If A - bmatrix 3 - 1 1-182lend bmatrix and I - begin bmatrix 180 10 - 1lend bmatrix - find k so that A 2-5 A - kI-

A^2 =[ 3 amp; 1; 1 amp; 2 ][ 3 amp; 1; 1 amp; 2 ]=[ 8 amp; 5; 5 amp; 2 ], 5A = [ 15 amp; 5; 5 amp; 10 ], kI =[ k ...

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Byju's Answer

Standard XII

Mathematics

Elementary Transformations

If A = bmatri...

Question

If $A = [\begin{matrix} 3 & 1 - 1 & 2 \end{matrix}]$ and $I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ , find k so that $A^{2} = 5 A + k I$ .

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Solution

$A^{2} = [\begin{matrix} 3 & 1 - 1 & 2 \end{matrix}] [\begin{matrix} 3 & 1 - 1 & 2 \end{matrix}] = [\begin{matrix} 8 & 5 - 5 & 2 \end{matrix}], 5 A = [\begin{matrix} 15 & 5 - 5 & 10 \end{matrix}], k I = [\begin{matrix} k & 0 0 & k \end{matrix}]$
Now $A^{2} = 5 A + k I$ so, $[\begin{matrix} 8 & 5 - 5 & 3 \end{matrix}] = [\begin{matrix} 15 & 5 - 5 & 10 \end{matrix}] + [\begin{matrix} k & 0 0 & k \end{matrix}] \Rightarrow [\begin{matrix} 8 & 5 - 5 & 3 \end{matrix}] = [\begin{matrix} 15 + k & 5 - 5 & 10 + k \end{matrix}]$
On comparing the corresponding elements in both matrices, we get:
8 =15 +k, 3 =10 +k $∴ k = - 7$ .

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If A=[31−12] and I=[1001], find k so that A2=5A+kI.

A2=[31−12][31−12]=[85−52],5A=[155−510],kI=[k00k] Now A2=5A+kI so, [85−53]=[155−510]+[k00k]⇒[85−53]=[15+k5−510+k] On comparing the corresponding elements in both matrices, we get: 8 =15 +k, 3 =10 +k ∴k=−7.

If $A = [\begin{matrix} 3 & 1 - 1 & 2 \end{matrix}]$ and $I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ , find k so that $A^{2} = 5 A + k I$ .