Question

If $A=\left[\begin{array}{cc}2& -5\\ 0& 1\end{array}\right]$ and $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ then find the matrix $X$ such that $4A-2X+I=0$

• A. $\left[\begin{array}{cc}0& \frac{1}{2}\\ -5& \frac{9}{2}\end{array}\right]$
• B. $\left[\begin{array}{cc}\frac{9}{2}& -1\\ -8& \frac{3}{2}\end{array}\right]$
• C. $\left[\begin{array}{cc}\frac{9}{2}& -10\\ 0& \frac{5}{2}\end{array}\right]$
• D. $\left[\begin{array}{cc}\frac{5}{2}& -6\\ 1& \frac{3}{2}\end{array}\right]$

Answer 1

The correct option is C $\left[\begin{array}{cc}\frac{9}{2}& -10\\ 0& \frac{5}{2}\end{array}\right]$

Let $X=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Given, $4A-2X+I=0$

$\Rightarrow 4\left[\begin{array}{cc}2& -5\\ 0& 1\end{array}\right]-2\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}8& -20\\ 0& 4\end{array}\right]-\left[\begin{array}{cc}2a& 2b\\ 2c& 2d\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}8-2a+1& -20-2b+0\\ 0-2c+0& 4-2d+1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}9-2a& -20-2b\\ -2c& 5-2d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow 9-2a=0\Rightarrow a=\frac{9}{2}$

And $-20-2b=0\Rightarrow b=-10$

And $-2c=0\Rightarrow c=0$

And $5-2d=0\Rightarrow d=\frac{5}{2}$

So, $X=\left[\begin{array}{cc}\frac{9}{2}& -10\\ 0& \frac{5}{2}\end{array}\right]$