First slide

Operations of matrix

Question

If $A = [\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{array}]$ , then $A^{2} - 5 A + 6 I$ is equal to

Easy

A

$[\begin{array}{ccc} 1 & - 1 & - 3 \\ - 1 & - 1 & - 10 \\ - 5 & 4 & 4 \end{array}]$
B

$[\begin{array}{ccc} 1 & 1 & - 5 \\ - 1 & - 1 & 4 \\ - 3 & - 10 & 4 \end{array}]$
C

0
D

I

Solution

$A^{2} = [\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{array}] [\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{array}] = [\begin{array}{ccc} 5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2 \end{array}]$
Now, $A^{2} - 5 A + 6 I$
$\begin{array}{l} = [\begin{array}{clc} 5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2 \end{array}] - 5 [\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{array}] + 6 [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ = [\begin{array}{ccc} 5 - 10 + 6 & - 1 - 0 + 0 & 2 - 5 + 0 \\ 9 - 10 + 0 & - 2 - 5 + 6 & 5 - 15 + 0 \\ 0 - 5 + 0 & - 1 + 5 + 0 & - 2 - 0 + 6 \end{array}] \end{array}$
$= [\begin{array}{ccc} 1 & - 1 & - 3 \\ - 1 & - 1 & - 10 \\ - 5 & 4 & 4 \end{array}]$

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