Answer
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Step 1:: Add the matrices together.
\left[\begin{array}{ll}3 & 1 \ 5 & 2 \ 10 & 4\end{array}\right]+\left[\begin{array}{ll}6 & 7 \ 0 & 8 \ 9 & 1\end{array}\right] = \left[\begin{array}{ll}3+6 & 1+7 \ 5+0 & 2+8 \ 10+9 & 4+1\end{array}\right] = \boxed{\left[\begin{array}{ll}9 & 8 \ 5 & 10 \ 19 & 5\end{array}\right]}
To do this, add the corresponding entries.
Step 2:: Add the matrices together.
\left[\begin{array}{ll}22 & 35 \ 48 & 62\end{array}\right]+\left[\begin{array}{ll}14 & 27 \ 55 & 28\end{array}\right] = \left[\begin{array}{ll}22+14 & 35+27 \ 48+55 & 62+28\end{array}\right] = \boxed{\left[\begin{array}{ll}36 & 62 \ 103 & 90\end{array}\right]}
To do this, add the corresponding entries.
Step 3:: Subtract the matrices.
\left[\begin{array}{lll}11 & 22 & 33 \ 44 & 55 & 66\end{array}\right]-\left[\begin{array}{lll}9 & 8 & 7 \ 6 & 5 & 2\end{array}\right] = \left[\begin{array}{lll}11-9 & 22-8 & 33-7 \ 44-6 & 55-5 & 66-2\end{array}\right] = \boxed{\left[\begin{array}{lll}2 & 14 & 26 \ 38 & 50 & 64\end{array}\right]}
To do this, subtract the corresponding entries.
Step 4:: Subtract the matrices.
However, we cannot subtract a 3x^3 matrix from a 2x^3 matrix, so there is no solution.
Step 5:: Multiply the matrix by the scalar.
5\left[\begin{array}{cc}4 & 8 \ -7 & 13\end{array}\right] = \left[\begin{array}{cc}5\times4 & 5\times8 \ 5\times-7 & 5\times13\end{array}\right] = \boxed{\left[\begin{array}{cc}20 & 40 \ -35 & 65\end{array}\right]}
To do this, multiply every entry in the matrix by the scalar.
Final Answer
5\left[\begin{array}{cc}4 & 8 \ - 7 & 13\end{array}\right] = \left[\begin{array}{cc}5\times^4 & 5\times^8 \ 5\times- 7 & 5\times^13\end{array}\right] = \boxed{\left[\begin{array}{cc}20 & 40 \ - 35 & 65\end{array}\right]}
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