$\left[\begin{array}{cc}n-1&6\\-19&m+3\end{array}\right] +\left[\begin{array}{cc}-1&0\\16&-8\end{array}\right] =\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]$ .

We must first simplify the Left Hand Side of the equation by adding corresponding entries.

$\left[\begin{array}{cc}n-1+-1&6+0\\-19+16&m+3-8\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]$ .

That is;

$\left[\begin{array}{cc}n-2&6\\-3&m-5\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]$ .

Since the two matrices are equal, their corresponding entries are also equal. we equate corresponding entries and solve for m and n.

This implies that;

$n-2=10$

We got this equation from row one-column one entry of both matrices.

$n=12$

Also, the row three-column three entries of both matrices will give us the equation;

$m-5=40$

$m=45$

Hence the correct answer is $m=45,n=12$ .

The correct option is option 2

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3 years ago