Multiplying a Matrix by a Scalar Test

Question 1 of 5: Find 2A:

$$A=\left[ \begin{array}{ccc}4 & -3 & 8\\ -7 & 1 & 0\end{array}\right]$$

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$$\left[ \begin{array}{ccc}-8 & 6 & -16\\ 14 & -2 & 0\end{array}\right]$$

$$\left[ \begin{array}{ccc}6 & -1 & 10\\ -5 & 3 & 2\end{array}\right]$$

$$\left[ \begin{array}{ccc}-7 & 12 & 5\\ -1 & 1 & 0\end{array}\right]$$

$$\left[ \begin{array}{ccc}8 & -6 & 16\\ -14 & 2 & 0\end{array}\right]$$

Does not exist

Question 2 of 5: Find -3A:

$$A=\left[ \begin{array}{ccc}-5 & 0 & -1\\ 9 & 19 & 3 \\ -7 & 6 & 31\end{array}\right]$$

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$$\left[ \begin{array}{ccc}15 & 0 & 3\\ -27 & -57 & -9 \\21 & -18 & -93\end{array}\right]$$

$$\left[ \begin{array}{ccc}3 & -1 & 6\\ -15 & -30 & 2 \\13 & 17 & 19\end{array}\right]$$

Does not exist

$$\left[ \begin{array}{ccc}-5 & -5 & 19\\ -48 & -93 & 11 \\29 & 5 & 14\end{array}\right]$$

$$\left[ \begin{array}{ccc}18 & -2 & -7\\ 18 & 26 & 11 \\22 & -3 & 19\end{array}\right]$$

Question 3 of 5: Find 4A - 2B:

$$A=\left[ \begin{array}{ccc}4 & -5 & 0\\ 6 & 7 & -1\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}-2 & 6 & -2\\ 1 & 0 & 5\end{array}\right]$$

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Does not exist

$$A=\left[ \begin{array}{ccc}12 & -8 & -4\\ 26 & 28 & 6\end{array}\right]$$

$$A=\left[ \begin{array}{ccc}14 & -1 & 2\\ 13 & 22 & 7\end{array}\right]$$

$$A=\left[ \begin{array}{ccc}12 & -8 & -2\\ 16 & 11 & 5\end{array}\right]$$

$$A=\left[ \begin{array}{ccc}20 & -32 & 4\\ 22 & 28 & -14\end{array}\right]$$

Question 4 of 5: Find X:

$$A=\left[ \begin{array}{cc}9 & 1\\ -7 & 0\end{array}\right]$$ $$B=\left[ \begin{array}{cc}6 & 4\\ 8 & -3\end{array}\right]$$ $$3X + A=B$$

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Does not exist

$$\left[ \begin{array}{cc}-1 & 1\\ 5 & -1\end{array}\right]$$

$$\left[ \begin{array}{cc}-11 & 17\\ 6 & 20\end{array}\right]$$

$$\left[ \begin{array}{cc}19 & 22\\ -17 & 37\end{array}\right]$$

$$\left[ \begin{array}{cc}22 & -15\\ 18 & 10\end{array}\right]$$

Question 5 of 5: Find X:

$$A=\left[ \begin{array}{ccc}2 & 1 & 8\\ 4 & -5 & 0\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}-7 & -9 & -6\\ 0 & 8 & 3\end{array}\right]$$ $$\frac{1}{2}X - A=B$$

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Does not exist

$$A=\left[ \begin{array}{ccc}-12 & -11 & 2\\ -4 & 6 & 5\end{array}\right]$$

$$A=\left[ \begin{array}{ccc}-10 & -16 & 4\\ 8 & 6 & 6\end{array}\right]$$

$$A=\left[ \begin{array}{ccc}5 & 7 & 9\\ 3 & -2 & -1\end{array}\right]$$

$$A=\left[ \begin{array}{ccc}12 & 11 & 18\\ -4 & 15 & 2\end{array}\right]$$

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