If is a non singular matrix, then find A–1 by using elementary row tranformation. Hence, find the inverse of

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

If $A = [\begin{array}{lll} x 0 0 \\ 0 y 0 \\ 0 0 z \end{array}]$ is a non singular matrix, then find A^–1 by using elementary row tranformation. Hence, find the inverse of $[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}]$

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Since A is a non-singular matrix, A^–1 exists. We write AA^–1 = 1
$∴ [\begin{array}{lll} x 0 0 \\ 0 y 0 \\ 0 0 z \end{array}] A^{- 1} = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}]$

By $(\frac{1}{x}) R_{1}, (\frac{1}{y}) R_{2} and (\frac{1}{z}) R_{3}$ , we get,
$[\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A^{- 1} = [\begin{array}{ccc} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{array}]$

$∴ A^{- 1} = [\begin{array}{ccc} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{array}]$
Comparing $[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}]$ with $[\begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}]$
We get, x = 2, y = 1, z = – 1
$∴ \frac{1}{x} = \frac{1}{2}, \frac{1}{y} = \frac{1}{1} = 1, \frac{1}{z} = \frac{1}{- 1} = - 1$
Hence, the inverse of
$[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}]$ is $[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}]$

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