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Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.2

Solution

We verify the identity:

(AB)T=BTAT(AB)^T=B^TA^T

by computing both sides.

(i)

A=[1−120−31],B=[11−3−201]A=\begin{bmatrix}1&-1&2\\0&-3&1\end{bmatrix},\quad B=\begin{bmatrix}1&1\\-3&-2\\0&1\end{bmatrix}

Compute ABAB:

AB=[1⋅1+(−1)(−3)+2⋅01⋅1+(−1)(−2)+2⋅10⋅1+(−3)(−3)+1⋅00⋅1+(−3)(−2)+1⋅1]=[4597]AB= \begin{bmatrix} 1\cdot 1+(-1)(-3)+2\cdot 0&1\cdot 1+(-1)(-2)+2\cdot 1\\ 0\cdot 1+(-3)(-3)+1\cdot 0&0\cdot 1+(-3)(-2)+1\cdot 1 \end{bmatrix} = \begin{bmatrix} 4&5\\ 9&7 \end{bmatrix}

So:

(AB)T=[4957](AB)^T=\begin{bmatrix}4&9\\5&7\end{bmatrix}

Now compute BTATB^TA^T.

BT=[1−301−21],AT=[10−1−321]B^T=\begin{bmatrix}1&-3&0\\1&-2&1\end{bmatrix},\qquad A^T=\begin{bmatrix}1&0\\-1&-3\\2&1\end{bmatrix} BTAT=[1−301−21][10−1−321]=[4957]B^TA^T= \begin{bmatrix}1&-3&0\\1&-2&1\end{bmatrix} \begin{bmatrix}1&0\\-1&-3\\2&1\end{bmatrix} = \begin{bmatrix}4&9\\5&7\end{bmatrix}

Hence:

(AB)T=BTAT\boxed{(AB)^T=B^TA^T}

(ii)

A=[121421],B=[1−3−21]A=\begin{bmatrix}1&2\\1&4\\2&1\end{bmatrix},\quad B=\begin{bmatrix}1&-3\\-2&1\end{bmatrix}

Compute ABAB:

AB=[121421][1−3−21]=[−3−1−710−5]AB= \begin{bmatrix} 1&2\\ 1&4\\ 2&1 \end{bmatrix} \begin{bmatrix} 1&-3\\ -2&1 \end{bmatrix} = \begin{bmatrix} -3&-1\\ -7&1\\ 0&-5 \end{bmatrix}

So:

(AB)T=[−3−70−11−5](AB)^T=\begin{bmatrix}-3&-7&0\\-1&1&-5\end{bmatrix}

Now:

BT=[1−2−31],AT=[112241]B^T=\begin{bmatrix}1&-2\\-3&1\end{bmatrix},\qquad A^T=\begin{bmatrix}1&1&2\\2&4&1\end{bmatrix} BTAT=[1−2−31][112241]=[−3−70−11−5]B^TA^T= \begin{bmatrix}1&-2\\-3&1\end{bmatrix} \begin{bmatrix}1&1&2\\2&4&1\end{bmatrix} = \begin{bmatrix}-3&-7&0\\-1&1&-5\end{bmatrix}

Thus:

(AB)T=BTAT\boxed{(AB)^T=B^TA^T}
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