Precalculus (6th Edition) Blitzer

The required matrix is, $\left[ \begin{array}{*{35}{r}} -1 & 2 \\ -5 & 4 \\ \end{array} \right]$
To solve the provided expression, follow the methods given below: Matrix addition: consider two matrices $X,Y$ $X=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]$ and $Y=\left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right]$ Now, \begin{align} & X+Y=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]+\left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} {{x}_{1}}+{{y}_{1}} & {{x}_{2}}+{{y}_{2}} \\ {{x}_{3}}+{{y}_{3}} & {{x}_{4}}+{{y}_{4}} \\ \end{matrix} \right] \end{align} Matrix multiplication with a scalar: Consider a matrix $Z$ and a scalar value a: $Z=\left[ \begin{matrix} {{z}_{1}} & {{z}_{2}} \\ {{z}_{3}} & {{z}_{4}} \\ \end{matrix} \right]$ Now, \begin{align} & aZ=a\left[ \begin{matrix} {{z}_{1}} & {{z}_{2}} \\ {{z}_{4}} & {{z}_{3}} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} a{{z}_{1}} & a{{z}_{2}} \\ a{{z}_{4}} & a{{z}_{3}} \\ \end{matrix} \right] \end{align} Using the method given above, solve the provided expression: \begin{align} & BC-3B=\left[ \begin{array}{*{35}{r}} 1 & -1 \\ 2 & 1 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} 1 & 2 \\ -1 & 3 \\ \end{array} \right]-3\left[ \begin{array}{*{35}{r}} 1 & -1 \\ 2 & 1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} 1+1 & 2-3 \\ 2-1 & 4+3 \\ \end{array} \right]-\left[ \begin{array}{*{35}{r}} 3 & -3 \\ 6 & 3 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} 2 & -1 \\ 1 & 7 \\ \end{array} \right]-\left[ \begin{array}{*{35}{r}} 3 & -3 \\ 6 & 3 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} -1 & 2 \\ -5 & 4 \\ \end{array} \right] \end{align} The matrix is, $BC-3B=\left[ \begin{array}{*{35}{r}} -1 & 2 \\ -5 & 4 \\ \end{array} \right]$.