Question
If $A=\left[\begin{array}{ll}9 & 1 \\ 7 & 8\end{array}\right], B=\left[\begin{array}{cc}1 & 5 \\ 7 & 12\end{array}\right]$ find matrix $C$ such that $5A + 5B + 2C$ is a null matrix.
✓
Answer
Let $C=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
We have $A =\left[\begin{array}{ll}9 & 1 \\ 7 & 8\end{array}\right], B =\left[\begin{array}{cc}1 & 5 \\ 7 & 12\end{array}\right]$
Now $5 A +3 B +2 C =0$
$ \Rightarrow 5\left[\begin{array}{ll} 9 & 1 \\ 7 & 8 \end{array}\right]+3\left[\begin{array}{cc}
1 & 5 \\ 7 & 12 \end{array}\right]+2\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\ 0 & 0\end{array}\right] $
$\Rightarrow\left[\begin{array}{cc} 45 & 5 \\ 35 & 40 \end{array}\right]+\left[\begin{array}{cc} 3 & 15 \\ 21 & 36
\end{array}\right]+\left[\begin{array}{ll} 2 a & 2 b \\ 2 c & 2 d \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\
0 & 0 \end{array}\right]$
$\Rightarrow\left[\begin{array}{cc} 45+3+2 a & 5+15+2 b \\ 35+21+2 c & 40+36+2 d \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] $
$\Rightarrow\left[\begin{array}{ll} 48+2 a & 20+2 b \\ 56+2 c & 76+2 d \end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\ 0 & 0 \end{array}\right] $
$\Rightarrow 48+2 a =0$
$ \Rightarrow 2 a =-48$
$ \Rightarrow a =-24= 20+2 b =0 $
$\Rightarrow 2 b =20 $
$\Rightarrow b =-10 = 56+2 c =0$
$ \Rightarrow 2 c =56 $
$\Rightarrow c =-28 = 76+2 d =0 $
$\Rightarrow 2 d =-76 $
$\Rightarrow d =-38$
Thus $C =\left[\begin{array}{ll} -24 & -10 \\ -28 & -38\end{array}\right] .$
We have $A =\left[\begin{array}{ll}9 & 1 \\ 7 & 8\end{array}\right], B =\left[\begin{array}{cc}1 & 5 \\ 7 & 12\end{array}\right]$
Now $5 A +3 B +2 C =0$
$ \Rightarrow 5\left[\begin{array}{ll} 9 & 1 \\ 7 & 8 \end{array}\right]+3\left[\begin{array}{cc}
1 & 5 \\ 7 & 12 \end{array}\right]+2\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\ 0 & 0\end{array}\right] $
$\Rightarrow\left[\begin{array}{cc} 45 & 5 \\ 35 & 40 \end{array}\right]+\left[\begin{array}{cc} 3 & 15 \\ 21 & 36
\end{array}\right]+\left[\begin{array}{ll} 2 a & 2 b \\ 2 c & 2 d \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\
0 & 0 \end{array}\right]$
$\Rightarrow\left[\begin{array}{cc} 45+3+2 a & 5+15+2 b \\ 35+21+2 c & 40+36+2 d \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] $
$\Rightarrow\left[\begin{array}{ll} 48+2 a & 20+2 b \\ 56+2 c & 76+2 d \end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\ 0 & 0 \end{array}\right] $
$\Rightarrow 48+2 a =0$
$ \Rightarrow 2 a =-48$
$ \Rightarrow a =-24= 20+2 b =0 $
$\Rightarrow 2 b =20 $
$\Rightarrow b =-10 = 56+2 c =0$
$ \Rightarrow 2 c =56 $
$\Rightarrow c =-28 = 76+2 d =0 $
$\Rightarrow 2 d =-76 $
$\Rightarrow d =-38$
Thus $C =\left[\begin{array}{ll} -24 & -10 \\ -28 & -38\end{array}\right] .$
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