Matrix A = begin{bmatrix} x & 3 & 2 1 & y & | 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 $A$,the property $A (adj A) = |A| I$ holds,where $I$ is the identity matrix of the same order.First,we calculat

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We know that for any square matrix $A$,the property $A (adj A) = |A| I$ holds,where $I$ is the identity matrix of the same order.First,we calculate the determinant $|A|$ of the matrix $A = \begin{bmatrix} x & 3 & 2 \ 1 & y & 4 \ 2 & 2 & z \end{bmatrix}$.$|A| = x(yz - 8) - 3(z - 8) + 2(2 - 2y)$$|A| = xyz - 8x - 3z + 24 + 4 - 4y$$|A| = xyz - (8x + 4y + 3z) + 28$Given $xyz = 60$ and $8x + 4y + 3z = 20$,we substitute these values into the expression for $|A|$:$|A| = 60 - 20 + 28 = 68$.Therefore,$A (adj A) = |A| I = 68 I = \begin{bmatrix} 68 & 0 & 0 \ 0 & 68 & 0 \ 0 & 0 & 68 \end{bmatrix}$.

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

Similar Questions

If $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$,$10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$ and $B$ is the inverse of $A$,then the value of $\alpha$ is

Matrices $A$ and $B$ will be inverse of each other only if

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$,$10B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$ and $B$ is the inverse of matrix $A$,then $\alpha$ is equal to . . . . . . .

If $A$ is a square matrix of order $3$ such that $\operatorname{det}(A)=3$ and $\operatorname{det}\left(\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2A)^{-1}\right)\right)\right)\right)\right)=2^{m} 3^{n}$,then $m+2n$ is equal to:

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 Test Series

Exam Paper Generator

Online Exam Module