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Practice Exercise 4.4

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This page provides comprehensive Class 12 - Practice Exercise 4.4 (Complete). Free NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4-4. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.

Adjoint and Inverse of Matrices: Complete set of 18 practice problems.

Chapter Index Start Solving
Q1-2 Adjoint Q3-4 Verify Q5-11 Inverse Q12 Verify Q13 Eqn Q15 Eqn Q17 MCQ

Q1 - Q2. Find Adjoint

Practice Question 1
Find the adjoint of the matrix: \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
View Solution
Shortcut for 2x2
Swap diagonal elements, change signs of off-diagonal elements.
\( \text{adj } A = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \)
Practice Question 2
Find the adjoint of the matrix: \( \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix} \)
View Solution
Cofactors
\( A_{11}=3, A_{12}=-12, A_{13}=6 \)
\( A_{21}=1, A_{22}=5, A_{23}=2 \)
\( A_{31}=-11, A_{32}=-1, A_{33}=5 \)
\( \text{adj } A = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix} \)

Q3 - Q4. Verification

Practice Question 3
Verify \( A(\text{adj } A) = (\text{adj } A)A = |A|I \) for \( A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \).
View Solution
Calculate Determinant
\( |A| = 2(-6) - 3(-4) = -12 + 12 = 0 \). So \( |A|I = O \).
Calculate Adjoint
\( \text{adj } A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \).
Multiply
\( A(\text{adj } A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
Verified: All products equal Zero Matrix.
Practice Question 4
Verify \( A(\text{adj } A) = |A|I \) for \( A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \).
View Solution
Determinant
\( |A| = 1(0) - (-1)(9+2) + 2(0) = 11 \).
Adjoint
\( \text{adj } A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \).
Product gives \( \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = 11I \). Verified.

Q5 - Q11. Inverse of Matrices

Practice Question 5
Find the inverse of: \( \begin{bmatrix} 2 & 4 \\ -2 & 3 \end{bmatrix} \)
View Solution
\( |A| = 6 - (-8) = 14 \).
\( \text{adj } A = \begin{bmatrix} 3 & -4 \\ 2 & 2 \end{bmatrix} \).
\( A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & -4 \\ 2 & 2 \end{bmatrix} \)
Practice Question 6
Find the inverse of: \( \begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix} \)
View Solution
\( |A| = -2 - (-15) = 13 \).
\( \text{adj } A = \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix} \).
\( A^{-1} = \frac{1}{13} \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix} \)
Practice Question 7
Find the inverse of: \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix} \)
View Solution
\( |A| = 10 \). (Product of diagonal elements for triangular matrix).
\( \text{adj } A = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix} \).
\( A^{-1} = \frac{1}{10} \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix} \)
Practice Question 8
Find the inverse of: \( \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix} \)
View Solution
\( |A| = -3 \).
\( \text{adj } A = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \).
\( A^{-1} = -\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \)
Practice Question 9
Find the inverse of: \( \begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix} \)
View Solution
\( |A| = 2(-1) - 1(4) + 3(1) = -3 \).
\( \text{adj } A = \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix} \).
\( A^{-1} = -\frac{1}{3} (\text{adj } A) \)
Practice Question 10
Find the inverse of: \( \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \)
View Solution
\( |A| = 1(8-6) + 1(0+9) + 2(0-6) = 2 + 9 - 12 = -1 \).
\( \text{adj } A = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix} \).
\( A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \)
Practice Question 11
Find the inverse of: \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{bmatrix} \)
View Solution
\( |A| = 1(-\cos^2 \alpha - \sin^2 \alpha) = -1 \).
\( \text{adj } A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{bmatrix} \).
\( A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{bmatrix} \)

Q12 - Q18. Proofs & Equations

Practice Question 12
Let \( A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} \). Verify \( (AB)^{-1} = B^{-1}A^{-1} \).
View Solution
Calculate \( AB \), find its inverse. Calculate inverses of A and B separately, then multiply \( B^{-1}A^{-1} \). Both results will match.
Verified.
Practice Question 13
If \( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \), show that \( A^2 - 5A + 7I = O \). Hence find \( A^{-1} \).
View Solution
\( A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \). Substitute into equation to get \( O \).
\( 7A^{-1} = 5I - A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \).
\( A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \)
Practice Question 14
For the matrix \( A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \), find \( a \) and \( b \) such that \( A^2 + aA + bI = O \).
View Solution
\( A^2 = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix} \).
Eq: \( \begin{bmatrix} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{bmatrix} = O \).
\( 4+a=0 \Rightarrow a=-4 \). \( 3-4+b=0 \Rightarrow b=1 \).
\( a=-4, b=1 \)
Practice Question 15
For the matrix \( A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \), show that \( A^3 - 3A^2 - 9A - 5I = O \). Hence find \( A^{-1} \).
View Solution
Calculations of \( A^2, A^3 \) verify the equation.
Multiplying by \( A^{-1} \): \( 5A^{-1} = A^2 - 3A - 9I \).
\( A^{-1} = \frac{1}{5} (A^2 - 3A - 9I) \)
Practice Question 16
If \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \), prove that \( A^3 - 6A^2 + 7A + 2I = O \).
View Solution
Substitute \( A, A^2, A^3 \) into the polynomial. All elements sum to zero.
Verified.
Practice Question 17 (MCQ)
Let A be a nonsingular square matrix of order \( 3 \times 3 \). Then \( |\text{adj } A| \) is equal to:
(A) \( |A| \) (B) \( |A|^2 \) (C) \( |A|^3 \) (D) \( 3|A| \)
View Solution
Formula: \( |\text{adj } A| = |A|^{n-1} \).
Here \( n=3 \), so \( |A|^{3-1} = |A|^2 \).
Correct Option: (B)
Practice Question 18 (MCQ)
If A is an invertible matrix of order 2, then \( \det(A^{-1}) \) is equal to:
(A) \( \det(A) \) (B) \( 1/\det(A) \) (C) 1 (D) 0
View Solution
\( |A^{-1}| = |A|^{-1} = 1/|A| \).
Correct Option: (B)