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Miscellaneous Exercise 4(B) · Q196

Q.If A=[2−2−4−1341−2−3]A=\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix} show that A2=AA^2=A.

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A=[2−2−4−1341−2−3]A=\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}.

Row1⋅\cdotA: [2(2)+(−2)(−1)+(−4)(1), 2(−2)+(−2)(3)+(−4)(−2), 2(−4)+(−2)(4)+(−4)(−3)]=[4+2−4, −4−6+8, −8−8+12]=[2,−2,−4][2(2)+(-2)(-1)+(-4)(1),\ 2(-2)+(-2)(3)+(-4)(-2),\ 2(-4)+(-2)(4)+(-4)(-3)]=[4+2-4,\ -4-6+8,\ -8-8+12]=[2,-2,-4].

Row2⋅\cdotA: [−1(2)+3(−1)+4(1), −1(−2)+3(3)+4(−2), −1(−4)+3(4)+4(−3)]=[−2−3+4, 2+9−8, 4+12−12]=[−1,3,4][-1(2)+3(-1)+4(1),\ -1(-2)+3(3)+4(-2),\ -1(-4)+3(4)+4(-3)]=[-2-3+4,\ 2+9-8,\ 4+12-12]=[-1,3,4]. …

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