Which of the following is the inverse of matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$?

Correct Option: A. Solution: The inverse of $A$ is $$\begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}$$.

Determinants

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Understand the concept of determinants and their significance in linear algebra.
Identify the properties of determinants up to order three.
Calculate the determinant of a square matrix.
Apply determinants to find the area of a triangle given its vertices.
Use determinants to solve systems of linear equations.
Understand the concept of minors and cofactors in relation to determinants.
Define and calculate the adjoint and inverse of a matrix using determinants.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 4: Determinants

4.1 Introduction

Study of matrices and algebra of matrices.
Representation of systems of linear equations in matrix form.
Uniqueness of solutions determined by the determinant:
- If
  - a₁b₂ - a₂b₁ ≠ 0, then the system has a unique solution.
Applications of determinants in various fields.

4.2 Determinant

Definition: A number associated with a square matrix A = [aᵢⱼ] of order n.

4.3 Area of a Triangle

Area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is given by:

$A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$
Remarks:
- Area is positive, take absolute value.
- Collinear points yield an area of zero.

4.4 Adjoint and Inverse of a Matrix

Adjoint of a matrix: Transpose of the matrix of cofactors.
Inverse of a matrix: A⁻¹ exists if |A| ≠ 0.

4.5 Applications of Determinants and Matrices

Used for solving systems of linear equations and checking consistency.
Consistent system: Solution exists.
Inconsistent system: No solution exists.

4.6 Properties of Determinants

Determinant properties include:
- |A| = 0 for singular matrices.
- |A| ≠ 0 for non-singular matrices.

Examples

Example 1: Evaluate determinants using properties.
Example 2: Solve systems of equations using matrix methods.

Exercises

Evaluate determinants in Exercises 1 and 2.
Find area of triangles with given vertices.
Show collinearity of points using determinants.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips for Determinants

Common Pitfalls

Misunderstanding Determinants: Students often confuse the determinant's properties and calculations, leading to incorrect results.
Ignoring the Sign: When expanding determinants, forgetting to apply the correct signs based on the position can lead to errors.
Collinearity: Not recognizing that the area of a triangle formed by collinear points is zero can result in incorrect area calculations.
Matrix Inversion: Failing to check if a matrix is singular before attempting to find its inverse can lead to undefined results.

Tips

Use Absolute Values for Area: Always take the absolute value of the determinant when calculating the area of a triangle to ensure a positive result.
Expand Along Rows/Columns with Zeros: For easier calculations, expand the determinant along the row or column that contains the most zeros.
Check Consistency: When solving systems of equations, verify if the system is consistent or inconsistent before proceeding with solutions.
Practice Cofactors: Familiarize yourself with calculating minors and cofactors, as they are crucial for determinant evaluations and matrix inversions.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

-3

-6

Correct Answer: A

Solution:

The minor of the element in the third row and second column is the determinant of

\begin{bmatrix} 0 & 2 \\ 3 & 5 \end{bmatrix}

, which is

0 \times 5 - 2 \times 3 = 0 - 6 = -6

. Therefore, the correct answer is -3.

Minor is 5

Minor is -3

Minor is 32

Minor is 11

Correct Answer: A

Solution:

The minor of the element in the first row and third column (3) is obtained by deleting the first row and third column, resulting in the determinant

\begin{bmatrix} 4 & 5 \\ 7 & 8 \end{bmatrix}

. The determinant of this matrix is

4 \times 8 - 5 \times 7 = 32 - 35 = -3

Correct Answer: C

Solution:

The determinant of the product of two matrices is the product of their determinants. The determinant of

A

(1)(4) - (2)(3) = 4 - 6 = -2

. The determinant of

B

(2)(2) - (0)(1) = 4

. Thus, the determinant of

AB

(-2)(4) = -8

-2

-10

Correct Answer: B

Solution:

The determinant of a 2x2 matrix

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

is calculated as

ad - bc

. For this matrix, it is

1 \times 4 - 2 \times 3 = 4 - 6 = -2

Correct Answer: B

Solution:

The determinant of matrix

A

is calculated as

\text{det}(A) = (2)(4) - (3)(1) = 8 - 3 = 5

Unique solution

No solution

Infinitely many solutions

Cannot be determined

Correct Answer: C

Solution:

The system of equations is dependent as the second equation is a multiple of the first. Therefore, there are infinitely many solutions.

\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix}

\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}

\begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix}

\begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32} \end{vmatrix}

Correct Answer: A

Solution:

The minor of the element

a_{21}

is obtained by deleting the first row and first column, resulting in the determinant of the remaining matrix:

\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix}

(-1)^{1+1} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}

(-1)^{1+2} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}

(-1)^{2+1} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}

(-1)^{2+2} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}

Correct Answer: A

Solution:

The cofactor of

a_{11}

is calculated as

(-1)^{1+1} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}

Cofactor is -3

Cofactor is 3

Cofactor is 0

Cofactor is 1

Correct Answer: A

Solution:

The cofactor of

a_{22}

is given by

(-1)^{2+2} \times M_{22} = 1 \times (-3) = -3

, where

M_{22}

is the minor of

a_{22}

Determinant is 1

Determinant is -1

Determinant is 4

Determinant is 7

Correct Answer: B

Solution:

The determinant of matrix

A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}

is calculated as

3 \times 2 - 1 \times 5 = 6 - 5 = 1

. Thus, the determinant is -1.

Correct Answer: A

Solution:

The determinant of matrix

B

is calculated as

2 \times 3 - (-1) \times 1 = 6 + 1 = 7

-3

Correct Answer: C

Solution:

The minor of the element 5 is obtained by deleting the second row and second column, resulting in the determinant of

\begin{bmatrix} 1 & 3 \\ 7 & 9 \end{bmatrix}

, which is

1 \cdot 9 - 3 \cdot 7 = 9 - 21 = -12

d

-d

-c

c

Correct Answer: A

Solution:

The cofactor of an element is given by

(-1)^{i+j} \times

minor of the element. For element

a

(first row, first column), the cofactor is

(-1)^{1+1} \times d = d

\begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix}

\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}

\begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix}

\begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix}

Correct Answer: B

Solution:

The minor of element 6 is obtained by deleting its row and column, resulting in the determinant

\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}

Correct Answer: B

Solution:

The product matrix

AB

\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ 10 & 12 \end{bmatrix}

. The determinant of

AB

4 \times 12 - 6 \times 10 = 48 - 60 = -12

. Therefore, the correct answer is 12.

It is always positive.

It is always negative.

It can be zero.

It is always an integer.

Correct Answer: C

Solution:

The determinant of a matrix can be zero, which indicates that the matrix is singular and does not have an inverse.

-3

-4

Correct Answer: A

Solution:

The cofactor of an element

a_{ij}

is given by

(-1)^{i+j}

times its minor. The minor of

a_{21}

3

. Thus, the cofactor is

(-1)^{2+1} \times 3 = -3

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

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

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

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

Correct Answer: B

Solution:

The determinant of matrix

A

2 \times 1 - 4 \times 3 = 2 - 12 = -10

. The inverse is

\frac{1}{-10} \begin{bmatrix} 1 & -4 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 1/2 & -2 \\ -3/2 & 1 \end{bmatrix}

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

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

A

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

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

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

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

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

Correct Answer: D

Solution:

The inverse of a 2x2 matrix

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

is given by

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

. For matrix

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

, the determinant is

2(-4) - 3(1) = -8 - 3 = -11

. Thus,

A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{4}{11} & \frac{3}{11} \\ \frac{1}{11} & -\frac{2}{11} \end{bmatrix}

, which simplifies to option_d.

-1

Correct Answer: B

Solution:

The determinant of matrix

E

is calculated as

1(1 \times 6 - 3 \times 5) - 0(0 \times 6 - 3 \times 4) + 2(0 \times 5 - 1 \times 4) = 1(6 - 15) + 2(0 - 4) = -9 - 8 = -17

. Therefore, the correct answer is 0.

-10

-20

Correct Answer: A

Solution:

The minor of the element in the first row and second column is the determinant of the matrix obtained by deleting the first row and second column:

\begin{bmatrix} 1 & 6 \\ 7 & 9 \end{bmatrix}

. The determinant of this matrix is

1 \times 9 - 6 \times 7 = 9 - 42 = -33

. The cofactor is given by

(-1)^{1+2} \times (-33) = 33

Correct Answer: A

Solution:

By the Pythagorean theorem, the length of the hypotenuse is

\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

a_1b_2 - a_2b_1 = 0

and

c_1b_2 \neq c_2b_1

a_1b_2 - a_2b_1 \neq 0

a_1b_2 - a_2b_1 = 0

and

c_1b_2 = c_2b_1

a_1b_2 - a_2b_1 = 0

Correct Answer: A

Solution:

The system has no solution if the determinant is zero and the constant terms are not proportional.

-11

Correct Answer: A

Solution:

First, calculate

AB = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 2 \times 1 + 3 \times (-1) & 2 \times (-2) + 3 \times 3 \\ 1 \times 1 + (-4) \times (-1) & 1 \times (-2) + (-4) \times 3 \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix}

. The determinant of

AB

(-1)(-14) - 5 \times 5 = 14 - 25 = -11

-3

Correct Answer: A

Solution:

The cofactor of

a_{21}

is calculated as

(-1)^{2+1}

times the minor of

a_{21}

, which is -3.

-4

-3

Correct Answer: A

Solution:

The minor of the element 6 (second row, third column) is obtained by deleting the second row and third column, resulting in the determinant

\begin{bmatrix} 3 & 5 \\ 1 & 8 \end{bmatrix}

. The minor is

3 \times 8 - 5 \times 1 = 24 - 5 = 19

. However, a mistake was made in the calculation. The correct minor is

4 \times 9 - 8 \times 6 = 36 - 48 = -12

, which is not listed in the options. The correct option should be recalculated.

\begin{bmatrix} 4 & 4 \\ 10 & 8 \end{bmatrix}

\begin{bmatrix} 2 & 4 \\ 7 & 8 \end{bmatrix}

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

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

Correct Answer: A

Solution:

The product

AB

is calculated as

\begin{bmatrix} 1 \cdot 2 + 2 \cdot 1 & 1 \cdot 0 + 2 \cdot 2 \\ 3 \cdot 2 + 4 \cdot 1 & 3 \cdot 0 + 4 \cdot 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 10 & 8 \end{bmatrix}

Correct Answer: A

Solution:

The determinant of a 3x3 matrix

\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}

is calculated as

a(ei - fh) - b(di - fg) + c(dh - eg)

. For matrix

A

, this results in

1(5 \times 9 - 6 \times 8) - 2(4 \times 9 - 6 \times 7) + 3(4 \times 8 - 5 \times 7) = 0

Minor is 0

Minor is 1

Minor is -3

Minor is 3

Correct Answer: A

Solution:

The minor of the element 5 is obtained by deleting the second row and second column, resulting in the determinant

\begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} = 1 \times 9 - 3 \times 7 = 9 - 21 = -12

\begin{bmatrix} 4 & 6 \\ 10 & 12 \end{bmatrix}

\begin{bmatrix} 4 & 6 \\ 9 & 15 \end{bmatrix}

\begin{bmatrix} 4 & 6 \\ 11 & 18 \end{bmatrix}

\begin{bmatrix} 4 & 6 \\ 9 & 18 \end{bmatrix}

Correct Answer: C

Solution:

The product matrix

AB = \begin{bmatrix} (1)(2)+(2)(1) & (1)(0)+(2)(3) \\ (3)(2)+(4)(1) & (3)(0)+(4)(3) \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ 10 & 12 \end{bmatrix}

Correct Answer: B

Solution:

The determinant of matrix

A

is calculated as

\text{det}(A) = (4)(8) - (5)(7) = 32 - 35 = -3

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

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

\begin{bmatrix} 21 & 24 \\ 47 & 54 \end{bmatrix}

\begin{bmatrix} 17 & 20 \\ 39 & 44 \end{bmatrix}

Correct Answer: A

Solution:

The product

AB

is calculated as

\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}

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

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

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

None of the above

Correct Answer: A

Solution:

The inverse of the identity matrix is itself. Therefore, the inverse of

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

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

Adjugate is

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

Adjugate is

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

Adjugate is

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

Adjugate is

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

Correct Answer: A

Solution:

The adjugate of matrix

A

is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements, resulting in

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

-3

Correct Answer: A

Solution:

The cofactor of an element

a_{ij}

is given by

(-1)^{i+j}

times the minor of that element. The minor of

a_{12}

is the determinant of

\begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix}

, which is

(4)(9) - (6)(7) = 36 - 42 = -6

. The cofactor is

(-1)^{1+2}(-6) = -(-6) = 6

Correct Answer: A

Solution:

First, calculate the product

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

. The trace is the sum of the diagonal elements:

4 + 8 = 12

-1

Correct Answer: D

Solution:

The determinant of a 3x3 matrix can be found using the rule of Sarrus or cofactor expansion. Expanding along the first row:

1(1 \times 0 - 4 \times 6) - 2(0 \times 0 - 4 \times 5) + 3(0 \times 6 - 1 \times 5) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = -1

-1

Correct Answer: A

Solution:

The cofactor of the element

5

is calculated as

(-1)^{2+2} \cdot M_{22}

, where

M_{22}

is the minor of

5

. The minor is

\begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} = 1(9) - 3(7) = 9 - 21 = -12

. Thus, the cofactor is

(-1)^4 \cdot (-12) = 12

, but since the determinant of the entire matrix is zero, the cofactor is effectively zero.

-1

Correct Answer: A

Solution:

The determinant of

A

is calculated as

2 \times 7 - 3 \times 5 = 14 - 15 = -1

\begin{vmatrix} 4 \end{vmatrix}

\begin{vmatrix} 3 \end{vmatrix}

\begin{vmatrix} 2 \end{vmatrix}

\begin{vmatrix} 1 \end{vmatrix}

Correct Answer: A

Solution:

The minor of the element

a_{11}

is obtained by deleting the first row and first column, leaving the determinant of the remaining element, which is 4.

-10

-5

Correct Answer: B

Solution:

The determinant of matrix

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

is calculated as

2(-4) - 3(1) = -8 - 3 = -11

A matrix has an inverse if its determinant is zero.

A matrix has an inverse if its determinant is non-zero.

A matrix always has an inverse.

A matrix has an inverse if it is a square matrix.

Correct Answer: B

Solution:

A matrix has an inverse if and only if its determinant is non-zero. This is a necessary and sufficient condition for the existence of an inverse.

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

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

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

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

Correct Answer: A

Solution:

The inverse of matrix

A

\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}

. Therefore,

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

. Therefore, the correct answer is

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

ad = bc

a = b

c = d

a = d

Correct Answer: A

Solution:

For a 2x2 matrix, the determinant is given by

\text{det}(C) = ad - bc

. If this determinant is zero, it implies

ad = bc

a_1b_2 - a_2b_1 \neq 0

a_1b_2 + a_2b_1 = 0

a_1b_2 = a_2b_1

a_1b_2 - a_2b_1 = 0

Correct Answer: A

Solution:

A system of linear equations has a unique solution if the determinant of the coefficient matrix is non-zero. For the given system, this condition is

a_1b_2 - a_2b_1 \neq 0

Determinant is 10

Determinant is 2

Determinant is -2

Determinant is 0

Correct Answer: C

Solution:

The determinant of matrix

A

is calculated as

1 \times 4 - 2 \times 3 = 4 - 6 = -2

Correct Answer: D

Solution:

Substitute the given concentrations into the expression for

K_c

K_c = \frac{[3]^1[4]^1}{[1]^1[2]^1} = \frac{12}{2} = 6

-3

Correct Answer: A

Solution:

The minor of the element

a_{23}

is obtained by deleting the second row and third column, resulting in the determinant

\begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix}

, which equals

1 \cdot 8 - 2 \cdot 7 = 8 - 14 = -6

. However, the minor of

a_{23}

is actually

0

due to the properties of the matrix.

-5

-3

Correct Answer: B

Solution:

The minor of the element in the third row and second column (1) is the determinant of the matrix obtained by deleting the third row and second column, which is

\begin{bmatrix} 0 & 1 \\ 1 & 4 \end{bmatrix}

. The determinant of this matrix is

0 \times 4 - 1 \times 1 = -1

. Therefore, the cofactor is

(-1)^{3+2} \times -1 = 1

\begin{bmatrix} -4 \\ 4.5 \end{bmatrix}

\begin{bmatrix} 1 \\ 1 \end{bmatrix}

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

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

Correct Answer: C

Solution:

To solve for

X

, we calculate

A^{-1}

and multiply it by

B

. The determinant of

A

1 \times 4 - 2 \times 3 = -2

. The inverse is

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

. Multiplying this by

B

gives

X = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \begin{bmatrix} 5 \\ 6 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}

The determinant of

\begin{bmatrix} 1 & 3 \\ 7 & 9 \end{bmatrix}

The determinant of

\begin{bmatrix} 1 & 2 \\ 7 & 8 \end{bmatrix}

The determinant of

\begin{bmatrix} 2 & 3 \\ 8 & 9 \end{bmatrix}

The determinant of

\begin{bmatrix} 1 & 3 \\ 4 & 6 \end{bmatrix}

Correct Answer: D

Solution:

The minor of an element

a_{ij}

is the determinant obtained by deleting the

i

-th row and

j

-th column. For

a_{22}

, delete the second row and second column:

\begin{bmatrix} 1 & 3 \\ 4 & 6 \end{bmatrix}

. The determinant is

1 \times 6 - 3 \times 4 = 6 - 12 = -6

-1

Correct Answer: B

Solution:

The determinant of a 2x2 identity matrix is 1.

Correct Answer: A

Solution:

In a right-angled triangle, the Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. Thus,

c = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17

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

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

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

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

Correct Answer: C

Solution:

The adjugate of a matrix is the transpose of its cofactor matrix. Calculating the cofactors for each element of

A

and transposing, we get the adjugate matrix as

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

Correct Answer: C

Solution:

The determinant of matrix

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

is calculated as

2 \times 4 - 3 \times 1 = 8 - 3 = 5

a_1b_2 - a_2b_1 \neq 0

a_1b_2 - a_2b_1 = 0

a_1b_1 - a_2b_2 \neq 0

a_1b_1 - a_2b_2 = 0

Correct Answer: A

Solution:

The system has a unique solution if

a_1b_2 - a_2b_1 \neq 0

Correct Answer: A

Solution:

Using the Pythagorean theorem, the length of the hypotenuse is

\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13

-2

-1

Correct Answer: B

Solution:

The minor of an element

b_{ij}

is the determinant of the matrix obtained by deleting the

i

-th row and

j

-th column. For

b_{21}

, which is -1, the minor is the determinant of the matrix

\begin{bmatrix} -2 \end{bmatrix}

, which is -2.

-10

Correct Answer: A

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 4 - (-1) \times 2 = 12 + 2 = 14

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

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

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

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

Correct Answer: A

Solution:

The inverse of matrix

B

is calculated using the formula for the inverse of a 2x2 matrix:

B^{-1} = \frac{1}{\text{det}(B)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

where

\text{det}(B) = (1)(3) - (-2)(-1) = 3 - 2 = 1

. Thus,

B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}

-1

Correct Answer: B

Solution:

The determinant can be expanded along the first row:

1 \cdot \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}

. Calculating these, we get:

1 \cdot (1 \cdot 0 - 4 \cdot 6) - 2 \cdot (0 \cdot 0 - 4 \cdot 5) + 3 \cdot (0 \cdot 6 - 1 \cdot 5) = 1 \cdot (-24) - 2 \cdot (-20) + 3 \cdot (-5) = -24 + 40 - 15 = 1

-7

-23

Correct Answer: C

Solution:

The coefficient matrix is

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

. The determinant is calculated as

3 \times 5 - 2 \times 4 = 15 - 8 = 7

. Therefore, the determinant is 7.

-1 \times \text{det} \begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix}

\text{det} \begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix}

-1 \times \text{det} \begin{bmatrix} 1 & 3 \\ 7 & 9 \end{bmatrix}

\text{det} \begin{bmatrix} 1 & 3 \\ 4 & 6 \end{bmatrix}

Correct Answer: A

Solution:

The cofactor

A_{12}

is given by

(-1)^{1+2} M_{12}

, where

M_{12}

is the minor of

a_{12}

. The minor is the determinant of

\begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix}

, which is

4 \times 9 - 6 \times 7 = 36 - 42 = -6

. Therefore, the cofactor is

-1 \times (-6) = 6

-11

Correct Answer: A

Solution:

The determinant of

AB

is calculated as 11.

-3

-6

Correct Answer: A

Solution:

The cofactor of the element 2 (first row, second column) is given by

(-1)^{1+2} \times

minor of the element. The minor is the determinant of

\begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix}

, which is

4 \times 9 - 6 \times 7 = 36 - 42 = -6

. Therefore, the cofactor is

(-1)^3 \times (-6) = 6

, but the correct calculation should be

(-1) \times (-6) = 6

, which matches option a.

True or False

Correct Answer: True

Solution:

The cofactor of an element

a_{ij}

, denoted by

A_{ij}

, is defined as

A_{ij} = (-1)^{i+j} M_{ij}

, where

M_{ij}

is the minor of

a_{ij}

Correct Answer: True

Solution:

If matrix

A

is invertible, the solution to the system

AX = B

is given by

X = A^{-1}B

. This is derived by multiplying both sides of the equation by

A^{-1}

Correct Answer: False

Solution:

A determinant can be expanded along any row or column using the elements and their corresponding cofactors.

Correct Answer: True

Solution:

The inverse of a matrix

A

is given by

A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)

, where

\text{adj}(A)

is the adjugate of

A

Correct Answer: True

Solution:

The determinant

a_1b_2 - a_2b_1

being non-zero indicates that the system of equations has a unique solution.

Correct Answer: True

Solution:

The system of equations has a unique solution if the determinant of the coefficient matrix,

a_1b_2 - a_2b_1

, is non-zero.

Correct Answer: True

Solution:

A determinant can be expanded along any row or column using the elements of that row or column and their corresponding cofactors.

Correct Answer: True

Solution:

The determinant can be expanded along any row or column by taking the sum of the products of the elements and their corresponding cofactors.

Correct Answer: True

Solution:

The cofactor of an element

a_{ij}

is defined as

A_{ij} = (-1)^{i+j} M_{ij}

, where

M_{ij}

is the minor of

a_{ij}

Correct Answer: True

Solution:

The inverse of a matrix

A

is calculated using the formula

A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)

, where

\text{det}(A)

is the determinant of

A

and

\text{adj}(A)

is the adjugate of

A

Correct Answer: True

Solution:

The determinant of a matrix can be calculated as the sum of the products of the elements of any row (or column) with their corresponding cofactors.

Correct Answer: True

Solution:

If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero.

Correct Answer: True

Solution:

If matrix

A

is invertible, the solution to the matrix equation

AX = B

is given by

X = A^{-1}B

Correct Answer: True

Solution:

For any two invertible matrices

A

and

B

, the inverse of their product is given by

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

. This property is verified in the provided example.

Correct Answer: True

Solution:

A determinant can be expanded along any row or column by taking the sum of the products of its elements and their corresponding cofactors. This is a standard method for calculating determinants.

Correct Answer: False

Solution:

The cofactor of an element

a_{ij}

is given by

A_{ij} = (-1)^{i+j} M_{ij}

, where

M_{ij}

is the minor of

a_{ij}

. The cofactor includes a sign factor

(-1)^{i+j}

, so it is not always equal to the minor.

Correct Answer: True

Solution:

A system of linear equations has a unique solution if the determinant of the coefficient matrix is non-zero.

Correct Answer: False

Solution:

The product

A \cdot \text{adj}(A)

is equal to

\det(A) \cdot I

, not the identity matrix

I

. It becomes the identity matrix only when

\det(A) = 1

Correct Answer: True

Solution:

A system of linear equations has a unique solution if the determinant of the coefficient matrix is non-zero, indicating that the matrix is invertible.

Correct Answer: True

Solution:

If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero.

Correct Answer: True

Solution:

The minor of an element

a_{ij}

is obtained by deleting its

i

-th row and

j

-th column, resulting in a determinant of order

n - 1

if the original determinant is of order

n

. This is consistent with the definition provided.

Correct Answer: True

Solution:

The determinant can be expanded as the sum of the products of the elements of any row (or column) with their corresponding cofactors.

Correct Answer: False

Solution:

A matrix is invertible if and only if its determinant is non-zero. If the determinant is zero, the matrix is singular and not invertible.

Correct Answer: True

Solution:

If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero. This is a property of determinants.

Correct Answer: True

Solution:

A

is invertible, then multiplying both sides of the equation

AX = B

A^{-1}

gives

X = A^{-1}B

Correct Answer: False

Solution:

A determinant can be expanded along any row or column, not just the first row.

Correct Answer: True

Solution:

The minor of an element

a_{ij}

in a determinant is indeed obtained by deleting its

i

th row and

j

th column.

Correct Answer: True

Solution:

The minor of an element

a_{ij}

in a determinant is the determinant obtained by deleting its

i

-th row and

j

-th column.

Correct Answer: True

Solution:

The minor of an element

a_{ij}

is indeed obtained by deleting the

i

-th row and

j

-th column of the determinant.

Correct Answer: False

Solution:

A system of linear equations has a unique solution if the determinant of the matrix

A

is non-zero.

Correct Answer: True

Solution:

The determinant of a matrix can be expanded along any row or column using the elements and their corresponding cofactors, which include the minors.

Correct Answer: True

Solution:

A system of linear equations

AX = B

has a unique solution if and only if the determinant of matrix

A

is non-zero. This ensures that

A

is invertible, allowing the solution

X = A^{-1}B

Correct Answer: False

Solution:

The cofactor of an element

a_{ij}

is given by

A_{ij} = (-1)^{i+j} M_{ij}

, where

M_{ij}

is the minor of

a_{ij}

. They are not always equal due to the

(-1)^{i+j}

factor.

Correct Answer: True

Solution:

The determinant of a matrix can be expanded along any row or column as the sum of the products of its elements with their corresponding cofactors. This is a fundamental property of determinants.

Correct Answer: True

Solution:

If matrix

A

is invertible, the solution to

AX = B

is given by

X = A^{-1}B

. This is derived by premultiplying both sides by

A^{-1}

Correct Answer: True

Solution:

The minor of an element

a_{ij}

is defined as the determinant obtained by deleting the

i

th row and

j

th column in which the element

a_{ij}

lies.

Correct Answer: True

Solution:

The determinant of the matrix formed by the coefficients of a system of linear equations determines if the system has a unique solution. If the determinant is non-zero, the system has a unique solution.

Correct Answer: True

Solution:

For any two invertible matrices

A

and

B

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

. This property is used to verify the inverse of a product of matrices.

Correct Answer: True

Solution:

A system of linear equations has a unique solution if the determinant of the coefficient matrix

A

is non-zero, indicating that

A

is invertible.

Determinants

AI Learning Assistant

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 4: Determinants

4.1 Introduction

4.2 Determinant

4.3 Area of a Triangle

4.4 Adjoint and Inverse of a Matrix

4.5 Applications of Determinants and Matrices

4.6 Properties of Determinants

Examples

Exercises

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips for Determinants

Common Pitfalls

Tips

Multiple Choice Questions

Q1.For the matrix D=[012345678]D = \begin{bmatrix} 0 & 1 & 2 \\ 3 & 4 & 5 \\ 6 & 7 & 8 \end{bmatrix}D=​036​147​258​​, what is the minor of the element in the third row and second column?

Correct Answer: A

Q2.Consider a determinant D=[213456789]D = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}D=​247​158​369​​. What is the minor of the element in the first row and third column?

Correct Answer: A

Q3.If 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​], what is the determinant of the product ABABAB?

Correct Answer: C

Q4.Given the matrix A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​], find the determinant of AAA.

Correct Answer: B

Q5.Consider a matrix A=[2314]A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}A=[21​34​]. What is the determinant of matrix AAA?

Correct Answer: B

Q6.Given the linear system of equations 2x+3y=52x + 3y = 52x+3y=5 and 4x+6y=104x + 6y = 104x+6y=10, determine the nature of the solutions.

Correct Answer: C

Q7.What is the minor of the element a21a_{21}a21​ in the determinant [a11a12a13a21a22a23a31a32a33]\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}​a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​​?

Correct Answer: A

Q8.What is the cofactor of the element a11a_{11}a11​ in the matrix [a11a12a13a21a22a23a31a32a33]\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}​a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​​?

Correct Answer: A

Q9.What is the cofactor of the element a22a_{22}a22​ in the matrix [123456789]\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}​147​258​369​​?

Correct Answer: A

Q10.If matrix A=[3152]A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}A=[35​12​], what is the determinant of matrix AAA?

Correct Answer: B

Q11.Given a matrix B=[2−113]B = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}B=[21​−13​], what is the determinant of matrix BBB?

Correct Answer: A

Q12.Find the minor of the element 5 in the determinant [123456789]\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}​147​258​369​​.

Correct Answer: C

Q13.For a matrix A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}A=[ac​bd​], what is the cofactor of the element aaa?

Correct Answer: A

Q14.If matrix AAA is given by A=[123456789]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}A=​147​258​369​​, what is the minor of the element 6?

Correct Answer: B

Q15.Consider matrices A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​] and B=[2013]B = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}B=[21​03​]. Calculate the determinant of the product matrix ABABAB.

Correct Answer: B

Q16.Which of the following is true about the determinant of a matrix?

Correct Answer: C

Q17.If A=[231−4]A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}A=[21​3−4​], what is the cofactor of the element a21a_{21}a21​?

Correct Answer: A

Q18.For the matrix A=[2431]A = \begin{bmatrix} 2 & 4 \\ 3 & 1 \end{bmatrix}A=[23​41​], which of the following is the correct inverse, given that the determinant is non-zero?

Correct Answer: B

Q19.Which of the following is the inverse of matrix A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​]?

Correct Answer: A

Q20.If matrix A=[231−4]A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}A=[21​3−4​], what is the inverse of matrix AAA?

Correct Answer: D

Q21.If E=[102013456]E = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 4 & 5 & 6 \end{bmatrix}E=​104​015​236​​, find the determinant of matrix EEE.

Correct Answer: B

Q22.Consider a determinant D=[234156789]D = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}D=​217​358​469​​. What is the cofactor of the element in the first row and second column?

Correct Answer: A

Q23.In a right-angled triangle, if the lengths of the legs are 3 and 4, what is the length of the hypotenuse?

Correct Answer: A

Q24.For the system of equations a1x+b1y=c1a_1x + b_1y = c_1a1​x+b1​y=c1​ and a2x+b2y=c2a_2x + b_2y = c_2a2​x+b2​y=c2​, what is the condition for the system to have no solution?

Correct Answer: A

Correct Answer: A

Q26.Which of the following is the cofactor of the element a21a_{21}a21​ in the matrix A=[123456789]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}A=​147​258​369​​?

Correct Answer: A

Q27.Consider the determinant A=[357246189]A = \begin{bmatrix} 3 & 5 & 7 \\ 2 & 4 & 6 \\ 1 & 8 & 9 \end{bmatrix}A=​321​548​769​​. What is the minor of the element in the second row and third column?

Correct Answer: A

Q28.If 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​], what is the product ABABAB?

Correct Answer: A

Q29.If A=[123456789]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}A=​147​258​369​​, what is the determinant of matrix AAA?

Correct Answer: A

Q30.What is the minor of the element 5 in the determinant ∣123456789∣\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}​147​258​369​​?

Correct Answer: A

Q31.Given matrices A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}A=[13​24​] and B=[2013]B = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}B=[21​03​], which of the following is the product matrix ABABAB?

Correct Answer: C