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## This Week

We solved some more linear systems and began a study of matrices. Updated notes.

## Exercise Solutions

I will give the solution to the first part of a question and then the answers to remaining parts.

### P. 21 Exercises

#### 1 (i)

Letting the first column be $x$, the second $y$ and the third $z$:

$x-y+6z=0$.

$y=3$.

$2x-y=1$.

#### (ii)

$2x-y=-1$

$-3x+2y+z=0$

$-y+z=3$

#### 2 (i)

$\left[\begin{array}{cc|c} 1 & 2 & 1\\ 3 & 4 & -1\end{array}\right]$

$\overset{{r}_2 -3 {r}_{1}}{\longrightarrow}\left[\begin{array}{cc|c} 1 & 2 & 1 \\ 0 & -2 & -4\end{array}\right]$

$\overset{r_2\times -1/2}{\longrightarrow}\left[\begin{array}{cc|c} 1 & 2 & 1\\ 0 & 1 & 2 \end{array}\right]$.

Hence $y=2$ and $x=1-2y=-3$.

(ii) $x=-17,y=13$ (iii) No solutions (iv) $x=1/9,\,y=10/9,\,z=-7/3$ (v) $x=-21-15t,\,y=-17-11t,\,z=t$ for $t\in\mathbb{R}$ (vi) $x=-7,\,y=-9,\,z=1$.

#### 3 (a)

No

$\left[\begin{array}{cc|c} 1&1&0\\ 0 & 0 & 0\end{array}\right]$

has non-zero solutions. For example $x=1,\,y=-1$.

#### (b)

No. The above example is a counter-example.

#### (c)

No.

$\left[\begin{array}{cc|c} 1 & 0 & 0\\ 0 & 1 & 0\end{array}\right]$

#### (d)

No.

$\left[\begin{array}{cc|c} 1&0&0 \\ 0&1&0\\ 0&0&0\end{array}\right]$

has the unique solution $x=y=0$.

### P. 29 Exercises

#### 1 (i)

Not possible — they are different sizes

(ii)  $\left(\begin{array}{cc}15&-5\\ 10 & 0\end{array}\right)$ (iii) $\left(\begin{array}{cc}-1&3\\ -2&-4\end{array}\right)$ (iv) Not possible. (v) $\left(\begin{array}{cc}5&2\\ 0 & -1\end{array}\right)$ (vi) Not possible.

#### 2 (i)

$2A=\left(\begin{array}{cc}1&0\\ 2&3\end{array}\right)-\left(\begin{array}{cc}5&2\\ 6&1\end{array}\right)$.

$\Rightarrow 2A=\left(\begin{array}{cc}-4&-2\\ -4&2\end{array}\right)$.

$\Rightarrow A=\left(\begin{array}{cc}-2&-1\\ -2 & 1\end{array}\right)$.

(b) $\left(4,1/2\right)^T$ (c) $\left(\begin{array}{cc}2 & 1 \\ 0 & -1\end{array}\right)$ (d) $\left(\begin{array}{cc}2& 7\\ -9/2 & -5\end{array}\right)$

#### Q. 3 (a)

These matrices are conformable:

$\left(\begin{array}{ccc}1 & -1 &2\\ 2 & 0 & 4\end{array}\right)\left(\begin{array}{ccc} 2 & 3 & 1\\ 1 & 9 & 7\\ -1 & 0 &2\end{array}\right)=\left(\begin{array}{ccc}2-1-2 & 3-9+0& 1-7+4\\ 4+0-4 & 6+0+0 & 2+0+8 \end{array}\right)$

$=\left(\begin{array}{ccc}-1& -6 & -2\\ 0 & 6 & 10\end{array}\right)$

(b) $(-3\,\,-15)$ (c) $\left(\begin{array}{cc}1&0\\ 0 & 1\end{array}\right)$ (d) $\left(\begin{array}{ccc}aa'&0&0\\ 0 & bb'&0\\ 0 & 0 & cc'\end{array}\right)$ (e) Non-conformable (f) $\left(\begin{array}{cc}4&10\\ -2&-1\end{array}\right)$

#### 4 (a)

Suppose that $A$ is an $m\times n$ matrix. Hence when we multiply $A\times A$ we need $m=n$ ($(m\times n)\times(m\times n)$). $A$ is a square matrix.

(b) One is $m\times n$ while the other is $n\times m$ (c) $3\times 5$