Question

If $\left[\begin{array}{ccc}-2&0\\4&5\end{array}\right] \times \left[\begin{array}{ccc}0&4\\-3&2\end{array}\right] -2M= \left[\begin{array}{ccc}6&-2\\3&5\end{array}\right]$ find the matrix.

Answer 1

here is your answer

Answer 2

$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,

$\left[\begin{array}{ccc}-2&0\\4&5\end{array}\right] \times \left[\begin{array}{ccc}0&4\\-3&2\end{array}\right] -2M= \left[\begin{array}{ccc}6&-2\\3&5\end{array}\right]$

$\underline{\large{\textsf{Step-1:}}}$
Consider LHS
$\left[\begin{array}{ccc}-2&0\\4&5\end{array}\right] \times \left[\begin{array}{ccc}0&4\\-3&2\end{array}\right] -2M$

$\underline{\large{\textsf{Step-2:}}}$
Multiply Matrices

$\left[\begin{array}{ccc}(-2)\times0)+(0\times(-3)&(-2\times4)+(0\times2)\\(4\times0)+(5\times(-3)&(4\times4)+(5\times2)\end{array}\right]- 2M$

$\left[\begin{array}{ccc}(0)+(0)&(-8)+(0)\\(0)+(-15)&(16)+(10)\end{array}\right]- 2M$

$\left[\begin{array}{ccc}0&-8\\-15&26\end{array}\right]- 2M$

$\underline{\large{\textsf{Step-3:}}}$
Equate it to RHS

$\left[\begin{array}{ccc}0&-8\\-15&26\end{array}\right]- 2M = \left[\begin{array}{ccc}6&-2\\3&5\end{array}\right]$

$\implies 2M = \left[\begin{array}{ccc}0&-8\\-15&26\end{array}\right] - \left[\begin{array}{ccc}6&-2\\3&5\end{array}\right]$

$\underline{\large{\textsf{Step-4:}}}$
Subtract the Matrices

$\implies 2M = \left[\begin{array}{ccc}0-6&-8-(-2)\\-15-3&26-5\end{array}\right]$

$\implies 2M = \left[\begin{array}{ccc}-6&-6\\-18&21\end{array}\right]$

$\underline{\large{\textsf{Step-5:}}}$
Divide with 2 on both sides

$\implies M = (\frac{1}{2})\left[\begin{array}{ccc}-6&-6\\-18&21\end{array}\right]$

$\implies M = \left[\begin{array}{ccc}\frac{-6}{2}&\frac{-6}{2}\\\frac{-18}{2}&\frac{21}{2}\end{array}\right]$

$\implies M =(\frac{3}{2}) \left[\begin{array}{ccc}-6&-6\\-6&7\end{array}\right]$