Question

answer only question 4 from the below picture matrices chapter

Answer 1

Solution

$\bf Given$

$\sf\implies \left[\begin{array}{cc}4&-2\\4&0\end{array}\right] +3A=\left[\begin{array}{cc}-2&-2\\1&-3\end{array}\right]$

$\bf We\: have\: to\: find\: value \:of\: A$

$\sf\implies 3A=\left[\begin{array}{cc}-2&-2\\1&-3\end{array}\right] -\left[\begin{array}{cc}4&-2\\4&0\end{array}\right]$

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

$\sf\implies 3A=\left[\begin{array}{cc}-6&0\\-3&-3\end{array}\right]$

$\sf\implies A=\dfrac{1}{3} \left[\begin{array}{cc}-6&0\\-3&-3\end{array}\right]$

$\sf\implies A=\left[\begin{array}{cc}\dfrac{-6}{3} &\dfrac{0}{3} \\\\\dfrac{-3}{3} &\dfrac{-3}{3} \end{array}\right]$

$\sf\implies A=\left[\begin{array}{cc}-2&0\\-1&-1\end{array}\right]$

$\bf Answer$

$\sf\implies A=\left[\begin{array}{cc}-2&0\\-1&-1\end{array}\right]$