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EXERCISE 1C Work out the following, where possible. 1 $A = \begin{pmatrix} 1 & 3 \\ 2 & 6 \\ 7 & 0 \\ 1 & 1 \end{pmatrix}$ $B = \begin{pmatrix} 2 & 2 \\ -1 & 6 \\ 2 & 3 \\ 3 & 3 \end{pmatrix}$ (a) A + B (b) A - B (c) A + A 2 $A = \begin{pmatrix} 2 & 2 & 5 & 4 \\ 1 & 2 & 1 & 2 \\ 0 & 9 & 1 & 7 \end{pmatrix}$ $B = \begin{pmatrix} 0 & 1 & 0 & 1 \\ 4 & 5 & 6 & 1 \\ 0 & -2 & 1 & -2 \end{pmatrix}$ (a) A + B (b) A - B (c) B + B + B 3 $A = \begin{pmatrix} 1 & 3 & 1 & 7 \end{pmatrix}$ $B = \begin{pmatrix} 7 & 6 & 5 & 4 \end{pmatrix}$ (a) A + B (b) A - B (c) A + B + A + B 4 $A = \begin{pmatrix} 6 & 1 & -6 & 8 \\ 0 & 2 & -1 & 5 \end{pmatrix}$ $B = \begin{pmatrix} 2 & 1 & -1 & 1 \\ 0 & 9 & -8 \end{pmatrix}$ $C = \begin{pmatrix} 3 & 5 & 3 & 9 \\ 5 & 1 & -2 & 1 \end{pmatrix}$ (a) A + C (b) B + C (c) A + B (d) A + B + C (e) A - B (f) A - C (g) B + B 1.4 Multiplying by a scalar You will see that there is another way of looking at the question 4(g). Exercise 1C. $\begin{pmatrix} 2 & 1 & -1 \\ 0 & 9 & 8 \end{pmatrix} + \begin{pmatrix} 2 & 1 & -1 \\ 0 & 9 & 8 \end{pmatrix} = \begin{pmatrix} 4 & 2 & -2 \\ 0 & 18 & 16 \end{pmatrix} + \begin{pmatrix} 2 & 1 & -1 \\ 0 & 9 \end{pmatrix}$ You can see that B + B = 2B, similarly B + B + B = 3B and so on. Multiplying a matrix by a scalar is the same as multiplying every term by that scalar.

          EXERCISE 1C
Work out the following, where possible.
1
$A = \begin{pmatrix} 1 & 3 \\ 2 & 6 \\ 7 & 0 \\ 1 & 1 \end{pmatrix}$ $B = \begin{pmatrix} 2 & 2 \\ -1 & 6 \\ 2 & 3 \\ 3 & 3 \end{pmatrix}$
(a) A + B
(b) A - B
(c) A + A
2
$A = \begin{pmatrix} 2 & 2 & 5 & 4 \\ 1 & 2 & 1 & 2 \\ 0 & 9 & 1 & 7 \end{pmatrix}$ $B = \begin{pmatrix} 0 & 1 & 0 & 1 \\ 4 & 5 & 6 & 1 \\ 0 & -2 & 1 & -2 \end{pmatrix}$
(a) A + B
(b) A - B
(c) B + B + B
3
$A = \begin{pmatrix} 1 & 3 & 1 & 7 \end{pmatrix}$ $B = \begin{pmatrix} 7 & 6 & 5 & 4 \end{pmatrix}$
(a) A + B
(b) A - B
(c) A + B + A + B
4
$A = \begin{pmatrix} 6 & 1 & -6 & 8 \\ 0 & 2 & -1 & 5 \end{pmatrix}$ $B = \begin{pmatrix} 2 & 1 & -1 & 1 \\ 0 & 9 & -8 \end{pmatrix}$ $C = \begin{pmatrix} 3 & 5 & 3 & 9 \\ 5 & 1 & -2 & 1 \end{pmatrix}$
(a) A + C
(b) B + C
(c) A + B
(d) A + B + C
(e) A - B
(f) A - C
(g) B + B
1.4 Multiplying by a scalar
You will see that there is another way of looking at the question 4(g).
Exercise 1C.
$\begin{pmatrix} 2 & 1 & -1 \\ 0 & 9 & 8 \end{pmatrix} + \begin{pmatrix} 2 & 1 & -1 \\ 0 & 9 & 8 \end{pmatrix} = \begin{pmatrix} 4 & 2 & -2 \\ 0 & 18 & 16 \end{pmatrix} + \begin{pmatrix} 2 & 1 & -1 \\ 0 & 9 \end{pmatrix}$
You can see that B + B = 2B, similarly B + B + B = 3B and so on.
Multiplying a matrix by a scalar is the same as multiplying every
term by that scalar.

EXERCISE 1C
Work out the following, where possible.
1
A =
< p m a t r i x > B =
< p m a t r i x >
(a) A + B
(b) A - B
(c) A + A
2
A =
< p m a t r i x > B =
< p m a t r i x >
(a) A + B
(b) A - B
(c) B + B + B
3
A =
< p m a t r i x > B =
< p m a t r i x >
(a) A + B
(b) A - B
(c) A + B + A + B
4
A =
< p m a t r i x > B =
< p m a t r i x > C =
< p m a t r i x >
(a) A + C
(b) B + C
(c) A + B
(d) A + B + C
(e) A - B
(f) A - C
(g) B + B
1.4 Multiplying by a scalar
You will see that there is another way of looking at the question 4(g).
Exercise 1C.
< p m a t r i x >
+
< p m a t r i x >
=
< p m a t r i x >
+
< p m a t r i x >
You can see that B + B = 2B, similarly B + B + B = 3B and so on.
Multiplying a matrix by a scalar is the same as multiplying every
term by that scalar.

Added by Chad B.

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