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Q3:
If $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$, find $(AB)^{-1}$.
Solution :
Step 1: Theoretical Foundation of Inverse of Matrix Products
To find the inverse of the product of two matrices, we utilize the property of reversal of order [Per the Property of Invertible Matrices]: $$(AB)^{-1} = B^{-1}A^{-1}$$ Given the variables: $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$
Step 2: Calculating $B^{-1}$
The inverse of a matrix $B$ is given by $B^{-1} = \frac{1}{|B|} \text{adj}(B)$.
First, calculate $|B|$ expanding along the first row: $|B| = 1(3(1) - 0(-2)) - 2(-1(1) - 0(0)) + (-2)(-1(-2) - 3(0))$ $|B| = 1(3) - 2(-1) - 2(2) = 3 + 2 - 4 = 1$
Next, find the matrix of cofactors $C_{ij}$:
| $C_{11} = +(3-0) = 3$ | $C_{12} = -(-1-0) = 1$ | $C_{13} = +(2-0) = 2$ |
| $C_{21} = -(2-4) = 2$ | $C_{22} = +(1-0) = 1$ | $C_{23} = -(-2-0) = 2$ |
| $C_{31} = +(0-(-6)) = 6$ | $C_{32} = -(0-2) = 2$ | $C_{33} = +(3-(-2)) = 5$ |
The Adjoint, $\text{adj}(B) = [C_{ij}]^T = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$. Since $|B| = 1$, $B^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$.
Step 3: Calculating $(AB)^{-1} = B^{-1}A^{-1}$
Perform the matrix multiplication: $(AB)^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$
Calculating each element $(i, j)$ using the dot product of the $i$-th row of $B^{-1}$ and $j$-th column of $A^{-1}$:
- Row 1: $(3)(3) + (2)(-15) + (6)(5) = 9 - 30 + 30 = 9$ $(3)(-1) + (2)(6) + (6)(-2) = -3 + 12 - 12 = -3$ $(3)(1) + (2)(-5) + (6)(2) = 3 - 10 + 12 = 5$
- Row 2: $(1)(3) + (1)(-15) + (2)(5) = 3 - 15 + 10 = -2$ $(1)(-1) + (1)(6) + (2)(-2) = -1 + 6 - 4 = 1$ $(1)(1) + (1)(-5) + (2)(2) = 1 - 5 + 4 = 0$
- Row 3: $(2)(3) + (2)(-15) + (5)(5) = 6 - 30 + 25 = 1$ $(2)(-1) + (2)(6) + (5)(-2) = -2 + 12 - 10 = 0$ $(2)(1) + (2)(-5) + (5)(2) = 2 - 10 + 10 = 2$
Visual Representation of Matrix Transformation
Final Solution: $(AB)^{-1} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}$
More Questions from Class 12 Mathematics Determinants EXERCISE Miscellaneous
- Q1: Prove that the determinant $\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$ is independent of $\theta$.
- Q2: Evaluate $\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$.
- Q4(i): Let $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that (i) $[adj A]^{-1} = adj (A^{-1})$
- Q4(ii): Let $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that (ii) $(A^{-1})^{-1} = A$
- Q5: Evaluate $\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$
- Q6: Evaluate $\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$
- Q7: Solve the system of equations $\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$, $\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$, $\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$
- Q8: If $x, y, z$ are nonzero real numbers, then the inverse of matrix $A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ is
- Q9: Let $A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}$, where $0 \leq \theta \leq 2\pi$. Then
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