a. If A=1 2 22 1 22 2 1 then show that A2−4A−5I=O (null matrix)b. A=1−2101−13−11 then show that A3−3A2−A−3I=O

Question

a. If A=1    2    22    1    22    2    1 then show that A2−4A−5I=O (null matrix)b. A=1−2101−13−11 then show that A3−3A2−A−3I=O

Accepted Answer

A=1    2    22    1    22    2    1A2=1    2    22    1    22    2    11    2    22    1    22    2    1=1+4+4    2+2+4    2+4+22+2+4    4+1+4    4+2+22+4+2    4+2+2    4+4+1=9    8    88    9    88    8    9L.H.S = A2−4A−5I=9    8    88    9    88    8    9−41    2    22    1    22    2    1−51    0    00    1    00    0    1=9    8    88    9    88    8    9−4    8    88    4    88    8    4−5    0    00    5    00    0    5=9−4−5    8−8−0    8−8−08−8−0    9−4−5    8−8−08−8−0    8−8−0    9−4−50    0    00    0    00    0    0=O3×3=R⋅H⋅SA2=A⋅A=1−2101−13−111−2101−13−11A2=1−0+3−2−2−11+2+10+0−30+1+10−1−13+0+3−6−1−13+1+1=4−54−32−26−85A3=A2⋅A=4−54−32−26−851−2101−13−11=4+0+12−8−5−44+5+4−3+0−66+2+2−3−2−26+0+15−12−8−56+8+5=16−1713−910−721−2519 Now3A2=34−54−32−26−85=12−1512−98−618−2415∴A3−3A2−A−3I=16−1713−910−721−2519−12−1512−96−618−2415−1−2101−13−11−3100010001=16−12−1−3−17−(−15)−(−2)−013−12−1−0−9−(−9)−0−010−6−1−3−7−(−6)−1(−1)−021−18−3−0−25−(−24)−(−1)−019−15−1−3=0    0    00    0    00    0    0=O

$a. If A = [\begin{array}{lll} 1 2 2 \\ 2 1 2 \\ 2 2 1 \end{array}]$ then show that $A^{2} - 4 A - 5 I = O$ (null matrix)
$b . A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 0 & 1 & - 1 \\ 3 & - 1 & 1 \end{array}]$ then show that $A^{3} - 3 A^{2} - A - 3 I = O$

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