If A = $\left[\begin{array}{lll}{2} & {4} & {4} \\ {4} & {2} & {4} \\ {4} & {4} & {2}\end{array}\right]$ show that $A^{2}

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written 5.2 years ago by

teamques10 ★ 68k

Solution:

$A^{2} - 8A$

$= A.A - 8A$

$=\left[\begin{array}{lll}{2} & {4} & {4} \\ {4} & {2} & {4} \\ {4} & {4} & {2}\end{array}\right]\left[\begin{array}{lll}{2} & {4} & {4} \\ {4} & {2} & {4} \\ {4} & {4} & {2}\end{array}\right]-8\left[\begin{array}{rrr}{2} & {4} & {4} \\ {4} & {2} & {4} \\ {4} & {4} & {2}\end{array}\right]$

$=\left[\begin{array}{ccc}{36} & {32} & {32} \\ {32} & {36} & {32} \\ {32} & {32} & {36}\end{array}\right]-\left[\begin{array}{ccc}{16} & {32} & {32} \\ {32} & {16} & {32} \\ {32} & {32} & {16}\end{array}\right]$

$=\left[\begin{array}{ccc}{20} & {0} & {0} \\ {0} & {20} & {0} \\ {0} & {0} & {20}\end{array}\right]$

$\therefore A^{2}-8 A is scalar matrix$

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