### About Multiplying a Matrix by a Scalar:

A scalar in matrix algebra is a real number that is not inside of a matrix. In order to multiply a matrix by a scalar, we will multiply the scalar by each and every element of the given matrix.

Test Objectives
• Demonstrate the ability to multiply a matrix by a scalar
• Demonstrate the ability to add matrices
• Demonstrate the ability to subtract matrices
• Demonstrate the ability to solve matrix equations
Multiplying a Matrix by a Scalar Practice Test:

#1:

Instructions: find 2A.

$$a)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}1 & 0\\ -4 & 9\end{array}\right]$$

Instructions: find (-1/2)A.

$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}6 & 11\\ -3 & 14\end{array}\right]$$

#2:

Instructions: find -4A.

$$a)\hspace{.2em}$$ $$A=\left[ \begin{array}{ccc}1 & 8 & -3\\ 0 & \frac{1}{2}& 9\\ -12 & -2 & -1\end{array}\right]$$

Instructions: find 3A - 2B.

$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{ccc}0 & 1 & 5\\ 8 & 1 & -2\\ 6 & 10 & -5\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}-2 & 1 & 14\\ -3 & 1 & 7\\ 9 & 11 & 10\end{array}\right]$$

#3:

Instructions: find (1/2)A - (1/4)B.

$$a)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}8 & 16\\ 0 & -4\\ 4 & 0\end{array}\right]$$ $$B=\left[ \begin{array}{cc}20 & 40\\ 8 & 0\\ 28 & 5\end{array}\right]$$

Instructions: find (-1/2)A - 4B.

$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{ccc}4 & 0 & -4\\ -10 & 19 & 6\\ 1 & 14 & -7\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}1 & -1 & 13\\ 7 & 0 & -6\\ -2 & 4 & 0\end{array}\right]$$

#4:

Instructions: if 2X + A = B, find X.

$$a)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}-1 & 7\\ 0 & 5\end{array}\right]$$ $$B=\left[ \begin{array}{cc}-2 & 1\\ 6 & 3\end{array}\right]$$

Instructions: if X + 3A = -B, find X.

$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}2 & -1\\ 4 & 6\end{array}\right]$$ $$B=\left[ \begin{array}{cc}7 & -2\\ \frac{1}{2}& 0\end{array}\right]$$

#5:

Instructions: if 2A + X = -3B, find X.

$$a)\hspace{.2em}$$ $$A=\left[ \begin{array}{ccc}5 & 2 & 3\\ 0 & 7 & -1\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}-7 & 0 & 4\\ -2 & 10 & 6\end{array}\right]$$

Instructions: if (1/2)A + X = -2B, find X.

$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{ccc}4 & 0 & 2\\ -2 & 0 & -6 \\ 7 & 16 & 0\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}10 & 8 & 1\\ 12 & -1 & 2 \\ 0 & 5 & 6\end{array}\right]$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}2 & 0\\ -8 & 18\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-3 & -\frac{11}{2}\\ \frac{3}{2}& -7\end{array}\right]$$

#2:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-4 & -32 & 12\\ 0 & -2 & -36 \\ 48 & 8 & 4\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}4 & 1 & -13\\ 30 & 1 & -20 \\0 & 8 & -35\end{array}\right]$$

#3:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-1 & -2\\ -2 & -2 \\ -5 & -\frac{5}{4}\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-6 & 4 & -50\\ -23 & -\frac{19}{2}& 21 \\ \frac{15}{2}& -23 & \frac{7}{2}\end{array}\right]$$

#4:

Solutions:

$$a)\hspace{.2em}$$ $$X=\left[ \begin{array}{cc}-\frac{1}{2}& -3\\ 3 & -1\end{array}\right]$$

$$b)\hspace{.2em}$$ $$X=\left[ \begin{array}{cc}-13 & 5\\ -\frac{25}{2}& -18\end{array}\right]$$

#5:

Solutions:

$$a)\hspace{.2em}$$ $$X=\left[ \begin{array}{ccc}11 & -4 & -18\\ 6 & -44 & -16\end{array}\right]$$

$$b)\hspace{.2em}$$ $$X=\left[ \begin{array}{ccc}-22 & -16 & -3\\ -23 & 2 & -1 \\ -\frac{7}{2}& -18 & -12\end{array}\right]$$