If A = begin{bmatrix} 3 & 4 5 & 7 end{bmatri | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that for any square matrix $A$ of order $n$,the property $A(adj A) = (adj A)A = |A|I$ holds true,where $I$ is the identity matrix of the s

Vedclass · Accepted Answer

$|A|I$

Vedclass · Answer

(C) We know that for any square matrix $A$ of order $n$,the property $A(adj A) = (adj A)A = |A|I$ holds true,where $I$ is the identity matrix of the same order.Given $A = \begin{bmatrix} 3 & 4 \ 5 & 7 \end{bmatrix}$.First,calculate the determinant $|A| = (3 	imes 7) - (4 	imes 5) = 21 - 20 = 1$.The adjoint of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:$adj A = \begin{bmatrix} 7 & -4 \ -5 & 3 \end{bmatrix}$.Now,calculate $A(adj A)$:$A(adj A) = \begin{bmatrix} 3 & 4 \ 5 & 7 \end{bmatrix} \begin{bmatrix} 7 & -4 \ -5 & 3 \end{bmatrix} = \begin{bmatrix} (21 - 20) & (-12 + 12) \ (35 - 35) & (-20 + 21) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Since $|A| = 1$,we have $I = |A|I$.Therefore,$A(adj A) = |A|I$.

If $A = \begin{bmatrix} 3 & 4 \\ 5 & 7 \end{bmatrix}$,then $A(adj A) = $

Similar Questions

Let the matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ and the matrix $B_{0} = A^{49} + 2A^{98}$. If $B_{n} = \text{Adj}(B_{n-1})$ for all $n \geq 1$,then $\det(B_{4})$ is equal to:

If $A$ is a square matrix of order $3$ and $|A|=5$,then $|A \text{ adj. } A|$ is

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$, then the matrix $A^{-3}$ when $\theta = \pi/6$ is equal to...

If $A$ is a non-singular matrix such that $(A-2I)(A-4I)=0$,then $A+8A^{-1} = \_\_\_\_$

Inverse of the matrix $\left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module