(i) Find a matrix X such that 2A + B + X = O, where
Question:

(i) Find a matrix $X$ such that $2 A+B+X=0$, where

$A=\left[\begin{array}{rr}-1 & 2 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{rr}3 & -2 \\ 1 & 5\end{array}\right]$

(ii) If $A=\left[\begin{array}{rr}8 & 0 \\ 4 & -2 \\ 3 & 6\end{array}\right]$ and $B=\left[\begin{array}{rr}2 & -2 \\ 4 & 2 \\ -5 & 1\end{array}\right]$, then find the matrix $X$ of order $3 \times 2$ such that $2 A+3 X=5 B$.

Solution:

(i)

$2 A+B+X=0$

$\Rightarrow 2\left[\begin{array}{cc}-1 & 2 \\ 3 & 4\end{array}\right]+\left[\begin{array}{cc}3 & -2 \\ 1 & 5\end{array}\right]+X=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}-2 & 4 \\ 6 & 8\end{array}\right]+\left[\begin{array}{cc}3 & -2 \\ 1 & 5\end{array}\right]+X=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}-2+3 & 4-2 \\ 6+1 & 8+5\end{array}\right]+X=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}1 & 2 \\ 7 & 13\end{array}\right]+X=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

$\therefore X=\left[\begin{array}{cc}-1 & -2 \\ -7 & -13\end{array}\right]$

(ii)

$2 A+3 X=5 B$

$\Rightarrow 2\left[\begin{array}{cc}8 & 0 \\ 4 & -2 \\ 3 & 6\end{array}\right]+3 X=5\left[\begin{array}{cc}2 & -2 \\ 4 & 2 \\ -5 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}16 & 0 \\ 8 & -4 \\ 6 & 12\end{array}\right]+3 X=\left[\begin{array}{cc}10 & -10 \\ 20 & 10 \\ -25 & 5\end{array}\right]$

$\Rightarrow 3 X=\left[\begin{array}{cc}10 & -10 \\ 20 & 10 \\ -25 & 5\end{array}\right]-\left[\begin{array}{cc}16 & 0 \\ 8 & -4 \\ 6 & 12\end{array}\right]$

$\Rightarrow 3 X=\left[\begin{array}{cc}10-16 & -10-0 \\ 20-8 & 10+4 \\ -25-6 & 5-12\end{array}\right]$

$\Rightarrow 3 X=\left[\begin{array}{cc}-6 & -10 \\ 12 & 14 \\ -31 & -7\end{array}\right]$

$\Rightarrow X=\frac{1}{3}\left[\begin{array}{cc}-6 & -10 \\ 12 & 14 \\ -31 & -7\end{array}\right]$

$\therefore\left[\begin{array}{cc}-2 & \frac{-10}{3} \\ 4 & \frac{14}{3} \\ \frac{-31}{3} & \frac{-7}{3}\end{array}\right]$

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