If A- begin bmatrix 1 - 2 - 10 - 1 -13 -1 - 1end bmatrix - then

The correct option is D A^3 3A^2 A + 9I_3 = 0[ 1 amp; 2 amp; 1; 0 amp; 1 amp; 1; 3 amp; 1 amp; 1 ] A^2 = A.A = [ 1 amp; 2 ...

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Byju's Answer

Standard X

Mathematics

Multiplication of Matrices

If A= begin b...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ , then

A^{3} + 3 A^{2} + A - 9 I_{3} = 0

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A^{3} - 3 A^{2} + A + 9 I_{3} = 0

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A^{3} + 3 A^{2} - A + 9 I_{3} = 0

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A^{3} - 3 A^{2} - A + 9 I_{3} = 0

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Solution

The correct option is D $A^{3} - 3 A^{2} - A + 9 I_{3} = 0$
$⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ A^{2} = A . A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 4 & 3 & 0 - 3 & 2 & - 2 6 & 4 & 5 \end{matrix} ⎤ ⎥ ⎦ A . A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 4 & 3 & 0 - 9 & - 2 & - 7 21 & 11 & 7 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow A^{3} - 3 A^{2} - A + 9 I_{3} = 0$

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If A=⎡⎢⎣12101−13−11⎤⎥⎦, then

The correct option is D A3−3A2−A+9I3=0⎡⎢⎣12101−13−11⎤⎥⎦A2=A.A=⎡⎢⎣12101−13−11⎤⎥⎦⎡⎢⎣12101−13−11⎤⎥⎦=⎡⎢⎣430−32−2645⎤⎥⎦A.A2=⎡⎢⎣12101−13−11⎤⎥⎦⎡⎢⎣430−9−2−721117⎤⎥⎦⇒A3−3A2−A+9I3=0

If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ , then