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Exercise 7.5 · Q24

Q.If A+I=[3−241]A + I = \begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}, then (A+I)(A−I)(A + I)(A - I) is equal to

(1) [−5−48−9]\begin{bmatrix} -5 & -4 \\ 8 & -9 \end{bmatrix}
(2) [−54−89]\begin{bmatrix} -5 & 4 \\ -8 & 9 \end{bmatrix}
(3) [5489]\begin{bmatrix} 5 & 4 \\ 8 & 9 \end{bmatrix}
(4) [−5−4−8−9]\begin{bmatrix} -5 & -4 \\ -8 & -9 \end{bmatrix}
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Find A−IA-I from A+IA+I, then multiply.

This tests matrix products (note AA and II commute, so (A+I)(A−I)=A2−I(A+I)(A-I)=A^2-I).

Step 1. Given A+I=[3−241]A + I = \begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}, so A−I=(A+I)−2I=[3−2−241−2]=[1−24−1]A - I = (A+I) - 2I = \begin{bmatrix} 3-2 & -2 \\ 4 & 1-2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -1 \end{bmatrix}.

Step 2. Multiply [3−241][1−24−1]\begin{bmatrix} 3 & -2 \\ 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 4 & -1 \end{bmatrix}: top-left =3−8=−5= 3 - 8 = -5; top-right =−6+2=−4= -6 + 2 = -4. …

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