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Exercise 7.1 · Q13

Q.Verify the property A(B+C)=AB+ACA(B+C)=AB+AC, when the matrices A,B,A, B, and CC are given by
[!FORMULA] A=(20−3145),B=(31−1042),C=(47211−1)A=\begin{pmatrix} 2 & 0 & -3 \\ 1 & 4 & 5\end{pmatrix},\quad B=\begin{pmatrix} 3 & 1 \\ -1 & 0 \\ 4 & 2\end{pmatrix},\quad C=\begin{pmatrix} 4 & 7 \\ 2 & 1 \\ 1 & -1\end{pmatrix}

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We compute A(B+C)A(B+C) directly, then compute ABAB and ACAC separately and add them, and show the two routes give the same 2×22\times2 matrix.

Step 1. Compute B+CB+C.

B+C=(3+41+7−1+20+14+12−1)=(781151)B+C=\begin{pmatrix}3+4&1+7\\-1+2&0+1\\4+1&2-1\end{pmatrix}=\begin{pmatrix}7&8\\1&1\\5&1\end{pmatrix}

Step 2. Compute A(B+C)A(B+C).

(1,1): 2(7)+0(1)+(−3)(5)=14+0−15=−1(1,1):\ 2(7)+0(1)+(-3)(5)=14+0-15=-1

(1,2): 2(8)+0(1)+(−3)(1)=16+0−3=13(1,2):\ 2(8)+0(1)+(-3)(1)=16+0-3=13

(2,1): 1(7)+4(1)+5(5)=7+4+25=36(2,1):\ 1(7)+4(1)+5(5)=7+4+25=36

(2,2): 1(8)+4(1)+5(1)=8+4+5=17(2,2):\ 1(8)+4(1)+5(1)=8+4+5=17

So A(B+C)=(−1133617)A(B+C)=\begin{pmatrix}-1&13\\36&17\end{pmatrix}.

Step 3. Compute ABAB separately.

(1,1): 2(3)+0(−1)+(−3)(4)=6+0−12=−6(1,1):\ 2(3)+0(-1)+(-3)(4)=6+0-12=-6

(1,2): 2(1)+0(0)+(−3)(2)=2+0−6=−4(1,2):\ 2(1)+0(0)+(-3)(2)=2+0-6=-4

(2,1): 1(3)+4(−1)+5(4)=3−4+20=19(2,1):\ 1(3)+4(-1)+5(4)=3-4+20=19

(2,2): 1(1)+4(0)+5(2)=1+0+10=11(2,2):\ 1(1)+4(0)+5(2)=1+0+10=11

AB=(−6−41911)AB=\begin{pmatrix}-6&-4\\19&11\end{pmatrix}

Step 4. Compute ACAC separately.

(1,1): 2(4)+0(2)+(−3)(1)=8+0−3=5(1,1):\ 2(4)+0(2)+(-3)(1)=8+0-3=5

(1,2): 2(7)+0(1)+(−3)(−1)=14+0+3=17(1,2):\ 2(7)+0(1)+(-3)(-1)=14+0+3=17 …

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