Matrix multiplication Problem set A Answers

1. (a) \(\begin{pmatrix}11 & -1 \\ 16 & -4\end{pmatrix}\)
2. (b) \(\begin{pmatrix}1\end{pmatrix}\) (a 1 × 1 matrix)
3. (c) \(\begin{pmatrix}4 & -8 & 12 \\ 0 & 0 & 0 \\ -1 & 2 & -3\end{pmatrix}\)
4. (d) \(\begin{pmatrix}7 & 9 \\ 9 & -8\end{pmatrix}\)
5. (e) Does not exist: 2 columns against 3 rows.
6. (f) \(\begin{pmatrix}2 & 1 \\ 4 & 3\end{pmatrix}\)

1. (a) 3 × 4
2. (b) Does not exist
3. (c) 2 × 3
4. (d) 4 × 2
5. (e) Does not exist
6. (f) Does not exist

1. (a) \(\begin{pmatrix}4 & 3 \\ 3 & 12\end{pmatrix}\)
2. (b) \(\begin{pmatrix}14 & -1 \\ 11 & 2\end{pmatrix}\), so \(AB \ne BA\)
3. (c) \(\begin{pmatrix}1 & -2 \\ 6 & -3\end{pmatrix}\)
4. (d) \(\begin{pmatrix}-3 & 1 \\ -3 & -1\end{pmatrix}\)

1. (a) \(x = 4,\; y = -2\)
2. (b) \(p = 3,\; q = -2\)

(a) \(P\begin{pmatrix}1.04 & 0 \\ 0 & 0.97\end{pmatrix} = \begin{pmatrix}2.496 & 2.522 \\ 4.68 & 4.074 \\ 1.872 & 1.94\end{pmatrix}\), to the penny \(\begin{pmatrix}2.50 & 2.52 \\ 4.68 & 4.07 \\ 1.87 & 1.94\end{pmatrix}\). (b) \(\begin{pmatrix}1.10 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0.95\end{pmatrix}P = \begin{pmatrix}2.64 & 2.86 \\ 4.5 & 4.2 \\ 1.71 & 1.9\end{pmatrix}\). Columns are scaled on the right, rows on the left.

\(\begin{pmatrix}120 & 45 \\ 80 & 110 \\ 65 & 20\end{pmatrix}\begin{pmatrix}9.5 \\ 6\end{pmatrix} = \begin{pmatrix}1410 \\ 1420 \\ 737.5\end{pmatrix}\)

\(PN = \begin{pmatrix}112 & 115 \\ 135 & 133 \\ 93 & 92\end{pmatrix}\). Second option: 135 + 133 = 268.

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\(k = -1\) (compare the top-right elements: \(14 = k + 15\)).

\(A^2 = \begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix},\; A^3 = \begin{pmatrix}1 & 3 \\ 0 & 1\end{pmatrix},\; A^n = \begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix}\)

\((A+B)^2 = A^2 + AB + BA + B^2\). \(AB\) and \(BA\) are usually different, so they cannot be collected as \(2AB\).