 3000 + 100 \times 1000 + 50 \times 10000 \end{bmatrix} = \begin{bmatrix} 340000 & 720000 \end{bmatrix}

. The total cost in city X is ₹3400 and in city Y is ₹7200. Matrices provide a structured way to handle large datasets and perform complex calculations efficiently.

Expected Answer:

The expected answer should describe the genotypic and phenotypic ratios observed in the F2 generation of pea plants, based on Mendel's monohybrid cross experiment.

Detailed Solution:

In Mendel's monohybrid cross experiment, the parental generation consists of a tall plant (TT) and a dwarf plant (tt). The F1 generation, resulting from this cross, is all heterozygous tall (Tt). When the F1 generation is self-pollinated (Tt x Tt), the F2 generation results in a genotypic ratio of 1 TT : 2 Tt : 1 tt. This corresponds to a phenotypic ratio of 3 tall plants to 1 dwarf plant.

Expected Answer:

The expected answer should use matrix equations to allocate funds between the two bonds to achieve the specified interest. It should discuss how the allocation affects the risk and return profile of the investment.

Detailed Solution:

Let

x

be the amount in the 5% bond and

y

in the 7% bond. Set up the equation

0.05x + 0.07y = 1800

with

x + y = 30000

. Solve the system to find

x = 18000

and

y = 12000

. This allocation balances the interest income with risk, as higher interest bonds might carry more risk.

Expected Answer:

First, compute

(A + B) = \begin{bmatrix} 12 & -3 \\ 3 & -12 \\ 4 & 12 \end{bmatrix} + \begin{bmatrix} 5 & 0 & 2 \\ 4 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 17 & -3 & 2 \\ 7 & -10 & 5 \\ 4 & 12 \end{bmatrix}

. Next, compute

(B - C) = \begin{bmatrix} 5 & 0 & 2 \\ 4 & 2 & 5 \end{bmatrix} - \begin{bmatrix} 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -3 & 0 \\ 3 & 4 & 2 \end{bmatrix}

. Then, verify

A + (B - C) = (A + B) - C

by computing both sides:

A + (B - C) = \begin{bmatrix} 12 & -3 \\ 3 & -12 \\ 4 & 12 \end{bmatrix} + \begin{bmatrix} 5 & -3 & 0 \\ 3 & 4 & 2 \end{bmatrix} = \begin{bmatrix} 17 & -6 & 0 \\ 6 & -8 & 2 \\ 4 & 12 \end{bmatrix}

and

(A + B) - C = \begin{bmatrix} 17 & -3 & 2 \\ 7 & -10 & 5 \\ 4 & 12 \end{bmatrix} - \begin{bmatrix} 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 17 & -6 & 0 \\ 6 & -8 & 2 \\ 4 & 12 \end{bmatrix}

. Both results are equal, verifying the property. Verifying such properties ensures the correctness of matrix operations, which is crucial in applications involving complex matrix manipulations.

Detailed Solution:

Compute

(A + B) = \begin{bmatrix} 17 & -3 & 2 \\ 7 & -10 & 5 \\ 4 & 12 \end{bmatrix}

and

(B - C) = \begin{bmatrix} 5 & -3 & 0 \\ 3 & 4 & 2 \end{bmatrix}

. Verify

A + (B - C) = (A + B) - C

A + (B - C) = \begin{bmatrix} 17 & -6 & 0 \\ 6 & -8 & 2 \\ 4 & 12 \end{bmatrix}

and

(A + B) - C = \begin{bmatrix} 17 & -6 & 0 \\ 6 & -8 & 2 \\ 4 & 12 \end{bmatrix}

. Both are equal, confirming the property. Verifying such properties is important for ensuring the accuracy of matrix operations in various applications.

Expected Answer:

The matrix

A^2

should be calculated and then used to verify the given matrix equation.

Detailed Solution:

First, calculate

A^2 = A \cdot A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = \begin{bmatrix} 30 & 36 & 42 \\ 66 & 81 & 96 \\ 102 & 126 & 150 \end{bmatrix}

. Now, calculate

5A = 5 \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = \begin{bmatrix} 5 & 10 & 15 \\ 20 & 25 & 30 \\ 35 & 40 & 45 \end{bmatrix}

and

6I = 6 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}

. Then, calculate

A^2 - 5A + 6I = \begin{bmatrix} 30 & 36 & 42 \\ 66 & 81 & 96 \\ 102 & 126 & 150 \end{bmatrix} - \begin{bmatrix} 5 & 10 & 15 \\ 20 & 25 & 30 \\ 35 & 40 & 45 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

, which verifies the equation.

Expected Answer:

For

AB

to be symmetric,

(AB)' = AB

. Since

A

and

B

are symmetric,

A' = A

and

B' = B

. Therefore,

(AB)' = B'A' = BA

. For

AB = BA

, both sides must be equal, implying

AB

is symmetric if

AB = BA

. This property is significant in linear transformations as it ensures the preservation of symmetry in compositions of transformations.

Detailed Solution:

To prove

AB

is symmetric if and only if

AB = BA

: Assume

AB

is symmetric, then

(AB)' = AB

. Since

A' = A

and

B' = B

(AB)' = B'A' = BA

. Hence,

AB = BA

. Conversely, if

AB = BA

, then

(AB)' = BA = AB

, so

AB

is symmetric. This result is important in linear transformations as it indicates when two transformations can be applied in any order without affecting symmetry.

Expected Answer:

The expected answer involves calculating the matrix products

AC

BC

, and

(A + B)C

, and verifying that

(A + B)C = AC + BC

Detailed Solution:

First, calculate

AC = \begin{bmatrix} -6 & 0 & 8 \\ 7 & -8 & 0 \end{bmatrix} \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 12 \\ -14 \end{bmatrix}

. Next, calculate

BC = \begin{bmatrix} 1 & 0 & 2 \\ -2 & 7 & 8 \\ 0 & 3 & 1 \end{bmatrix} \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ 16 \\ 3 \end{bmatrix}

. Then, calculate

(A + B)C = \begin{bmatrix} -5 & 0 & 10 \\ 8 & -6 & 0 \end{bmatrix} \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 10 \\ -12 \end{bmatrix}

. Finally, verify that

AC + BC = \begin{bmatrix} 12 \\ -14 \end{bmatrix} + \begin{bmatrix} -2 \\ 16 \\ 3 \end{bmatrix} = \begin{bmatrix} 10 \\ -12 \end{bmatrix}

, confirming that

(A + B)C = AC + BC

Expected Answer:

Mendel's monohybrid cross experiment involves crossing two homozygous parents, one dominant (TT) and one recessive (tt). The F1 generation is all heterozygous (Tt) and phenotypically dominant. Self-pollinating F1 results in an F2 generation with a genotypic ratio of 1:2:1 (TT:Tt:tt) and a phenotypic ratio of 3:1 (dominant:recessive). This demonstrates the segregation of alleles and the predictable patterns of inheritance.

Detailed Solution:

In Mendel's experiment, the parental generation consists of TT (tall) and tt (dwarf) plants. The F1 generation, resulting from this cross, is all Tt and tall. When these F1 plants self-pollinate, the F2 generation consists of genotypes TT, Tt, Tt, and tt, giving a genotypic ratio of 1:2:1 and a phenotypic ratio of 3:1. The Punnett square visually represents these ratios, illustrating Mendel's laws of segregation and dominance. These principles are foundational in understanding genetic inheritance patterns.

Expected Answer:

In the F2 generation, the genotypic ratio is 1 TT : 2 Tt : 1 tt, and the phenotypic ratio is 3 Tall : 1 Dwarf. This experiment demonstrates Mendel's law of segregation, which states that alleles segregate independently during gamete formation, and each gamete carries only one allele for each trait. The F1 generation, being all Tt, shows the dominance of the tall trait, while the F2 generation reveals the segregation of alleles, resulting in a 3:1 phenotypic ratio.

Detailed Solution:

The F2 generation results from self-pollinating the F1 hybrids (Tt x Tt). The Punnett square for the F2 generation shows the combinations: TT, Tt, Tt, tt. The genotypic ratio is 1 TT : 2 Tt : 1 tt, and the phenotypic ratio is 3 Tall : 1 Dwarf. Mendel's laws are demonstrated as the alleles segregate independently, and the dominant allele masks the presence of the recessive allele in the heterozygous condition.

Expected Answer:

The total revenue in each market can be calculated by multiplying the sales matrix

S

with the price matrix

P = \begin{bmatrix} 2.5 \\ 1.5 \\ 1.0 \end{bmatrix}

. The resulting revenue matrix

R = S \cdot P = \begin{bmatrix} 10000 & 6000 \\ 2000 & 20000 \\ 18000 & 8000 \end{bmatrix} \cdot \begin{bmatrix} 2.5 \\ 1.5 \\ 1.0 \end{bmatrix} = \begin{bmatrix} 10000 \times 2.5 + 2000 \times 1.5 + 18000 \times 1.0 & 6000 \times 2.5 + 20000 \times 1.5 + 8000 \times 1.0 \end{bmatrix} = \begin{bmatrix} 64000 & 53000 \end{bmatrix}

. Thus, the total revenue in market I is ₹64000 and in market II is ₹53000. Matrices simplify financial calculations by organizing data and allowing for efficient computation of total revenues across multiple markets.

Detailed Solution:

Perform matrix multiplication:

R = S \cdot P = \begin{bmatrix} 10000 \times 2.5 + 2000 \times 1.5 + 18000 \times 1.0 & 6000 \times 2.5 + 20000 \times 1.5 + 8000 \times 1.0 \end{bmatrix} = \begin{bmatrix} 64000 & 53000 \end{bmatrix}

. The total revenue in market I is ₹64000 and in market II is ₹53000. Using matrices allows for efficient handling of large datasets and simplifies the computation of total revenues.

Expected Answer:

The transpose of a matrix is obtained by swapping rows with columns. For matrices

A

and

B

of appropriate dimensions,

(A + B)' = A' + B'

holds because transposition is a linear operation.

Detailed Solution:

First, compute

A + B = \begin{bmatrix} 4+1 & 3+2 & \sqrt{3}+4 \\ 2+2 & 0-1 & 2+2 \end{bmatrix} = \begin{bmatrix} 5 & 5 & \sqrt{3}+4 \\ 4 & -1 & 4 \end{bmatrix}

. The transpose of this matrix is

(A + B)' = \begin{bmatrix} 5 & 4 \\ 5 & -1 \\ \sqrt{3}+4 & 4 \end{bmatrix}

. Now,

A' = \begin{bmatrix} 4 & 2 \\ 3 & 0 \\ \sqrt{3} & 2 \end{bmatrix}

and

B' = \begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 4 & 2 \end{bmatrix}

. Adding these gives

A' + B' = \begin{bmatrix} 4+1 & 2+2 \\ 3+2 & 0-1 \\ \sqrt{3}+4 & 2+2 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 5 & -1 \\ \sqrt{3}+4 & 4 \end{bmatrix}

. Thus,

(A + B)' = A' + B'

, verifying the property.

Expected Answer:

The transpose of the sum of two matrices is equal to the sum of their transposes.

Detailed Solution:

To verify

(A + B)' = A' + B'

, first calculate

A + B

and then find its transpose. Similarly, calculate the transpose of

A

and

B

separately and add them. Both should yield the same result.

Expected Answer:

The resulting matrix after performing the scalar multiplication and subtraction.

Detailed Solution:

Multiply matrix

A

by 3 and matrix

B

by 5. Subtract the resulting matrix of

5B

from

3A

to get the final matrix.

Expected Answer:

To verify the properties, we start by calculating the transpose of the matrices. The transpose of a matrix is obtained by swapping its rows and columns. For property 1, calculate

A + B

and then find its transpose,

(A + B)'

. Separately, find

A'

and

B'

and then compute

A' + B'

. Show that these results are equal. For property 2, compute

kB

and then its transpose,

(kB)'

. Separately, find

B'

and then

kB'

. Show that these results are equal. Discuss how transposition is used in applications such as rotating objects in computer graphics or transforming datasets in machine learning.

Detailed Solution:

Calculate $A + B$ : $A + B = \begin{bmatrix} 4 & 3 \\ \sqrt{3} & 2 \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 4 \\ 2 & -1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 5 & 4 \\ \sqrt{3} + 2 & 1 & 2 \\ 0 & 2 & 2 \end{bmatrix}$ Now, find $(A + B)'$ : $(A + B)' = \begin{bmatrix} 5 & \sqrt{3} + 2 & 0 \\ 5 & 1 & 2 \\ 4 & 2 & 2 \end{bmatrix}$ Calculate $A'$ and $B'$ : $A' = \begin{bmatrix} 4 & \sqrt{3} & 0 \\ 3 & 2 & 2 \end{bmatrix}$ $B' = \begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 4 & 2 \end{bmatrix}$ Now, find $A' + B'$ : $A' + B' = \begin{bmatrix} 4 & \sqrt{3} & 0 \\ 3 & 2 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 5 & \sqrt{3} + 2 & 0 \\ 5 & 1 & 2 \\ 4 & 2 & 2 \end{bmatrix}$ Thus, $(A + B)' = A' + B'$ . 2) For $(kB)' = kB'$ , let $k = 3$ . Compute $3B$ : $3B = \begin{bmatrix} 3 & 6 & 12 \\ 6 & -3 & 6 \end{bmatrix}$ Now, find $(3B)'$ : $(3B)' = \begin{bmatrix} 3 & 6 \\ 6 & -3 \\ 12 & 6 \end{bmatrix}$ Compute $3B'$ : $3B' = 3 \begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 6 & -3 \\ 12 & 6 \end{bmatrix}$ Thus, $(3B)' = 3B'$ . These properties demonstrate the linearity of transposition, which is crucial in simplifying complex matrix operations in various applications such as computer graphics for object transformations and data preprocessing in machine learning.

Expected Answer:

The expected answer should include the allocation of ₹30,000 between the two bonds to achieve the desired annual interest using matrix multiplication.

Detailed Solution:

Let

x

be the amount invested in the first bond and

y

be the amount invested in the second bond. We have the equations

x + y = 30,000

and

0.05x + 0.07y = 1,800

. Solving these equations, we express them in matrix form:

\begin{bmatrix} 1 & 1 \\ 0.05 & 0.07 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 30,000 \\ 1,800 \end{bmatrix}

. Solving this system, we find

x = 20,000

and

y = 10,000

. Therefore, ₹20,000 should be invested in the first bond and ₹10,000 in the second bond.

Expected Answer:

In Mendel's monohybrid cross, the F1 generation is obtained by crossing homozygous tall (TT) and dwarf (tt) plants, resulting in all heterozygous tall (Tt) offspring. Self-pollination of F1 (Tt x Tt) results in an F2 generation with a phenotypic ratio of 3 tall:1 dwarf and a genotypic ratio of 1 TT:2 Tt:1 tt. The Punnett square shows these combinations.

Detailed Solution:

The F1 generation consists of heterozygous tall plants (Tt). When these are self-pollinated, the gametes are T and t. The Punnett square is:

\begin{array}{c|c|c} & T & t \\ \hline T & TT & Tt \\ \hline t & Tt & tt \end{array}

. The phenotypic ratio (3 tall:1 dwarf) arises because TT and Tt are tall, while tt is dwarf. The genotypic ratio (1 TT:2 Tt:1 tt) is directly from the Punnett square.

Expected Answer:

The expected answer should demonstrate that the product of symmetric matrices

AB - BA

results in a skew-symmetric matrix.

Detailed Solution:

To prove that

AB - BA

is skew-symmetric, we need to show that

(AB - BA)^T = -(AB - BA)

. Since

A

and

B

are symmetric, we have

A^T = A

and

B^T = B

. Therefore,

(AB - BA)^T = B^T A^T - A^T B^T = BA - AB = -(AB - BA)

. Thus,

AB - BA

is skew-symmetric.

Expected Answer:

The transpose of the sum of matrices

A

and

B

is equal to the sum of the transposes of

A

and

B

Detailed Solution:

First, calculate

A + B

A + B = \begin{bmatrix} 4+1 & 3+2 & \sqrt{3}+4 \\ 2+2 & 0-1 & 2+2 \end{bmatrix} = \begin{bmatrix} 5 & 5 & \sqrt{3}+4 \\ 4 & -1 & 4 \end{bmatrix}

. Then, find the transpose:

(A + B)' = \begin{bmatrix} 5 & 4 \\ 5 & -1 \\ \sqrt{3}+4 & 4 \end{bmatrix}

. Now, calculate

A'

and

B'

A' = \begin{bmatrix} 4 & 2 \\ 3 & 0 \\ \sqrt{3} & 2 \end{bmatrix}

and

B' = \begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 4 & 2 \end{bmatrix}

. Sum the transposes:

A' + B' = \begin{bmatrix} 4+1 & 2+2 \\ 3+2 & 0-1 \\ \sqrt{3}+4 & 2+2 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 5 & -1 \\ \sqrt{3}+4 & 4 \end{bmatrix}

. Thus,

(A + B)' = A' + B'

Expected Answer:

To verify

(A')' = A

, we start by computing the transpose of

A

, which is

A' = \begin{bmatrix} 4 & 2 \\ 3 & 0 \\ \sqrt{3} & 2 \end{bmatrix}

. The transpose of

A'

, denoted as

(A')'

, is calculated as

\begin{bmatrix} 4 & 3 & \sqrt{3} \\ 2 & 0 & 2 \end{bmatrix}

, which is identical to the original matrix

A

. This property, that the transpose of the transpose of a matrix returns the original matrix, is fundamental in linear algebra as it ensures consistency in operations involving transpositions, such as in the computation of symmetric matrices or the adjoint of matrices.

Detailed Solution:

Compute the transpose of

A

A' = \begin{bmatrix} 4 & 2 \\ 3 & 0 \\ \sqrt{3} & 2 \end{bmatrix}

. Then, compute the transpose of

A'

(A')' = \begin{bmatrix} 4 & 3 & \sqrt{3} \\ 2 & 0 & 2 \end{bmatrix}

. Since

(A')' = A

, the property is verified. This property is significant because it simplifies the manipulation of matrices, particularly in proofs and derivations involving matrix equations.

Expected Answer:

The product

AB = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix}

and

BA = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix}

are equal in this case because

B

is the identity matrix. Generally, matrix multiplication is not commutative unless specific conditions are met, such as both matrices being diagonal or one being the identity matrix.

Detailed Solution:

To calculate

AB

, multiply

A

and

B

AB = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix}

. Similarly, for

BA

BA = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix}

. The products are equal because

B

is the identity matrix, which does not alter the other matrix. In general,

AB = BA

only if

A

and

B

are both diagonal or one is the identity matrix.

Expected Answer:

The values of

x

y

, and

z

that make the product of

A

and its transpose equal to the identity matrix.

Detailed Solution:

Compute

A'

and then

A'A

. Set the resulting matrix equal to the identity matrix and solve for

x

y

, and

z

Expected Answer:

The expected answer should show the step-by-step calculation of

A^3

23A

, and

40I

, followed by their subtraction to arrive at the zero matrix. The significance lies in demonstrating a property of matrices where a polynomial expression of the matrix results in the zero matrix, indicating a characteristic polynomial.

Detailed Solution:

First, calculate

A^2 = A \cdot A

, then

A^3 = A^2 \cdot A

. Compute

23A

and

40I

. Subtract these from

A^3

to show the result is the zero matrix. This demonstrates a matrix polynomial identity, useful in understanding eigenvalues and characteristic polynomials.

Expected Answer:

The expected answer includes the combined sales matrix and the decrease in sales matrix.

Detailed Solution:

(i) Add matrices

A

and

B

to find the combined sales. (ii) Subtract matrix

B

from matrix

A

to find the decrease in sales.

Expected Answer:

The total revenue in Market I is ₹57,000 and in Market II is ₹49,000.

Detailed Solution:

For Market I, the revenue is

10,000 \times 2.50 + 2,000 \times 1.50 + 18,000 \times 1.00 = ₹57,000

. For Market II, the revenue is

6,000 \times 2.50 + 20,000 \times 1.50 + 8,000 \times 1.00 = ₹49,000

Expected Answer:

The product

AC

is calculated by taking the dot product of rows of

A

with columns of

C

. For example, the first element of

AC

(-6)(-2) + (0)(12) + (8)(3) = 12 + 24 = 36

. This process is repeated for each element of the resulting matrix. The verification of

(A + B)C = AC + BC

involves calculating

(A + B)

first, then multiplying by

C

, and comparing with the sum of

AC

and

BC

Detailed Solution:

To calculate

AC

, perform the following matrix multiplication:

AC = \begin{bmatrix} -6 & 0 & 8 \\ 7 & -8 & 0 \end{bmatrix} \begin{bmatrix} -2 & 0 \\ 12 & 0 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} (-6)(-2) + (0)(12) + (8)(3) & (-6)(0) + (0)(0) + (8)(2) \\ (7)(-2) + (-8)(12) + (0)(3) & (7)(0) + (-8)(0) + (0)(2) \end{bmatrix} = \begin{bmatrix} 36 & 16 \\ -110 & 0 \end{bmatrix}.

To verify

(A + B)C = AC + BC

, compute

(A + B)

, multiply by

C

, and compare with

AC + BC

. This property shows the distributive law of matrix multiplication, which is fundamental in linear algebra.

Matrices

AI Learning Assistant

Summary

Summary of Chapter 3: Matrices

Learning Objectives

Detailed Notes

Matrices

Introduction

Matrix Basics

Matrix Operations

Special Matrices

Invertible Matrices

Examples

Applications

Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Practice & Assessment

Multiple Choice Questions

Q1.If A=[1001]A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A=[10​01​], what is the inverse of matrix AAA?

Correct Answer: A

Q2.If A=[3003]A = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}A=[30​03​], what is the determinant of matrix A?

Correct Answer: C

Q3.If A is a square matrix such that A2=AA^2 = AA2=A, what is the value of (I+A)3−7A(I + A)^3 - 7A(I+A)3−7A?

Correct Answer: B

Q4.If A=[2112]A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}A=[21​12​], what is the eigenvalue of matrix AAA?

Correct Answer: A

Q5.A trust fund has ₹30,000 to be invested in two bonds. The first bond offers 5% interest per annum, and the second offers 7% interest. If the total annual interest must be ₹1800, how much should be invested in the first bond?

Correct Answer: D

Q6.If A=[2345]A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}A=[24​35​], find the determinant of AAA.

Correct Answer: C

Q7.In Mendel's monohybrid cross experiment, if the F2 generation shows a phenotypic ratio of 3 tall plants to 1 dwarf plant, what is the correct genotypic ratio for this generation?

Correct Answer: A

Q8.If A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​] and B=[0110]B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}B=[01​10​], what is the product ABABAB?

Correct Answer: A

Q9.Given the matrix A=[332−1]A = \begin{bmatrix} 3 & \sqrt{3} \\ 2 & -1 \end{bmatrix}A=[32​3​−1​], what is the transpose of matrix A?

Correct Answer: A

Q10.Consider the matrices A=[4332]A = \begin{bmatrix} 4 & 3 \\ \sqrt{3} & 2 \end{bmatrix}A=[43​​32​] and B=[1234]B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}B=[13​24​]. If the transpose of a matrix is equal to its inverse, which of the following matrices satisfies this property?

Correct Answer: B

Correct Answer: A

Q12.If the unit costs of three commodities are ₹2.00, ₹1.00, and ₹0.50 respectively, and their unit sale prices are ₹2.50, ₹1.50, and ₹1.00 respectively, what is the gross profit from selling 10,000 units of each commodity?

Correct Answer: A

Q13.If a matrix A=[3−21421100]A = \begin{bmatrix} 3 & -2 & 1 \\ 4 & 2 & 1 \\ 1 & 0 & 0 \end{bmatrix}A=​341​−220​110​​, show that A3−23A−40I=OA^3 - 23A - 40I = OA3−23A−40I=O. What does this imply about the eigenvalues of A?

Correct Answer: D

Q14.If A and B are symmetric matrices, which of the following is true about the matrix AB - BA?

Correct Answer: B

Q15.If A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​], what is the trace of matrix A?

Correct Answer: B

Q16.If A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​], what is the inverse of matrix A?

Correct Answer: A

Q17.Consider the matrices A=[2314]A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}A=[21​34​] and B=[1001]B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}B=[10​01​]. If C=A2−2A+IC = A^2 - 2A + IC=A2−2A+I, where III is the identity matrix, what is the determinant of matrix CCC?

Correct Answer: B

Q18.Given matrices A=[2314]A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}A=[21​34​] and B=[0110]B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}B=[01​10​], compute A−BA - BA−B and find the determinant of the resulting matrix.

Correct Answer: C

Q19.Consider the matrices A=[2314]A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}A=[21​34​] and B=[1001]B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}B=[10​01​]. If C=A2−5A+6BC = A^2 - 5A + 6BC=A2−5A+6B, what is the matrix CCC?

Correct Answer: A

Q20.Consider the matrices A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​] and B=[2012]B = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}B=[21​02​]. Calculate the matrix product ABABAB and find the trace of the resulting matrix.

Correct Answer: A

Q21.Consider the matrices A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​] and B=[5678]B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}B=[57​68​]. Find the matrix XXX such that X⋅A=BX \cdot A = BX⋅A=B.

Correct Answer: A

Q22.If A=[3−21421]A = \begin{bmatrix} 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}A=[34​−22​11​], verify that A3−23A−40I=OA^3 - 23A - 40I = OA3−23A−40I=O. What is the value of A3A^3A3?

Correct Answer: A

Q23.If a matrix AAA is such that A=[3−21421123]A = \begin{bmatrix} 3 & -2 & 1 \\ 4 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}A=​341​−222​113​​, verify whether A3−23A−40I=OA^3 - 23A - 40I = OA3−23A−40I=O.

Correct Answer: A

Q24.If A=[1001]A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A=[10​01​] and B=[0110]B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}B=[01​10​], what is ABABAB?

Correct Answer: A

Correct Answer: A

Q26.If A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​] and B=[5678]B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}B=[57​68​], what is the sum of matrices A and B?

Correct Answer: A

Q27.In Mendel's monohybrid cross experiment, what is the phenotypic ratio in the F2 generation?

Correct Answer: A

Q28.If A=[5005]A = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}A=[50​05​], what is the inverse of matrix A?

Correct Answer: A

Q29.If a matrix AAA satisfies the equation A2−5A+6I=0A^2 - 5A + 6I = 0A2−5A+6I=0, what is the possible eigenvalue of matrix AAA?

Correct Answer: D

Q30.If matrices A and B are symmetric, which of the following statements is true about the matrix product AB?

Correct Answer: C

Q31.Consider the matrices A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​] and B=[2012]B = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}B=[21​02​]. If C=A+BC = A + BC=A+B, what is the determinant of matrix CCC?

Correct Answer: A

Q32.If A=[123456]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}A=[14​25​36​] and B=[789101112]B = \begin{bmatrix} 7 & 8 \\ 9 & 10 \\ 11 & 12 \end{bmatrix}B=​7911​81012​​, what is the product ABABAB?