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Mathematics

Given A = [−360−9]\begin{bmatrix}[r] -3 & 6 \ 0 & -9 \end{bmatrix} and At is its transpose matrix. Find :

(i) 2A + 3At

(ii) 2At - 3A

(iii) 12A−13At\dfrac{1}{2}A - \dfrac{1}{3}A^t

(iv) At−13AA^t - \dfrac{1}{3}A

Matrices

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Answer

A=[−360−9] and At=[−306−9]A = \begin{bmatrix}[r] -3 & 6 \ 0 & -9 \end{bmatrix} \text{ and } A^t = \begin{bmatrix}[r] -3 & 0 \ 6 & -9 \end{bmatrix}.

(i) Substituting value of A and At in 2A + 3At we get,

⇒2A+3At=2[−360−9]+3[−306−9]=[−6120−18]+[−9018−27]=[−6+(−9)12+00+18−18+(−27)]=[−151218−45].\Rightarrow 2A + 3A^t = 2\begin{bmatrix}[r] -3 & 6 \ 0 & -9 \end{bmatrix} + 3\begin{bmatrix}[r] -3 & 0 \ 6 & -9 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -6 & 12 \ 0 & -18 \end{bmatrix} + \begin{bmatrix}[r] -9 & 0 \ 18 & -27 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -6 + (-9) & 12 + 0 \ 0 + 18 & -18 + (-27) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -15 & 12 \ 18 & -45 \end{bmatrix}.

Hence, 2A + 3At = [−151218−45].\begin{bmatrix}[r] -15 & 12 \ 18 & -45 \end{bmatrix}.

(ii) Substituting value of A and At in 2At - 3A we get,

⇒2At−3A=2[−306−9]−3[−360−9]=[−6012−18]−[−9180−27]=[−6−(−9)0−1812−0−18−(−27)]=[3−18129].\Rightarrow 2A^t - 3A = 2\begin{bmatrix}[r] -3 & 0 \ 6 & -9 \end{bmatrix} - 3\begin{bmatrix}[r] -3 & 6 \ 0 & -9 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -6 & 0 \ 12 & -18 \end{bmatrix} - \begin{bmatrix}[r] -9 & 18 \ 0 & -27 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -6 - (-9) & 0 - 18 \ 12 - 0 & -18 - (-27) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 3 & -18 \ 12 & 9 \end{bmatrix}.

Hence, 2At−3A=[3−18129].2A^t - 3A = \begin{bmatrix}[r] 3 & -18 \ 12 & 9 \end{bmatrix}.

(iii) Substituting value of A and At in 12A−13At\dfrac{1}{2}A - \dfrac{1}{3}A^t we get,

⇒12A−13At=12[−360−9]−13[−306−9]=[−3230−92]−[−102−3]=[−32−(−1)3−00−2−92−(−3)]=[−32+13−2−92+3]=[−123−2−32]\Rightarrow \dfrac{1}{2}A - \dfrac{1}{3}A^t = \dfrac{1}{2}\begin{bmatrix}[r] -3 & 6 \ 0 & -9 \end{bmatrix} - \dfrac{1}{3}\begin{bmatrix}[r] -3 & 0 \ 6 & -9 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -\dfrac{3}{2} & 3 \ 0 & -\dfrac{9}{2} \end{bmatrix} - \begin{bmatrix}[r] -1 & 0 \ 2 & -3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -\dfrac{3}{2} - (-1) & 3 - 0 \ 0 - 2 & -\dfrac{9}{2} - (-3) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -\dfrac{3}{2} + 1 & 3 \ -2 & -\dfrac{9}{2} + 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -\dfrac{1}{2} & 3 \ -2 & -\dfrac{3}{2} \end{bmatrix}

Hence, 12A−13At=[−123−2−32].\dfrac{1}{2}A - \dfrac{1}{3}A^t = \begin{bmatrix}[r] -\dfrac{1}{2} & 3 \ -2 & -\dfrac{3}{2} \end{bmatrix}.

(iv) Substituting value of A and At in At−13AA^t - \dfrac{1}{3}A we get,

⇒At−13A=[−306−9]−13[−360−9]=[−306−9]−[−120−3]=[−3−(−1)0−26−0−9−(−3)]=[−2−26−6].\Rightarrow A^t - \dfrac{1}{3}A = \begin{bmatrix}[r] -3 & 0 \ 6 & -9 \end{bmatrix} - \dfrac{1}{3}\begin{bmatrix}[r] -3 & 6 \ 0 & -9 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -3 & 0 \ 6 & -9 \end{bmatrix} - \begin{bmatrix}[r] -1 & 2 \ 0 & -3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -3 - (-1) & 0 - 2 \ 6 - 0 & -9 - (-3) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -2 & -2 \ 6 & -6 \end{bmatrix}.

Hence, At−13A=[−2−26−6].A^t - \dfrac{1}{3}A = \begin{bmatrix}[r] -2 & -2 \ 6 & -6 \end{bmatrix}.

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