If A- begin bmatrix 1 - 0 - 2 1 10 - 2 - 1 -2 - 0 - 311lend bmatrix - prove that A3-6 A2-7 A-2 I-0

A^2 =A × A=[ 1 amp; 0 amp; 2; 0 amp; 2 amp; 1; 2 amp; 0 amp; 3 ][ 1 amp; 0 amp; 2; 0 amp; 2 amp; 1; 2 amp; 0 amp; 3; ] =[ ...

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Byju's Answer

Standard XII

Mathematics

Transpose of a Matrix

If A=⟨ begin ...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦$ , prove that $A^{3} - 6 A^{2} + 7 A + 2 I = 0.$

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Solution

$A^{2} = A \times A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 + 0 + 4 & 0 + 0 + 0 & 2 + 0 + 6 0 + 0 + 2 & 0 + 4 + 0 & 0 + 2 + 3 2 + 0 + 6 & 0 + 0 + 0 & 4 + 0 + 9 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 5 & 0 & 8 2 & 4 & 5 8 & 0 & 13 \end{matrix} ⎤ ⎥ ⎦ A^{3} = A^{2} \times A = ⎡ ⎢ ⎣ \begin{matrix} 5 & 0 & 8 2 & 4 & 5 8 & 0 & 13 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 5 + 0 + 16 & 0 + 0 + 0 & 10 + 0 + 24 2 + 0 + 10 & 0 + 8 + 0 & 4 + 4 + 15 8 + 0 + 26 & 0 + 0 + 0 & 16 + 0 + 39 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 21 & 0 & 34 12 & 8 & 23 34 & 0 & 55 \end{matrix} ⎤ ⎥ ⎦ ∴ A^{3} - 6 A^{2} + 7 A + 2 I = ⎡ ⎢ ⎣ \begin{matrix} 21 & 0 & 34 12 & 8 & 23 34 & 0 & 55 \end{matrix} ⎤ ⎥ ⎦ - 6 ⎡ ⎢ ⎣ \begin{matrix} 5 & 0 & 8 2 & 4 & 5 8 & 0 & 13 \end{matrix} ⎤ ⎥ ⎦ + 7 ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 0 & 2 & 1 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ + 2 ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 21 & 0 & 34 12 & 8 & 23 34 & 0 & 55 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 30 & 0 & 48 12 & 24 & 30 48 & 0 & 78 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 7 & 0 & 14 0 & 14 & 7 14 & 0 & 21 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 0 0 & 2 & 0 0 & 0 & 2 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 21 - 30 + 7 + 2 & 0 - 0 + 0 + 0 & 34 - 48 + 14 + 0 12 - 12 + 0 + 0 & 8 - 24 + 14 + 2 & 23 - 30 + 7 + 0 34 - 48 + 14 + 0 & 0 - 0 + 0 + 0 & 55 - 78 + 21 + 2 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 0 & 0 & 0 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎦ = 0$
Hence proved.

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