If A = [ 3 - 2 4 - 2 ] and I = [ 1 0 0 1 ] , find k so that A 2 = k A - 2 I .

Question

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

Smart Achievers · Accepted Answer

(2)

Ans. 1

Explanation :

$A = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}]$

$∴$ $A^{2} = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] = [\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}]$

$∵$ $A^{2} = k A - 2 I$

$[\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] = [\begin{matrix} 3 k & - 2 k \\ 4 k & - 2 k \end{matrix}] - [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}]$

on comparing $3 k - 2 = 1$

$3 k = 3$ $\Rightarrow k = 1$

Solution

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

1 0

2 1

3 2

4 - 1

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Solution

Q. If A=[3-24-2] and I=[1001], find k so that A2=kA-2I.

1 0

2 1

3 2

4 - 1

Ans. (2) Ans. 1 Explanation : A=[3-24-2] ∴ A2=[3-24-2][3-24-2]=[1-24-4] ∵ A2=kA-2I [1-24-4]=[3k-2k4k-2k]-[2002] on comparing 3k-2=1 3k=3⇒k=1

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