Let $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(AB)^{-1} = B^{-1} A^{-1}$.
(A) Given $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$.
$|A| = (3 \times 5) - (7 \times 2) = 15 - 14 = 1$.
$adj(A) = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} adj(A) = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.
Given $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$.
$|B| = (6 \times 9) - (8 \times 7) = 54 - 56 = -2$.
$adj(B) = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$.
$B^{-1} = \frac{1}{|B|} adj(B) = -\frac{1}{2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix}$.
Now,$B^{-1} A^{-1} = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} -\frac{45}{2} - 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & -\frac{49}{2} - 9 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \dots (1)$.
Now,$AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 18+49 & 24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$.
$|AB| = (67 \times 61) - (87 \times 47) = 4087 - 4089 = -2$.
$adj(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$.
$(AB)^{-1} = \frac{1}{|AB|} adj(AB) = -\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \dots (2)$.
From $(1)$ and $(2)$,$(AB)^{-1} = B^{-1} A^{-1}$ is verified.
$|A| = (3 \times 5) - (7 \times 2) = 15 - 14 = 1$.
$adj(A) = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} adj(A) = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.
Given $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$.
$|B| = (6 \times 9) - (8 \times 7) = 54 - 56 = -2$.
$adj(B) = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$.
$B^{-1} = \frac{1}{|B|} adj(B) = -\frac{1}{2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix}$.
Now,$B^{-1} A^{-1} = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} -\frac{45}{2} - 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & -\frac{49}{2} - 9 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \dots (1)$.
Now,$AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 18+49 & 24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$.
$|AB| = (67 \times 61) - (87 \times 47) = 4087 - 4089 = -2$.
$adj(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$.
$(AB)^{-1} = \frac{1}{|AB|} adj(AB) = -\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \dots (2)$.
From $(1)$ and $(2)$,$(AB)^{-1} = B^{-1} A^{-1}$ is verified.
Explore More
Similar Questions
Let $A = \begin{bmatrix} 2k - 1 & 1 & 1 \\ 0 & 2k - 1 & 1 \\ 0 & 0 & 2k - 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2k - 1 & 1 \\ 1 - 2k & 0 & k \\ -1 & -k & 0 \end{bmatrix}$ where $k$ is a real number. If $\det(\text{adj } A) + \det(\text{adj } B) = 11^6$, then the value of $k - 5$ is equal to...
DifficultMHT CET 2026
View SolutionIf $A$ is a non-singular matrix and $(A+I)(A-I)=0$,then $A+A^{-1} = \dots$
Inverse of the matrix $\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$ is
Easy
View SolutionIf $A = \begin{bmatrix} 1 & \cot \frac{\theta}{2} \\ -\cot \frac{\theta}{2} & 1 \end{bmatrix}$,then $A^{-1} =$
If $M$ is any square matrix of order $3$ over $\mathbb{R}$ and if $M^{\prime}$ is the transpose of $M$, then $\text{adj}(M^{\prime}) - (\text{adj } M)^{\prime}$ is equal to
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo