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Home  >>  CBSE XII  >>  Math  >>  Determinants
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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

$\begin{array}{l \hspace{40 mm} l} (A) \, \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}\; \quad& (B)\, xyz \,\begin{bmatrix} x^{-1} & 0&0 \\ 0& y^{-1} &0 \\ 0 & 0 & z^{-1} \end{bmatrix} \\[1em] (C) \, \frac{1}{xyz} \,\begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \quad & (D)\, \frac{1}{xyz} \,\begin{bmatrix} 1 & 0&0 \\ 0& 1 &0 \\ 0 & 0 & 1 \end{bmatrix} \end{array} $
Can you answer this question?
 
 

1 Answer

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Toolbox:
  • A matrix is said to be singular if |A|=0.
  • A matrix is said to be invertible if |A|$\neq 0.$
  • $A^{-1}=\frac{1}{|A|}(adj A)$
Let $A=\begin{vmatrix}x & 0 & 0\\0 &y &0\\0 & 0 & z\end{vmatrix}$
 
Let us find the value of determinant A by expanding along $R_1$
 
$|A|=x(yz-0)-0+0.$
 
$\Delta=xyz\neq 0$
 
$A^{-1}\;exists.$
 
Let us find the adjoint of A
 
The cofactors of matrix A are,
 
$M_{11}=\begin{vmatrix}y& 0\\0 & z\end{vmatrix}=yz.$
 
$M_{12}=\begin{vmatrix}0& 0\\0 & z\end{vmatrix}=0.$
 
$M_{13}=\begin{vmatrix}0& y\\0 & 0\end{vmatrix}=0.$
 
$M_{21}=\begin{vmatrix}0& 0\\0 & z\end{vmatrix}=0.$
 
$M_{22}=\begin{vmatrix}x& 0\\0 & z\end{vmatrix}=xz.$
 
$M_{23}=\begin{vmatrix}x& 0\\0 & 0\end{vmatrix}=0.$
 
$M_{31}=\begin{vmatrix}& 0\\y & 0\end{vmatrix}=0.$
 
$M_{32}=\begin{vmatrix}x& 0\\0 & 0\end{vmatrix}=0.$
 
$M_{33}=\begin{vmatrix}x& 0\\0 & y\end{vmatrix}=xy.$
 
$A_{11}=(-1)^{1+1}.yz=yz.$
 
$A_{12}=0.$
 
$A_{13}=0.$
 
$A_{21}=0.$
 
$A_{22}=(-1)^{2+2}.xz=xz.$
 
$A_{23}=0.$
 
$A_{31}=0.$
 
$A_{32}=0.$
 
$A_{33}=(-1)^{3+3}.xy=xy.$
 
adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}yz & 0 & 0\\0 & xz & 0\\0 & 0 & xy\end{bmatrix}$
 
We know $A^{-1}=\frac{1}{|A|}adj A$,|A|=xyz.
 
$A^{-1}=\frac{1}{xyz}\begin{bmatrix}yz & 0 & 0\\0 & xz & 0\\0 & 0 & xy\end{bmatrix}$
 
$\;\;\;=\begin{bmatrix}yz /xyz& 0 & 0\\0 & xz/xyz & 0\\0 & 0 & xy/xyz\end{bmatrix}=\begin{bmatrix}1/x & 0 & 0\\0 & 1/y & 0\\0 & 0 &1/z\end{bmatrix}$
 
$\;\;\;=\begin{bmatrix}x^{-1} & 0 & 0\\0 & y ^{-1} & 0\\0 & 0 &z^{-1}\end{bmatrix}$
 
Hence the correct answer is A.

 

answered Mar 4, 2013 by sreemathi.v
 

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