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Exercise 7.1 · Q7

Q.If A=(42−1x)A=\begin{pmatrix} 4 & 2 \\ -1 & x\end{pmatrix} and such that (A−2I)(A−3I)=O(A-2I)(A-3I)=O, find the value of xx.

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We compute the matrix product (A−2I)(A−3I)(A-2I)(A-3I) symbolically in terms of xx and force it to equal the zero matrix, which pins down xx.

Step 1. Form the two shifted matrices.

A−2I=(4−22−1x−2)=(22−1x−2)A-2I=\begin{pmatrix}4-2&2\\-1&x-2\end{pmatrix}=\begin{pmatrix}2&2\\-1&x-2\end{pmatrix}

A−3I=(4−32−1x−3)=(12−1x−3)A-3I=\begin{pmatrix}4-3&2\\-1&x-3\end{pmatrix}=\begin{pmatrix}1&2\\-1&x-3\end{pmatrix}

Step 2. Multiply the two matrices.

(A−2I)(A−3I)=(22−1x−2)(12−1x−3)(A-2I)(A-3I)=\begin{pmatrix}2&2\\-1&x-2\end{pmatrix}\begin{pmatrix}1&2\\-1&x-3\end{pmatrix}

(1,1): 2(1)+2(−1)=0(1,1):\ 2(1)+2(-1)=0

(1,2): 2(2)+2(x−3)=4+2x−6=2x−2(1,2):\ 2(2)+2(x-3)=4+2x-6=2x-2

(2,1): −1(1)+(x−2)(−1)=−1−x+2=1−x(2,1):\ -1(1)+(x-2)(-1)=-1-x+2=1-x

(2,2): −1(2)+(x−2)(x−3)=−2+x2−5x+6=x2−5x+4(2,2):\ -1(2)+(x-2)(x-3)=-2+x^2-5x+6=x^2-5x+4

So (A−2I)(A−3I)=(02x−21−xx2−5x+4)(A-2I)(A-3I)=\begin{pmatrix}0 & 2x-2\\ 1-x & x^2-5x+4\end{pmatrix}. …

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