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Exercise 4.6 · Q128

Q.If A=[48−2−4]A=\begin{bmatrix}4&8\\-2&-4\end{bmatrix}, prove that A2=0A^2=0.

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Given A=[48−2−4]A=\begin{bmatrix}4&8\\-2&-4\end{bmatrix}.

A2=A⋅A=[48−2−4][48−2−4]A^2=A\cdot A=\begin{bmatrix}4&8\\-2&-4\end{bmatrix}\begin{bmatrix}4&8\\-2&-4\end{bmatrix}

(A2)11=4(4)+8(−2)=16−16=0(A^2)_{11}=4(4)+8(-2)=16-16=0

(A2)12=4(8)+8(−4)=32−32=0(A^2)_{12}=4(8)+8(-4)=32-32=0

(A2)21=−2(4)+(−4)(−2)=−8+8=0(A^2)_{21}=-2(4)+(-4)(-2)=-8+8=0

(A2)22=−2(8)+(−4)(−4)=−16+16=0(A^2)_{22}=-2(8)+(-4)(-4)=-16+16=0 …

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