Matrix multiplication Problem set B Answers

1. (a) \(\begin{pmatrix}4 & -5 \\ -2 & 15\end{pmatrix}\)
2. (b) \(\begin{pmatrix}-3\end{pmatrix}\) (a 1 × 1 matrix)
3. (c) \(\begin{pmatrix}3 & 12 & -6 \\ -1 & -4 & 2 \\ 2 & 8 & -4\end{pmatrix}\)
4. (d) \(\begin{pmatrix}13 & 16 \\ -7 & -5\end{pmatrix}\)
5. (e) Does not exist: 2 columns against 3 rows.
6. (f) \(\begin{pmatrix}5 & 3 \\ 2 & 7\end{pmatrix}\)

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

1. (a) \(\begin{pmatrix}10 & 0 \\ -4 & 2\end{pmatrix}\)
2. (b) \(\begin{pmatrix}8 & 2 \\ 6 & 4\end{pmatrix}\), so \(AB \ne BA\)
3. (c) \(\begin{pmatrix}7 & 3 \\ -6 & -2\end{pmatrix}\)
4. (d) \(\begin{pmatrix}4 & 1 \\ -2 & 1\end{pmatrix}\)

1. (a) \(x = 5,\; y = -1\)
2. (b) \(p = 2,\; q = -3\)

(a) \(P\begin{pmatrix}1.05 & 0 \\ 0 & 0.98\end{pmatrix} = \begin{pmatrix}3.36 & 3.43 \\ 5.25 & 4.704 \\ 2.52 & 2.548\end{pmatrix}\), to the cent \(\begin{pmatrix}3.36 & 3.43 \\ 5.25 & 4.70 \\ 2.52 & 2.55\end{pmatrix}\). (b) \(\begin{pmatrix}1.08 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0.9\end{pmatrix} P = \begin{pmatrix}3.456 & 3.78 \\ 5 & 4.8 \\ 2.16 & 2.34\end{pmatrix}\), to the cent \(\begin{pmatrix}3.46 & 3.78 \\ 5.00 & 4.80 \\ 2.16 & 2.34\end{pmatrix}\). Columns are scaled on the right, rows on the left.

\(\begin{pmatrix}110 & 35 \\ 90 & 60 \\ 75 & 40\end{pmatrix}\begin{pmatrix}11.5 \\ 7\end{pmatrix} = \begin{pmatrix}1510 \\ 1455 \\ 1142.5\end{pmatrix}\)

\(PN = \begin{pmatrix}88 & 96 \\ 172 & 168 \\ 100 & 96\end{pmatrix}\). Second club: 172 + 168 = 340.

Wholesaler 1: €372.00, wholesaler 2: €361.50. Wholesaler 2 is cheaper by €10.50.

\(k = -9\) (compare the top-right elements: \(15 = k + 24\)).

\(A^2 = \begin{pmatrix}1 & 6 \\ 0 & 1\end{pmatrix},\; A^3 = \begin{pmatrix}1 & 9 \\ 0 & 1\end{pmatrix},\; A^n = \begin{pmatrix}1 & 3n \\ 0 & 1\end{pmatrix}\)

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