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
#1:
Instructions: find 2A.
$$a)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}1 & 0\\ -4 & 9\end{array}\right]$$
Instructions: find (-1/2)A.
$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}6 & 11\\ -3 & 14\end{array}\right]$$
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#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.
$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{ccc}0 & 1 & 5\\ 8 & 1 & -2\\ 6 & 10 & -5\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}-2 & 1 & 14\\ -3 & 1 & 7\\ 9 & 11 & 10\end{array}\right]$$
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#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.
$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{ccc}4 & 0 & -4\\ -10 & 19 & 6\\ 1 & 14 & -7\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}1 & -1 & 13\\ 7 & 0 & -6\\ -2 & 4 & 0\end{array}\right]$$
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#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.
$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}2 & -1\\ 4 & 6\end{array}\right]$$ $$B=\left[ \begin{array}{cc}7 & -2\\ \frac{1}{2}& 0\end{array}\right]$$
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#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.
$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{ccc}4 & 0 & 2\\ -2 & 0 & -6 \\ 7 & 16 & 0\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}10 & 8 & 1\\ 12 & -1 & 2 \\ 0 & 5 & 6\end{array}\right]$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}2 & 0\\ -8 & 18\end{array}\right]$$
$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-3 & -\frac{11}{2}\\ \frac{3}{2}& -7\end{array}\right]$$
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#2:
Solutions:
$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-4 & -32 & 12\\ 0 & -2 & -36 \\ 48 & 8 & 4\end{array}\right]$$
$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}4 & 1 & -13\\ 30 & 1 & -20 \\0 & 8 & -35\end{array}\right]$$
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#3:
Solutions:
$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-1 & -2\\ -2 & -2 \\ -5 & -\frac{5}{4}\end{array}\right]$$
$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-6 & 4 & -50\\ -23 & -\frac{19}{2}& 21 \\ \frac{15}{2}& -23 & \frac{7}{2}\end{array}\right]$$
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#4:
Solutions:
$$a)\hspace{.2em}$$ $$X=\left[ \begin{array}{cc}-\frac{1}{2}& -3\\ 3 & -1\end{array}\right]$$
$$b)\hspace{.2em}$$ $$X=\left[ \begin{array}{cc}-13 & 5\\ -\frac{25}{2}& -18\end{array}\right]$$
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#5:
Solutions:
$$a)\hspace{.2em}$$ $$X=\left[ \begin{array}{ccc}11 & -4 & -18\\ 6 & -44 & -16\end{array}\right]$$
$$b)\hspace{.2em}$$ $$X=\left[ \begin{array}{ccc}-22 & -16 & -3\\ -23 & 2 & -1 \\ -\frac{7}{2}& -18 & -12\end{array}\right]$$