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Q12:
Let $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(AB)^{-1} = B^{-1} A^{-1}$.
Solution :
Step 1: Compute the Product Matrix $AB$
Given $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$.
The product $AB$ is calculated as:
$AB = \begin{bmatrix} (3 \times 6) + (7 \times 7) & (3 \times 8) + (7 \times 9) \\ (2 \times 6) + (5 \times 7) & (2 \times 8) + (5 \times 9) \end{bmatrix}$
$AB = \begin{bmatrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$
Step 2: Calculate $(AB)^{-1}$
For a matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $M^{-1} = \frac{1}{\det(M)} \text{adj}(M)$, where $\det(M) = ad - bc$ and $\text{adj}(M) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ [By the Property of Matrix Inverses].
$\det(AB) = (67 \times 61) - (87 \times 47) = 4087 - 4089 = -2$.
$\text{adj}(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$.
$(AB)^{-1} = -\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -30.5 & 43.5 \\ 23.5 & -33.5 \end{bmatrix}$.
Step 3: Calculate $A^{-1}$ and $B^{-1}$
For $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$, $\det(A) = (3 \times 5) - (7 \times 2) = 15 - 14 = 1$.
$A^{-1} = \frac{1}{1} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.
For $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$, $\det(B) = (6 \times 9) - (8 \times 7) = 54 - 56 = -2$.
$B^{-1} = -\frac{1}{2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -4.5 & 4 \\ 3.5 & -3 \end{bmatrix}$.
Step 4: Calculate $B^{-1} A^{-1}$ and Verify Equality
$B^{-1} A^{-1} = \begin{bmatrix} -4.5 & 4 \\ 3.5 & -3 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$
$B^{-1} A^{-1} = \begin{bmatrix} (-4.5 \times 5) + (4 \times -2) & (-4.5 \times -7) + (4 \times 3) \\ (3.5 \times 5) + (-3 \times -2) & (3.5 \times -7) + (-3 \times 3) \end{bmatrix}$
$B^{-1} A^{-1} = \begin{bmatrix} -22.5 - 8 & 31.5 + 12 \\ 17.5 + 6 & -24.5 - 9 \end{bmatrix} = \begin{bmatrix} -30.5 & 43.5 \\ 23.5 & -33.5 \end{bmatrix}$
Final Solution: Since $(AB)^{-1} = \begin{bmatrix} -30.5 & 43.5 \\ 23.5 & -33.5 \end{bmatrix}$ and $B^{-1} A^{-1} = \begin{bmatrix} -30.5 & 43.5 \\ 23.5 & -33.5 \end{bmatrix}$, it is verified that $(AB)^{-1} = B^{-1} A^{-1}$.
More Questions from Class 12 Mathematics Determinants EXERCISE 4.4
- Q1: Find adjoint of the matrix $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
- Q2: Find adjoint of the matrix $\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$
- Q3: Verify $A (adj A) = (adj A) A = |A| I$ for the matrix $\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$
- Q4: Verify $A (adj A) = (adj A) A = |A| I$ for the matrix $\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$
- Q5: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$
- Q6: Find the inverse of the matrix (if it exists): $\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$
- Q7: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$
- Q8: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$
- Q9: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$
- Q10: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$
- Q11: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$
- Q13: If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 – 5A + 7I = O$. Hence find $A^{-1}$.
- Q14: For the matrix $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$, find the numbers $a$ and $b$ such that $A^2 + aA + bI = O$.
- Q15: For the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$ Show that $A^3– 6A^2 + 5A + 11 I = O$. Hence, find $A^{-1}$.
- Q16: If $A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$ Verify that $A^3 – 6A^2 + 9A – 4I = O$ and hence find $A^{-1}$
- Q17: Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|adj A|$ is equal to
- Q18: If $A$ is an invertible matrix of order 2, then det $(A^{-1})$ is equal to
CBSE Solutions for Class 12 Mathematics Determinants
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