Given A = begin{bmatrix} x & 3 & 2 1 & y & 4 | vedclass

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

Question

(C) The property of matrices states that $A \cdot (	ext{adj } A) = |A| I$,where $I$ is the identity matrix of order $3 	imes 3$.First,we calculate t

Vedclass · Accepted Answer

$\begin{bmatrix} 68 & 0 & 0 \ 0 & 68 & 0 \ 0 & 0 & 68 \end{bmatrix}$

Vedclass · Answer

(C) The property of matrices states that $A \cdot (	ext{adj } A) = |A| I$,where $I$ is the identity matrix of order $3 	imes 3$.First,we calculate the determinant of matrix $A$:$|A| = x(yz - 8) - 3(z - 8) + 2(2 - 2y)$$|A| = xyz - 8x - 3z + 24 + 4 - 4y$$|A| = xyz - (8x + 4y + 3z) + 28$Substituting the given values $xyz = 60$ and $8x + 4y + 3z = 20$:$|A| = 60 - 20 + 28 = 68$Therefore,$A \cdot (	ext{adj } A) = 68 I = \begin{bmatrix} 68 & 0 & 0 \ 0 & 68 & 0 \ 0 & 0 & 68 \end{bmatrix}$.

Given $A = \begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix}$,$xyz = 60$ and $8x + 4y + 3z = 20$,then $A \cdot (\text{adj } A)$ is equal to

Similar Questions

If $A = \begin{bmatrix} 2 & 2 \\ 9 & 4 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,then $10 A^{-1}$ is equal to

If $A$ is a square matrix of order $3 \times 3$,then $|\operatorname{adj} A| =$ . . . . . . .

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix}$,then the inverse of $(2A^2 + 5A)$ is

If $A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$,then $A \cdot \operatorname{adj}(A)$ is equal to

If $A^2-A+I=0$, then the inverse of the matrix $A$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module