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Question

Compute the products AA BB and BB AA, if possible, for the following:

(a) AA

=(0231),B=\left(\begin{array}{rr}0 & -2 \\ 3 & 1\end{array}\right), B

=(1415)=\left(\begin{array}{rr}-1 & 4 \\ 1 & 5\end{array}\right)

(b) AA

=(832104),B=\left(\begin{array}{rrr}8 & 3 & -2 \\ 1 & 0 & 4\end{array}\right), B

=(224315)=\left(\begin{array}{rr}2 & -2 \\ 4 & 3 \\ 1 & -5\end{array}\right)

(c) AA

=(024),B=\left(\begin{array}{r}0 \\ -2 \\ 4\end{array}\right), B

=(0,2,3)=\left(\begin{array}{lll}0, & -2, & 3\end{array}\right)

(d) AA

=(1024),B=\left(\begin{array}{rr}-1 & 0 \\ 2 & 4\end{array}\right), \quad B

=(311102)=\left(\begin{array}{rr}3 & 1 \\ -1 & 1 \\ 0 & 2\end{array}\right)

Solution

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Answered 1 year ago
Answered 1 year ago
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(a) AB\boldsymbol{A}\boldsymbol{B} is a 2×22\times2 matrix obtained by multiplying each row of A\boldsymbol{A} by the corresponding column of B\boldsymbol{B}.

AB=(0231)(1415)=(0(1)2104253(1)+1134+15)=(210217)\begin{aligned} \boldsymbol{A}\boldsymbol{B}&= \begin{pmatrix} 0&-2\\3&1 \end{pmatrix} \begin{pmatrix} -1&4\\1&5 \end{pmatrix}\\ &=\begin{pmatrix} 0\cdot(-1)-2\cdot1&0\cdot4-2\cdot5\\3\cdot(-1)+1\cdot1&3\cdot4+1\cdot5 \end{pmatrix}\\ &=\begin{pmatrix} -2 & -10\\-2&17 \end{pmatrix}\\ \end{aligned}

BA\boldsymbol{B}\boldsymbol{A} is a 2×22\times2 matrix obtained by multiplying each row of B\boldsymbol{B} by the corresponding column of A\boldsymbol{A}.

BA=(1415)(0231)=(10+431(2)+4110+531(2)+51)=(126153)\begin{aligned} \boldsymbol{B}\boldsymbol{A}&= \begin{pmatrix} -1&4\\1&5 \end{pmatrix} \begin{pmatrix} 0&-2\\3&1 \end{pmatrix}\\ &=\begin{pmatrix} -1\cdot 0+4\cdot3 & -1\cdot(-2)+4\cdot1\\1\cdot0+5\cdot3&1\cdot(-2)+5\cdot1 \end{pmatrix}\\ &=\begin{pmatrix} 12&6\\15&3 \end{pmatrix}\\ \end{aligned}

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