If A - begin bmatrix 38-2 4 -2end bmatrix and I - bmatrix 1 - 1 110 - 0 1end bmatrix - then find k so that A 2- kA -2 I-

Given A^2 =kA 2I ⇒ AA=kA 2I ⇒[ 3 amp; 2; 4 amp; 2; ][ 3 amp; 2; 4 amp; 2; ]=k [ 3 amp; 2; 4 amp; 2; ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Matrix

If A = begin ...

Question

If $A = [\begin{matrix} 3 & - 2 4 & - 2 \end{matrix}] a n d I = [\begin{matrix} 1 & 1 0 & 0 \end{matrix}]$ , then find k so that $A^{2} = k A - 2 I .$

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Solution

Given $A^{2} = k A - 2 I \Rightarrow A A = k A - 2 I$
$\Rightarrow [\begin{matrix} 3 & - 2 4 & - 2 \end{matrix}] [\begin{matrix} 3 & - 2 4 & - 2 \end{matrix}] = k [\begin{matrix} 3 & - 2 4 & - 2 \end{matrix}] - 2 [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 9 - 8 & - 6 + 4 12 - 8 & - 8 + 4 \end{matrix}] = [\begin{matrix} 3 k & - 2 k 4 k & - 2 k \end{matrix}] - [\begin{matrix} 2 & 0 0 & 2 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 2 4 & - 4 \end{matrix}] = [\begin{matrix} 3 k - 2 & - 2 k 4 k & - 2 k - 2 \end{matrix}]$
By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get
$3 k - 2 = 1 \Rightarrow k = 1 - 2 k = - 2 \Rightarrow k = 1 4 k = 4 \Rightarrow k = 1$
$- 4 = - 2 k - 2 \Rightarrow k = 1$ Hence, k=1

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If A=[3−24−2]andI=[1100], then find k so that A2=kA−2I.

If $A = [\begin{matrix} 3 & - 2 4 & - 2 \end{matrix}] a n d I = [\begin{matrix} 1 & 1 0 & 0 \end{matrix}]$ , then find k so that $A^{2} = k A - 2 I .$