Question

Let $A+B+C=\left[\begin{array}{cc}4& -1\\ 0& 1\end{array}\right]$, $4A+2B+C=\left[\begin{array}{cc}0& -1\\ -3& 2\end{array}\right]$ and $9A+3B+C=\left[\begin{array}{cc}0& -2\\ 2& 1\end{array}\right]$ then find A.

Answer 1

We are given

$A+B+C=\left[\begin{array}{cc}4& -1\\ 0& 1\end{array}\right]\to \left(1\right),4A+2B+C=\left[\begin{array}{cc}0& -1\\ -3& 2\end{array}\right]\to \left(2\right)$

& $9A+3B+C=\left[\begin{array}{cc}0& -2\\ 2& 1\end{array}\right]$. $\to \left(3\right)$

let subtracts equation $\left(1\right)$ by equation $\left(3\right)$

we get,

$(9A+3B+C)-(A+B+C)=\left[\begin{array}{cc}0& -2\\ 2& 1\end{array}\right]-\left[\begin{array}{cc}4& -1\\ 0& 1\end{array}\right]$

$8A+2B=\left[\begin{array}{cc}0-4& -2+1\\ 2-0& 1-1\end{array}\right]$

$8A+2B=\left[\begin{array}{cc}-4& -1\\ 2& 0\end{array}\right]⟶\left(i\right)$

Similarly subtracts equation $\left(2\right)$ by equation $\left(3\right)$

We get,

$(9A+3B+C)-(4A+2B+C)=\left[\begin{array}{cc}0& -2\\ 2& 1\end{array}\right]-\left[\begin{array}{cc}0& -1\\ -3& 2\end{array}\right]$

$5A+B=\left[\begin{array}{cc}0& -2+1\\ 2+3& 1-2\end{array}\right]$

$5A+B=\left[\begin{array}{cc}0& -1\\ 5& -1\end{array}\right]⟶\left(ii\right)$

from $\left(i\right)$ & $\left(ii\right)$ we get,

$8A=\left[\begin{array}{cc}-4& -1\\ 2& 0\end{array}\right]-2B$

Where $B=\left[\begin{array}{cc}0& -1\\ 5& -1\end{array}\right]-5A$

So, $8A=\left[\begin{array}{cc}-4& -1\\ 2& 0\end{array}\right]-2\{\left[\begin{array}{cc}0& -1\\ 5& -1\end{array}\right]-5A\}$

$8A=\left[\begin{array}{cc}-4& -1\\ 2& 0\end{array}\right]+\left[\begin{array}{cc}0& 2\\ -10& 2\end{array}\right]+10A$.

So, $-2A=\left[\begin{array}{cc}-4+0& -1+2\\ 2-10& 0+2\end{array}\right]$

$-2A=\left[\begin{array}{cc}-4& 1\\ -8& 2\end{array}\right]$

$A=\left[\begin{array}{cc}2& -1/2\\ 4& -1\end{array}\right]$

$A=\left[\begin{array}{cc}2& -\frac{1}{2}\\ 4& -1\end{array}\right]$