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Exercise 7.1 · Q9

Q.If A=(102021203)A=\begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{pmatrix} and A3−6A2+7A+kI=OA^3-6A^2+7A+kI=O, find the value of kk.

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We compute A2A^2, then A3=A2⋅AA^3=A^2\cdot A, substitute both into A3−6A2+7AA^3-6A^2+7A, and see that the whole expression collapses to a scalar multiple of II — which fixes kk.

Step 1. Compute A2=A⋅AA^2=A\cdot A.

Row 1⋅1\cdotCol 1=1(1)+0(0)+2(2)=51=1(1)+0(0)+2(2)=5;\ Row 1⋅1\cdotCol 2=1(0)+0(2)+2(0)=02=1(0)+0(2)+2(0)=0;\ Row 1⋅1\cdotCol 3=1(2)+0(1)+2(3)=83=1(2)+0(1)+2(3)=8

Row 2⋅2\cdotCol 1=0(1)+2(0)+1(2)=21=0(1)+2(0)+1(2)=2;\ Row 2⋅2\cdotCol 2=0(0)+2(2)+1(0)=42=0(0)+2(2)+1(0)=4;\ Row 2⋅2\cdotCol 3=0(2)+2(1)+1(3)=53=0(2)+2(1)+1(3)=5

Row 3⋅3\cdotCol 1=2(1)+0(0)+3(2)=81=2(1)+0(0)+3(2)=8;\ Row 3⋅3\cdotCol 2=2(0)+0(2)+3(0)=02=2(0)+0(2)+3(0)=0;\ Row 3⋅3\cdotCol 3=2(2)+0(1)+3(3)=133=2(2)+0(1)+3(3)=13

A2=(5082458013)A^2=\begin{pmatrix}5&0&8\\2&4&5\\8&0&13\end{pmatrix}

Step 2. Compute A3=A2⋅AA^3=A^2\cdot A.

Row 11: 5(1)+0(0)+8(2)=21; 5(0)+0(2)+8(0)=0; 5(2)+0(1)+8(3)=345(1)+0(0)+8(2)=21;\ 5(0)+0(2)+8(0)=0;\ 5(2)+0(1)+8(3)=34

Row 22: 2(1)+4(0)+5(2)=12; 2(0)+4(2)+5(0)=8; 2(2)+4(1)+5(3)=232(1)+4(0)+5(2)=12;\ 2(0)+4(2)+5(0)=8;\ 2(2)+4(1)+5(3)=23

Row 33: 8(1)+0(0)+13(2)=34; 8(0)+0(2)+13(0)=0; 8(2)+0(1)+13(3)=558(1)+0(0)+13(2)=34;\ 8(0)+0(2)+13(0)=0;\ 8(2)+0(1)+13(3)=55

A3=(210341282334055)A^3=\begin{pmatrix}21&0&34\\12&8&23\\34&0&55\end{pmatrix}

Step 3. Form A3−6A2+7AA^3-6A^2+7A. …

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