Question
If $A=\left[\begin{array}{ll}9 & 1 \\ 5 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 5 \\ 7 & -11\end{array}\right]$, find matrix $X$ such that $3A + 5B - 2X = 0.$
✓
Answer
Let $X =\left[\begin{array}{ll}x & y \\ z & u\end{array}\right]$
We have $A=\left[\begin{array}{ll}9 & 1 \\ 5 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 5 \\ 7 & -11\end{array}\right]$
$ 3 A =3\left[\begin{array}{ll} 9 & 1 \\ 5 & 3 \end{array}\right]=\left[\begin{array}{ll} 27 & 3 \\
15 & 9 \end{array}\right] $
$5 B =5\left[\begin{array}{cc} 1 & 5 \\ 7 & -11 \end{array}\right]=\left[\begin{array}{cc} 5 & 25 \\ 35 & -55
\end{array}\right] $
Now $3 A+5 B-2 X=0$
$\Rightarrow\left[\begin{array}{ll}27 & 3 \\ 15 & 9\end{array}\right]+\left[\begin{array}{cc}5 & 25 \\ 35 & -55\end{array}\right]+\left[\begin{array}{cc}-2 x & -2 y \\ -2 z & -2 u\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}27+5-2 x & 3+25-2 y \\ 15+35-2 z & 9-55-2 u\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}32-2 x & 28-2 y \\ 50-2 z & -46-2 u\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow 32-2 x=0 $
$\Rightarrow 2 x-32 $
$\Rightarrow x=16= 28-2 y=0 $
$\Rightarrow 2 y=28 $
$\Rightarrow y=14 = 50-2 z=0 $
$\Rightarrow 2 z=50$
$ \Rightarrow z=25-46-2 u=0 $
$\Rightarrow 2 u=-46$
$ \Rightarrow u=-23$
We have $A=\left[\begin{array}{ll}9 & 1 \\ 5 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 5 \\ 7 & -11\end{array}\right]$
$ 3 A =3\left[\begin{array}{ll} 9 & 1 \\ 5 & 3 \end{array}\right]=\left[\begin{array}{ll} 27 & 3 \\
15 & 9 \end{array}\right] $
$5 B =5\left[\begin{array}{cc} 1 & 5 \\ 7 & -11 \end{array}\right]=\left[\begin{array}{cc} 5 & 25 \\ 35 & -55
\end{array}\right] $
Now $3 A+5 B-2 X=0$
$\Rightarrow\left[\begin{array}{ll}27 & 3 \\ 15 & 9\end{array}\right]+\left[\begin{array}{cc}5 & 25 \\ 35 & -55\end{array}\right]+\left[\begin{array}{cc}-2 x & -2 y \\ -2 z & -2 u\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}27+5-2 x & 3+25-2 y \\ 15+35-2 z & 9-55-2 u\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}32-2 x & 28-2 y \\ 50-2 z & -46-2 u\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow 32-2 x=0 $
$\Rightarrow 2 x-32 $
$\Rightarrow x=16= 28-2 y=0 $
$\Rightarrow 2 y=28 $
$\Rightarrow y=14 = 50-2 z=0 $
$\Rightarrow 2 z=50$
$ \Rightarrow z=25-46-2 u=0 $
$\Rightarrow 2 u=-46$
$ \Rightarrow u=-23$
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