Worked solutions - Quiz 3¶
This page contains sketch notes to some of the quiz questions from Quiz 3. It is not comprehensive. We have tried to include most of the questions where it may not be obvious how to get to the correct answer.
Question 6¶
\[\begin{split}
\begin{align*}
A = \begin{pmatrix} 1 &4 \\ 5 &1 \\ 2 &3 \end{pmatrix}
\end{align*}
\end{split}\]
\[\begin{split}
\begin{align*}
B = \begin{pmatrix} 4 &2 &1 \\ 3 &6 &3 \\ 2 & 5 & 5 \end{pmatrix}
\end{align*}
\end{split}\]
\[\begin{split}
\begin{align*}
C = \begin{pmatrix} 2&4 \\ 3 &1 \\ 6 &0 \end{pmatrix}
\end{align*}
\end{split}\]
\[\begin{split}
\begin{align*}
A + C = \begin{pmatrix} 1 &4 \\ 5 &1 \\ 2 &3 \end{pmatrix} + \begin{pmatrix} 2&4 \\ 3 &1 \\ 6 &0 \end{pmatrix} =
\begin{pmatrix} 1+2 &4+4 \\ 5+3 &1+1 \\ 2+6 &3+0 \end{pmatrix} =
\begin{pmatrix} 3 &8 \\ 8 &2 \\ 8 &3 \end{pmatrix}
\end{align*}
\end{split}\]
\[\begin{split}
\begin{align*}
A - C = \begin{pmatrix} 1 &4 \\ 5 &1 \\ 2 &3 \end{pmatrix} - \begin{pmatrix} 2&4 \\ 3 &1 \\ 6 &0 \end{pmatrix} =
\begin{pmatrix} 1-2 &4-4 \\ 5-3 &1-1 \\ 2-6 &3-0 \end{pmatrix} =
\begin{pmatrix} -1 &0 \\ 2 &0 \\ -4 &3 \end{pmatrix}
\end{align*}
\end{split}\]
\[\begin{split}
\begin{align*}
A^T + C^T = (A+C)^T = \begin{pmatrix} 3 &8 \\ 8 &2 \\ 8 &3 \end{pmatrix} ^2 = \begin{pmatrix} 3 &8 &8 \\ 8 &2 &3 \end{pmatrix}
\end{align*}
\end{split}\]
Question 7¶
\[\begin{split}
\begin{align*}
2 A = 2 \times \begin{pmatrix} 1 &4 \\ 5 &1 \\ 2 &3 \end{pmatrix} = \begin{pmatrix} 2 \times 1 &2 \times 4 \\ 2 \times 5 &2 \times 1 \\ 2 \times2 &2 \times 3 \end{pmatrix} = \begin{pmatrix} 2 &8 \\ 10 &2 \\ 4 &6 \end{pmatrix}
\end{align*}
\end{split}\]
\[\begin{split}
\begin{align*}
-1(A-C)3 = -3 \times (A-C) = -3 \times \begin{pmatrix} -1 &0 \\ 2 &0 \\ -4 &3 \end{pmatrix} = \begin{pmatrix} -3 \times -1 &-3 \times 0 \\ -3 \times 2 &-3 \times 0 \\ -3 \times -4 &-3 \times 3 \end{pmatrix} = \begin{pmatrix} 3 &0 \\ -6 & 0 \\ 12 &-9 \end{pmatrix}
\end{align*}
\end{split}\]
Question 9¶
\[\begin{split}
\begin{align*}
D = \begin{pmatrix} 3 \\ -1 \end{pmatrix}
\end{align*}
\end{split}\]
\[\begin{split}
\begin{align*}
A D = \begin{pmatrix} 1 &4 \\ 5 &1 \\ 2 &3 \end{pmatrix} \times \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 1\times 3 + 4\times -1 \\ 5 \times 3 + 1 \times -1 \\ 2 \times 3 + 3 \times -1 \end{pmatrix} =\begin{pmatrix} 3 - 4 \\ 15 -1 \\ 6 - 3 \end{pmatrix} =\begin{pmatrix} -1 \\ 14 \\ 3 \end{pmatrix}
\end{align*}
\end{split}\]
Question 10¶
\[\begin{split}
\begin{align*}
B A = \begin{pmatrix} 4 &2 &1 \\ 3 &6 &3 \\ 2 & 5 & 5 \end{pmatrix} \begin{pmatrix} 1 &4 \\ 5 &1 \\ 2 &3 \end{pmatrix} & = \begin{pmatrix} 4 \times 1 + 2\times 5 + 1\times 2 & 4 \times 4 + 2\times 1 + 1\times 3 \\ 3\times 1 + 6\times 5+ 3\times 2 &3\times 4 + 6\times 1 + 3\times 3 \\ 2\times 1 + 5\times 5 + 5\times 2 &2\times 4 + 5\times 1 + 5\times 3 \end{pmatrix} \\ & = \begin{pmatrix} 4 + 10 + 2 & 16 + 2 + 3 \\ 3 + 30 + 6 & 12 + 6 + 9 \\ 2 + 25 + 10 &8 + 5 + 15 \end{pmatrix} \\ & = \begin{pmatrix} 16 &21 \\ 39 &27 \\ 37 &28 \end{pmatrix}
\end{align*}
\end{split}\]
\(AB\) is undefined.
Question 12¶
\[\begin{split}
\begin{align*}
E = \begin{pmatrix} 2 &1 \\ -3 &0 \end{pmatrix}
\end{align*}
\end{split}\]
\[
det(E) = 2 \times 0 - (1 \times -3) = 3
\]
Question 13¶
\[\begin{split}
\begin{align*}
E^{-1} = \frac{1}{det(E)}\begin{pmatrix} 0 &-1 \\ 3 &2 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 0 &-1 \\ 3 &2 \end{pmatrix} = \begin{pmatrix} 0 &-1/3 \\ 1 &2/3 \end{pmatrix}
\end{align*}
\end{split}\]
Check:
\[\begin{split}
\begin{align*}
E E^{-1} = \begin{pmatrix} 2 &1 \\ -3 &0 \end{pmatrix} \begin{pmatrix} 0 &-1/3 \\ 1 &2/3 \end{pmatrix} & = \begin{pmatrix} 2 \times 0 + 1 \times 1 &2 \times -1/3 + 1\times 2/3 \\ -3 \times 0 + 0 \times 1 &-3 \times -1/3 + 0 \times 2/3 \end{pmatrix} \\ & = \begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix} = I
\end{align*}
\end{split}\]
as expected.
Question 14¶
\[\begin{split}
\begin{align*}
E^2 = \begin{pmatrix} 2 &1 \\ -3 &0 \end{pmatrix}\begin{pmatrix} 2 &1 \\ -3 &0 \end{pmatrix} = \begin{pmatrix} 2 \times 2 + 1 \times -3 & 2 \times 1 + 1 \times 0 \\ -3 \times 2 + 0 \times -3 &-3 \times 1 + 0 \times 0 \end{pmatrix} = \begin{pmatrix} 1 &2 \\ -6 &-3 \end{pmatrix} \neq E
\end{align*}
\end{split}\]
so \(E\) is not idempotent.
\[\begin{split}
\begin{align*}
E^T = \begin{pmatrix} 2 &1 \\ -3 &0 \end{pmatrix}^T = \begin{pmatrix} 2 &-3 \\ 1 &0 \end{pmatrix} \neq E^{-1}
\end{align*}
\end{split}\]
so \(E\) is not orthogonal.