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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]

has a unique solution x=y=0.

(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)

Adding to both sides:

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

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