If A = \begin{bmatrix} -3 &6 \\-2 &4 \end{bmatrix}  then show that A^3=A

 

 

 

 
 
 
 
 

Answers (1)

A = \begin{bmatrix} -3 &6 \\-2 &4 \end{bmatrix}

A ^2= \begin{bmatrix} -3 &6 \\-2 &4 \end{bmatrix}\begin{bmatrix} -3 &6 \\-2 &4 \end{bmatrix}

\Rightarrow \begin{bmatrix} 9-12 &24-18 \\6-8 &-12+16 \end {bmatrix}\Rightarrow \begin{bmatrix} -3 &6 \\-2 &4 \end{bmatrix}

A ^3= \begin{bmatrix} -3 &6 \\-2 &4 \end{bmatrix}\begin{bmatrix} -3 &6 \\-2 &4 \end{bmatrix}= \begin{bmatrix} -3 &6 \\-2 &4 \end{bmatrix}

LHS=RHS=A

Hence Proved.

 

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