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Mathematics

If A = [0−112] and B=[12−11]\begin{bmatrix}[r] 0 & -1 \ 1 & 2 \end{bmatrix} \text{ and B} = \begin{bmatrix}[r] 1 & 2 \ -1 & 1 \end{bmatrix}, find the matrix X if

(i) 3A + X = B

(ii) X - 3B = 2A.

Matrices

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Answer

(i) 3A + X = B

⇒ X = B - 3A

⇒X=[12−11]−3[0−112]=[12−11]−[0−336]=[1−02−(−3)−1−31−6]=[15−4−5]\Rightarrow X = \begin{bmatrix}[r] 1 & 2 \ -1 & 1 \end{bmatrix} - 3\begin{bmatrix}[r] 0 & -1 \ 1 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 2 \ -1 & 1 \end{bmatrix} - \begin{bmatrix}[r] 0 & -3 \ 3 & 6 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 - 0 & 2 - (-3) \ -1 - 3 & 1 - 6 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 5 \ -4 & -5 \end{bmatrix}

Hence, matrix X = [15−4−5].\begin{bmatrix}[r] 1 & 5 \ -4 & -5 \end{bmatrix}.

(ii) X - 3B = 2A

⇒ X = 2A + 3B

X=2[0−112]+3[12−11]=[0−224]+[36−33]=[0+3−2+62+(−3)4+3]=[34−17]X = 2\begin{bmatrix}[r] 0 & -1 \ 1 & 2 \end{bmatrix} + 3\begin{bmatrix}[r] 1 & 2 \ -1 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 & -2 \ 2 & 4 \end{bmatrix} + \begin{bmatrix}[r] 3 & 6 \ -3 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 + 3 & -2 + 6 \ 2 + (-3) & 4 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 3 & 4 \ -1 & 7 \end{bmatrix}

Hence, matrix X = [34−17].\begin{bmatrix}[r] 3 & 4 \ -1 & 7 \end{bmatrix}.

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