Question

Choose the correct answer.
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

• A. \(\begin{bmatrix}x^{-1}& 0& 0\\ 0& y^{-}1& 0\\0&0& z^{-1}\end{bmatrix}\)
• B. xyz\(\begin{bmatrix}x^{-1}& 0& 0\\ 0& y^{-}1& 0\\0&0& z^{-1}\end{bmatrix}\)
• C. \(\frac{1}{xyz}\begin{bmatrix}x& 0& 0\\ 0& y& 0\\0&0& z\end{bmatrix}\)
• D. \(\frac{1}{xyz}\begin{bmatrix}1& 0& 0\\ 0& 1& 0\\0&0& 1\end{bmatrix}\)

Answer 1

The Correct Option is A

A=\(\begin{bmatrix}x& 0& 0\\ 0& y& 0\\0&0& z\end{bmatrix}\)
\(∴|A|=x(yz-0)=xyz≠0\)

Now,
\(A_{11}=yz,A_{12}=0,A_{13}=0\)
\(A_{21}=0,A_{22}=xz,A_{23}=0\)
\(A_{31}=0,A_{32}=0,A_{33}=xy\)

\(∴adjA\)=\(\begin{bmatrix}yz& 0& 0\\ 0& xz& 0\\0&0& xy\end{bmatrix}\)

\(∴A^{-1}\)=\(\frac{1}{|A|}\)\(adjA\)
=\(\frac{1}{xyz}\begin{bmatrix}yz& 0& 0\\ 0& xz& 0\\0&0& xy\end{bmatrix}\)
=\(\begin{bmatrix}\frac{yz}{xyz}& 0& 0\\ 0& \frac{xz}{xyz}& 0\\0&0& \frac{xy}{xyz}\end{bmatrix}\)
=\(\begin{bmatrix}\frac{1}{x}& 0& 0\\ 0& \frac{1}{y}& 0\\0&0& \frac{1}{z}\end{bmatrix}\)
=\(\begin{bmatrix}x^{-1}& 0& 0\\ 0& y^{-1}& 0\\0&0& z^{-1}\end{bmatrix}\)

The correct answer is A.