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Expansion of (x^2 - y^2)

If A=[[ 4 -1;...

Question

If $A = [\begin{matrix} 4 & - 1 x & - 3 \end{matrix}]$ is an idempotent matrix, then the value of $x$ is

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Solution

Given $A = [\begin{matrix} 4 & - 1 x & - 3 \end{matrix}]$ is an idempotent matrix.
We know that for an idempotent matrix, $A^{2} = A$
$A^{2} = [\begin{matrix} 4 & - 1 x & - 3 \end{matrix}] [\begin{matrix} 4 & - 1 x & - 3 \end{matrix}]$
$= [\begin{matrix} 16 - x & - 1 x & 9 - x \end{matrix}] = [\begin{matrix} 4 & - 1 x & - 3 \end{matrix}]$
Equating the terms, we get
$16 - x = 4 \Rightarrow x = 12$
or,
$9 - x = - 3 \Rightarrow x = 12$
$∴ x = 12$

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Expansion of (x^2 - y^2)

Standard VIII Mathematics

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