If x, y, z are nonzero real numbers,then the | vedclass

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

Question

(A) Given the diagonal matrix $A = \begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}$.The determinant of $A$ is $|A| = x(yz - 0) - 0 +

Vedclass · Accepted Answer

$\begin{bmatrix} x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1} \end{bmatrix}$

Vedclass · Answer

(A) Given the diagonal matrix $A = \begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}$.The determinant of $A$ is $|A| = x(yz - 0) - 0 + 0 = xyz$.Since $x, y, z 
eq 0$,$|A| = xyz 
eq 0$,so $A^{-1}$ exists.The inverse of a diagonal matrix $A = 	ext{diag}(x, y, z)$ is given by $A^{-1} = 	ext{diag}(x^{-1}, y^{-1}, z^{-1})$.Calculating the adjugate matrix:$adj(A) = \begin{bmatrix} yz & 0 & 0 \ 0 & xz & 0 \ 0 & 0 & xy \end{bmatrix}$.Thus,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{xyz} \begin{bmatrix} yz & 0 & 0 \ 0 & xz & 0 \ 0 & 0 & xy \end{bmatrix} = \begin{bmatrix} \frac{yz}{xyz} & 0 & 0 \ 0 & \frac{xz}{xyz} & 0 \ 0 & 0 & \frac{xy}{xyz} \end{bmatrix} = \begin{bmatrix} \frac{1}{x} & 0 & 0 \ 0 & \frac{1}{y} & 0 \ 0 & 0 & \frac{1}{z} \end{bmatrix} = \begin{bmatrix} x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1} \end{bmatrix}$.

If $x, y, z$ are nonzero real numbers,then the inverse of matrix $A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ is

Similar Questions

If $\begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} \begin{bmatrix} 1 & \tan \theta \\ -\tan \theta & 1 \end{bmatrix}^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$,then

If $A=\begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \operatorname{adj} A = A A^{T}$,then $2a + 3b$ is

If $P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$ is the adjoint of a matrix $A$ and $\det(A) = 4$,then the value of $\alpha$ is

Find the inverse of the matrix $A = \left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]$,if it exists.

If $A = \left[\begin{array}{cc}1+2 i & i \\ -i & 1-2 i\end{array}\right]$ where $i=\sqrt{-1}$,then $A (\operatorname{adj} A )=\ldots$. (in $I$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module