If A=-23-11, then A3 is equal to

Given,A=-23-11A2=A.A=-23-11.-23-11=(-2)(-2)+3(-1)(-2)(3)+3(1)(-1)(-2)+1(-1)(-1)(3)+(1)(1) ⇒ A2=1-31-2 A3=A2.A=1-31-2.-23-11=(1)(-2)+(-3)(-1)(1)(3)+(-3)(1)(1)(-2)+(-2)(-1)(1)(3)+(-2)(1) ⇒ A3=3-23-3-2+23-2=1001=I ⇒ A3=IAnswer: (d) I

If A=-23-11, then A3 is equal to

Given,A=-23-11A2=A.A=-23-11.-23-11=(-2)(-2)+3(-1)(-2)(3)+3(1)(-1)(-2)+1(-1)(-1)(3)+(1)(1) ⇒ A2=1-31-2 A3=A2.A=1-31-2.-23-11=(1)(-2)+(-3)(-1)(1)(3)+(-3)(1)(1)(-2)+(-2)(-1)(1)(3)+(-2)(1) ⇒ A3=3-23-3-2+23-2=1001=I ⇒ A3=IAnswer: (d) I

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If $A = [\begin{matrix} - 2 & 3 \\ - 1 & 1 \end{matrix}]$ , then $A^{3}$ is equal to

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$2 A$

$A$

$2 I$

$I$

answer is D.

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Detailed Solution

Given,

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

$A^{2} = A . A = [\begin{matrix} - 2 & 3 \\ - 1 & 1 \end{matrix}] . [\begin{matrix} - 2 & 3 \\ - 1 & 1 \end{matrix}] = [\begin{matrix} (- 2) (- 2) + 3 (- 1) & (- 2) (3) + 3 (1) \\ (- 1) (- 2) + 1 (- 1) & (- 1) (3) + (1) (1) \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & - 3 \\ 1 & - 2 \end{matrix}] A^{3} = A^{2} . A = [\begin{matrix} 1 & - 3 \\ 1 & - 2 \end{matrix}] . [\begin{matrix} - 2 & 3 \\ - 1 & 1 \end{matrix}] = [\begin{matrix} (1) (- 2) + (- 3) (- 1) & (1) (3) + (- 3) (1) \\ (1) (- 2) + (- 2) (- 1) & (1) (3) + (- 2) (1) \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 3 - 2 & 3 - 3 \\ - 2 + 2 & 3 - 2 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I \Rightarrow A^{3} = I$