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Exercise 4.6 · Q133

Q.If A=[1−256]A=\begin{bmatrix}1&-2\\5&6\end{bmatrix}, B=[3−137]B=\begin{bmatrix}3&-1\\3&7\end{bmatrix}, Find AB-2I, where I is unit matrix of order 2.

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Given A=[1−256]A=\begin{bmatrix}1&-2\\5&6\end{bmatrix}, B=[3−137]B=\begin{bmatrix}3&-1\\3&7\end{bmatrix}, I=[1001]I=\begin{bmatrix}1&0\\0&1\end{bmatrix}.

Step 1 — compute AB:

(AB)11=1(3)+(−2)(3)=3−6=−3(AB)_{11}=1(3)+(-2)(3)=3-6=-3

(AB)12=1(−1)+(−2)(7)=−1−14=−15(AB)_{12}=1(-1)+(-2)(7)=-1-14=-15

(AB)21=5(3)+6(3)=15+18=33(AB)_{21}=5(3)+6(3)=15+18=33

(AB)22=5(−1)+6(7)=−5+42=37(AB)_{22}=5(-1)+6(7)=-5+42=37

AB=[−3−153337]AB=\begin{bmatrix}-3&-15\\33&37\end{bmatrix}

Step 2 — form 2I:

2I=[2002]2I=\begin{bmatrix}2&0\\0&2\end{bmatrix} …

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