Let $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$. Verify that $A(\text{adj } A) = (\text{adj } A) A = |A| I$.
(A) Given $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$.
First,calculate the determinant $|A|$:
$|A| = (2)(-6) - (3)(-4) = -12 + 12 = 0$.
Thus,$|A| I = 0 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Next,find the adjoint of $A$ $(\text{adj } A)$:
The cofactors are $C_{11} = -6, C_{12} = 4, C_{21} = -3, C_{22} = 2$.
Therefore,$\text{adj } A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}$.
Now,calculate $A(\text{adj } A)$:
$A(\text{adj } A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Finally,calculate $(\text{adj } A) A$:
$(\text{adj } A) A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} = \begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Since $A(\text{adj } A) = (\text{adj } A) A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = |A| I$,the identity is verified.
First,calculate the determinant $|A|$:
$|A| = (2)(-6) - (3)(-4) = -12 + 12 = 0$.
Thus,$|A| I = 0 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Next,find the adjoint of $A$ $(\text{adj } A)$:
The cofactors are $C_{11} = -6, C_{12} = 4, C_{21} = -3, C_{22} = 2$.
Therefore,$\text{adj } A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}$.
Now,calculate $A(\text{adj } A)$:
$A(\text{adj } A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Finally,calculate $(\text{adj } A) A$:
$(\text{adj } A) A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} = \begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Since $A(\text{adj } A) = (\text{adj } A) A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = |A| I$,the identity is verified.
Explore More
Similar Questions
Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=81$. If $S =\{ n \in \mathbb{Z} :(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^2}{2}}=|A|^{(3n^2-5n-4)}\}$,then $\sum_{n \in S}|A^{(n^2+n)}|$ is equal to
For matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$,$(A^{-1})^2 = $ . . . . . .
If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \operatorname{adj} A = A^T$,then $5a + b$ is equal to
Let $\alpha, \beta, \gamma$ be real numbers. If $A=\begin{bmatrix} 7 & 3 & \alpha \\ \beta & 1 & -11 \\ -5 & \gamma & 19 \end{bmatrix}$ is a $3 \times 3$ matrix satisfying $A\begin{bmatrix} 5 \\ -13 \\ 11 \end{bmatrix}=\begin{bmatrix} -290 \\ -119 \\ 210 \end{bmatrix}$, then $(\operatorname{adj} A)^{-1}+\operatorname{adj} A^{-1}=$
DifficultTS EAMCET 2022
View SolutionIf $A = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}$ and $\text{adj } A = \begin{bmatrix} 5 & x & -2 \\ 1 & 1 & 0 \\ -2 & -2 & y \end{bmatrix}$,then the value of $x+y$ is:
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