Exercise 4.4, Question 1
Find adj A for \(A = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\).
Show solution
- Swap 1 and 4, change the signs of 2 and 3.
Answer: \[\begin{bmatrix}4 & -2 \\ -3 & 1\end{bmatrix}\]
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4
Exercise 4.4: Adjoint and inverse of a matrix. adj A is the transpose of the cofactor matrix; \(A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|I\). A is invertible exactly when \(|A| \ne 0\), and then \(A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A\). For \(2 \times 2\): swap the diagonal, change the signs of the other two. If a matrix satisfies a polynomial equation, multiply by \(A^{-1}\) to get the inverse from it.
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Exercise 4.4 questions and solutions
Exercise 4.4, Question 2
Find adj A for \(A = \begin{bmatrix}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{bmatrix}\).
Show solution
- Cofactors: \(A_{11} = 3, A_{12} = -12, A_{13} = 6\); \(A_{21} = 1, A_{22} = 5, A_{23} = 2\); \(A_{31} = -11, A_{32} = -1, A_{33} = 5\).
- adj A is the transpose of this cofactor matrix.
Answer: \[\begin{bmatrix}3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{bmatrix}\]
Exercise 4.4, Question 3
Verify \(A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|I\) for \(A = \begin{bmatrix}2 & 3 \\ -4 & -6\end{bmatrix}\).
Show solution
- \(|A| = -12 + 12 = 0\); \[\operatorname{adj} A = \begin{bmatrix}-6 & -3 \\ 4 & 2\end{bmatrix}\]
- \[\begin{aligned}A(\operatorname{adj} A) &= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} \\ &= O\end{aligned}\], and \((\operatorname{adj} A)A = O\) likewise.
Answer: Both products are \(O = 0 \cdot I = |A|I\)
Exercise 4.4, Question 4
Verify \(A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|I\) for \(A = \begin{bmatrix}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{bmatrix}\).
Show solution
- \(|A| = 11\) (expand along column 2: \(-(-1)(9 + 2)\)).
- \[\operatorname{adj} A = \begin{bmatrix}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{bmatrix}\]; multiplying either way gives \(11I\).
Answer: Both products equal \[\begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix} = |A|I\]
Exercise 4.4, Question 5
\(A = \begin{bmatrix}2 & -2 \\ 4 & 3\end{bmatrix}\)
Show solution
- Find \(|A|\); if it is non-zero, \[A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A\], where adj A is the transpose of the matrix of cofactors.
- \(|A| = 6 + 8 = 14\); \[\operatorname{adj} A = \begin{bmatrix}3 & 2 \\ -4 & 2\end{bmatrix}\]
Answer: \(|A| = 14\), \[A^{-1} = \dfrac{1}{14}\begin{bmatrix}3 & 2 \\ -4 & 2\end{bmatrix}\]
Exercise 4.4, Question 6
\(A = \begin{bmatrix}-1 & 5 \\ -3 & 2\end{bmatrix}\)
Show solution
- Find \(|A|\); if it is non-zero, \[A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A\], where adj A is the transpose of the matrix of cofactors.
- \(|A| = -2 + 15 = 13\); \[\operatorname{adj} A = \begin{bmatrix}2 & -5 \\ 3 & -1\end{bmatrix}\]
Answer: \(|A| = 13\), \[A^{-1} = \dfrac{1}{13}\begin{bmatrix}2 & -5 \\ 3 & -1\end{bmatrix}\]
Exercise 4.4, Question 7
\(A = \begin{bmatrix}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{bmatrix}\)
Show solution
- Find \(|A|\); if it is non-zero, \[A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A\], where adj A is the transpose of the matrix of cofactors.
- Upper triangular: \(|A| = 1 \cdot 2 \cdot 5 = 10\).
- Cofactors: \[\begin{aligned}A_{11} &= 10, A_{12} \\ &= 0, A_{13} \\ &= 0, A_{21} \\ &= -10, A_{22} \\ &= 5, A_{23} \\ &= 0, A_{31} \\ &= 2, A_{32} \\ &= -4, A_{33} \\ &= 2\end{aligned}\]; transpose for adj A.
Answer: \(|A| = 10\), \[A^{-1} = \dfrac{1}{10}\begin{bmatrix}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{bmatrix}\]
Exercise 4.4, Question 8
\(A = \begin{bmatrix}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{bmatrix}\)
Show solution
- Find \(|A|\); if it is non-zero, \[A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A\], where adj A is the transpose of the matrix of cofactors.
- Lower triangular: \(|A| = 1 \cdot 3 \cdot (-1) = -3\).
- Cofactors: \[\begin{aligned}A_{11} &= -3, A_{12} \\ &= 3, A_{13} \\ &= -9, A_{21} \\ &= 0, A_{22} \\ &= -1, A_{23} \\ &= -2, A_{31} \\ &= 0, A_{32} \\ &= 0, A_{33} \\ &= 3\end{aligned}\]; transpose for adj A.
Answer: \(|A| = -3\), \[A^{-1} = \dfrac{1}{-3}\begin{bmatrix}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{bmatrix}\]
Exercise 4.4, Question 9
\(A = \begin{bmatrix}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{bmatrix}\)
Show solution
- Find \(|A|\); if it is non-zero, \[A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A\], where adj A is the transpose of the matrix of cofactors.
- \[\begin{aligned}|A| &= 2(-1 - 0) - 1(4 - 0) + 3(8 - 7) \\ &= -2 - 4 + 3 \\ &= -3\end{aligned}\]
- Cofactors: \[\begin{aligned}A_{11} &= -1, A_{12} \\ &= -4, A_{13} \\ &= 1, A_{21} \\ &= 5, A_{22} \\ &= 23, A_{23} \\ &= -11, A_{31} \\ &= 3, A_{32} \\ &= 12, A_{33} \\ &= -6\end{aligned}\]; transpose for adj A.
Answer: \(|A| = -3\), \[A^{-1} = \dfrac{1}{-3}\begin{bmatrix}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{bmatrix}\]
Exercise 4.4, Question 10
\(A = \begin{bmatrix}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{bmatrix}\)
Show solution
- Find \(|A|\); if it is non-zero, \[A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A\], where adj A is the transpose of the matrix of cofactors.
- \[\begin{aligned}|A| &= 1(8 - 6) + 1(0 + 9) + 2(0 - 6) \\ &= 2 + 9 - 12 \\ &= -1\end{aligned}\]
- Cofactors: \[\begin{aligned}A_{11} &= 2, A_{12} \\ &= -9, A_{13} \\ &= -6, A_{21} \\ &= 0, A_{22} \\ &= -2, A_{23} \\ &= -1, A_{31} \\ &= -1, A_{32} \\ &= 3, A_{33} \\ &= 2\end{aligned}\]; adj A is the transpose.
Answer: \(|A| = -1\), \[A^{-1} = -\begin{bmatrix}2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2\end{bmatrix}\]
Exercise 4.4, Question 11
\(A = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos{\left(\alpha \right)} & \sin{\left(\alpha \right)} \\ 0 & \sin{\left(\alpha \right)} & - \cos{\left(\alpha \right)}\end{bmatrix}\)
Show solution
- Find \(|A|\); if it is non-zero, \[A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A\], where adj A is the transpose of the matrix of cofactors.
- \[\begin{aligned}|A| &= 1(-\cos^2\alpha - \sin^2\alpha) \\ &= -1\end{aligned}\]
- \[\operatorname{adj} A = \begin{bmatrix}-1 & 0 & 0 \\ 0 & - \cos{\left(\alpha \right)} & - \sin{\left(\alpha \right)} \\ 0 & - \sin{\left(\alpha \right)} & \cos{\left(\alpha \right)}\end{bmatrix}\]; divide by \(-1\).
Answer: \[\begin{aligned}A^{-1} &= \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos{\left(\alpha \right)} & \sin{\left(\alpha \right)} \\ 0 & \sin{\left(\alpha \right)} & - \cos{\left(\alpha \right)}\end{bmatrix} \\ &= A\end{aligned}\]
Exercise 4.4, Question 12
\(A = \begin{bmatrix}3 & 7 \\ 2 & 5\end{bmatrix},\ B = \begin{bmatrix}6 & 8 \\ 7 & 9\end{bmatrix}\). Verify \((AB)^{-1} = B^{-1}A^{-1}\).
Show solution
- \[AB = \begin{bmatrix}67 & 87 \\ 47 & 61\end{bmatrix}\], \(|AB| = 4087 - 4089 = -2\), so \[(AB)^{-1} = -\tfrac12\begin{bmatrix}61 & -87 \\ -47 & 67\end{bmatrix}\]
- \[A^{-1} = \begin{bmatrix}5 & -7 \\ -2 & 3\end{bmatrix}\] (\(|A| = 1\)), \[B^{-1} = -\tfrac12\begin{bmatrix}9 & -8 \\ -7 & 6\end{bmatrix}\] (\(|B| = -2\)).
- \[\begin{aligned}B^{-1}A^{-1} &= -\tfrac12\begin{bmatrix}9 & -8 \\ -7 & 6\end{bmatrix}\begin{bmatrix}5 & -7 \\ -2 & 3\end{bmatrix} \\ &= -\tfrac12\begin{bmatrix}61 & -87 \\ -47 & 67\end{bmatrix}\end{aligned}\]
Answer: Both equal \[\begin{bmatrix}- \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2}\end{bmatrix}\]
Exercise 4.4, Question 13
\(A = \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}\). Show \(A^2 - 5A + 7I = O\) and hence find \(A^{-1}\).
Show solution
- \[A^2 = \begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix}\]; \[\begin{aligned}A^2 - 5A + 7I &= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} \\ &= O\end{aligned}\]
- Multiply \(A^2 - 5A + 7I = O\) by \(A^{-1}\): \(A - 5I + 7A^{-1} = O\), so \(A^{-1} = \tfrac17(5I - A)\).
Answer: \[A^{-1} = \dfrac17\begin{bmatrix}2 & -1 \\ 1 & 3\end{bmatrix}\]
Exercise 4.4, Question 14
For \(A = \begin{bmatrix}3 & 2 \\ 1 & 1\end{bmatrix}\), find a and b with \(A^2 + aA + bI = O\).
Show solution
- \[A^2 = \begin{bmatrix}11 & 8 \\ 4 & 3\end{bmatrix}\]
- \[\begin{bmatrix}3 a + b + 11 & 2 a + 8 \\ a + 4 & a + b + 3\end{bmatrix} = O\]: \(4 + a = 0 \Rightarrow a = -4\), then \(3 - 4 + b = 0 \Rightarrow b = 1\) (the other two entries agree).
Answer: \(a = -4,\ b = 1\)
Exercise 4.4, Question 15
\(A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{bmatrix}\). Show \(A^3 - 6A^2 + 5A + 11I = O\) and hence find \(A^{-1}\).
Show solution
- \[A^2 = \begin{bmatrix}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{bmatrix}\], \[A^3 = \begin{bmatrix}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{bmatrix}\]; substituting gives O.
- Multiply by \(A^{-1}\): \(A^2 - 6A + 5I + 11A^{-1} = O\), so \(A^{-1} = -\tfrac{1}{11}(A^2 - 6A + 5I)\).
- \[A^2 - 6A + 5I = \begin{bmatrix}3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1\end{bmatrix}\]
Answer: \[A^{-1} = \dfrac{1}{11}\begin{bmatrix}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{bmatrix}\]
Exercise 4.4, Question 16
\(A = \begin{bmatrix}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{bmatrix}\). Verify \(A^3 - 6A^2 + 9A - 4I = O\) and hence find \(A^{-1}\).
Show solution
- \[A^2 = \begin{bmatrix}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{bmatrix}\], \[A^3 = \begin{bmatrix}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{bmatrix}\]; substituting gives O.
- Multiply by \(A^{-1}\): \(A^2 - 6A + 9I = 4A^{-1}\).
- \[A^2 - 6A + 9I = \begin{bmatrix}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{bmatrix}\]
Answer: \[A^{-1} = \dfrac14\begin{bmatrix}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{bmatrix}\]
Exercise 4.4, Question 17
A is a non-singular \(3 \times 3\) matrix. Then \(|\operatorname{adj} A|\) is: (A) \(|A|\) (B) \(|A|^2\) (C) \(|A|^3\) (D) \(3|A|\)
Show solution
- \(A(\operatorname{adj} A) = |A|I_3\), so \(|A| \cdot |\operatorname{adj} A| = |A|^3\).
- Divide by \(|A| \ne 0\): \(|\operatorname{adj} A| = |A|^{n-1}\) with \(n = 3\).
Answer: (B) \(|A|^2\)
Exercise 4.4, Question 18
A is invertible of order 2. Then \(\det(A^{-1})\) is: (A) \(\det A\) (B) \(\dfrac{1}{\det A}\) (C) 1 (D) 0
Show solution
- \[\begin{aligned}&AA^{-1} = I \\ \Rightarrow\ &|A| \cdot |A^{-1}| = 1\end{aligned}\]
Answer: (B)
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