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Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.1

Solution

Given:

A=[0−12321−104],B=[21−1124−121],C=[10−2−15034−1]A=\begin{bmatrix}0&-1&2\\3&2&1\\-1&0&4\end{bmatrix},\quad B=\begin{bmatrix}2&1&-1\\1&2&4\\-1&2&1\end{bmatrix},\quad C=\begin{bmatrix}1&0&-2\\-1&5&0\\3&4&-1\end{bmatrix}

(i) A−BA-B

Subtract corresponding entries:

A−B=[0−2−1−12−(−1)3−12−21−4−1−(−1)0−24−1]=[−2−2320−30−23]A-B= \begin{bmatrix} 0-2&-1-1&2-(-1)\\ 3-1&2-2&1-4\\ -1-(-1)&0-2&4-1 \end{bmatrix} = \begin{bmatrix} -2&-2&3\\ 2&0&-3\\ 0&-2&3 \end{bmatrix}

(ii) B−CB-C

B−C=[2−11−0−1−(−2)1−(−1)2−54−0−1−32−41−(−1)]=[1112−34−4−22]B-C= \begin{bmatrix} 2-1&1-0&-1-(-2)\\ 1-(-1)&2-5&4-0\\ -1-3&2-4&1-(-1) \end{bmatrix} = \begin{bmatrix} 1&1&1\\ 2&-3&4\\ -4&-2&2 \end{bmatrix}

(iii) (A−B)−C(A-B)-C

Using part (i):

(A−B)−C=[−2−2320−30−23]−[10−2−15034−1]=[−3−253−5−3−3−64](A-B)-C= \begin{bmatrix} -2&-2&3\\ 2&0&-3\\ 0&-2&3 \end{bmatrix} - \begin{bmatrix} 1&0&-2\\ -1&5&0\\ 3&4&-1 \end{bmatrix} = \begin{bmatrix} -3&-2&5\\ 3&-5&-3\\ -3&-6&4 \end{bmatrix}

(iv) A−(B−C)A-(B-C)

Using part (ii):

A−(B−C)=[0−12321−104]−[1112−34−4−22]=[−1−2115−3322]A-(B-C)= \begin{bmatrix} 0&-1&2\\ 3&2&1\\ -1&0&4 \end{bmatrix} - \begin{bmatrix} 1&1&1\\ 2&-3&4\\ -4&-2&2 \end{bmatrix} = \begin{bmatrix} -1&-2&1\\ 1&5&-3\\ 3&2&2 \end{bmatrix}
A−B=[−2−2320−30−23]\boxed{A-B=\begin{bmatrix}-2&-2&3\\2&0&-3\\0&-2&3\end{bmatrix}} B−C=[1112−34−4−22]\boxed{B-C=\begin{bmatrix}1&1&1\\2&-3&4\\-4&-2&2\end{bmatrix}} (A−B)−C=[−3−253−5−3−3−64]\boxed{(A-B)-C=\begin{bmatrix}-3&-2&5\\3&-5&-3\\-3&-6&4\end{bmatrix}} A−(B−C)=[−1−2115−3322]\boxed{A-(B-C)=\begin{bmatrix}-1&-2&1\\1&5&-3\\3&2&2\end{bmatrix}}
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