|
Solutions
to matrices
Clement Radcliffe, Contributor
 |
| Students
watch and cheer on artistes who were part of the STAR 'No Violence in Schools
Tour', at Spanish Town High School, last December. - Ricardo Makyn/Staff Photographer |
The review
of matrices should prove interesting as the problems set on the topic provide
very few challenges. We will continue this exercise with the solution to the following
problems. Given
the vectors: | M
= (4 0) N | =
(-1 2) | (0
2) | (0
3) |
Evaluate
(a) M + N (b) M - 2N Solution
| M
+ N = | (4
0) + | (-1
2) | =
(3 2) | | | (0
2) | (0
3) | (0
5) | | | | | | | :
M + N = | (3
2) | | | | | (0
5) | | |
| M
- 2N = | (4
0) | -2 | (-1
2) = | (4
0) - | (-2
4) | =
(6 -4) | | | (0
2) | | (0
3) | (0
2) | (0
6) | (0
-4) | | | | | | | | | | :
M - 2N = | (6
-4) | | | | | | | | (0
-4) | | | | | |
2.
Given that: | (4
-2) + | (y
6) = |
(3 4) | | (3
x) | (2
-3) | (5
-1) |
Find
the values of x and y. Solution
| (4
-2) + | (y
6) = |
(3 4) | | (3
x) | (2
-3) | (5-
-1) |
| Then
| (4
+ y 4) |
= | (3
4) | Equation
terms | | | (5
x -3) | | (5
-1) | |
:
4 + y = 3 : y = -1
X
- 3 = -1 : X = 2 Answers:
y = 1 and x = 2 I
do hope that you noticed that corresponding values are the same in equal matrices.
Please note also that from the above, given the matrix | M
= | (4
-2) |
then k x M + | (4K
-2K) | | | | (3
x) | | (3k
xk) | |
where
K is constant. MULTIPLICATION
OF MATRICES The
matrix Ax x Y refers to the Matrix A with orders X x Y, that is the matrix with
x rows and y columns. It
is important that you consider their orders when multiplying two matrices. The
orders are reviewed to determine: - If
multiplication is possible
- The
order of the product (matrix)
Given
the matrices Ax x y X By x the product can be found since the number of columns
of A is the same as the number of rows of b, that is, y in this case. The
order of the answer is X x Z It
is important to follow this procedure, especially if you are not comfortable with
the topic. Having
established that both matrices can be multiplied, let us attempt the following:
Find
the product of Using
the approach indicated previously to consider the orders of both, then A 1 X 2
X B2 X 1. They can be multipled and the order of the product is 1 X 1. The
product is found as follows: | (2
3) | (-5) |
= (2 X 5 + 3 X 1) = (-10 + 3) = (-7) | | | (1) | |
This
forms the basis of matrix multiplication where you multiply row by column. This
is repeated to other rows and columns in matrices. Now
let us attempt the following together. Example
| Given
A = | (2
1) |
and b = | (3
1) | | | | (0
3) | | (1
4) | |
Evaluate
A x B The
product of two 2 X 2 matrices has order 2 X 2. Solution
A
X B = (2 1) X (3 1) = (2 X 3 = 1 X 1 2 X 1 + 1 X 4) (
0 3) (1 4) (0 X 3 + 3 X 1 0 X 1 + 3 X 4) You
are encouraged to practise as many examples of the multiplication of two 2x2 matrices
as possible, as these provide the most challenge with respect to this topic. The
following is a typical exam-type problem, | Given
that P = | (3
1) |
and S = | (2
1) | | | | (-2
0) | | (-1
-2) | |
Find
the value of: 91) 2P - S (11) S2 Solution:
Given
the matrices P and S, then: | 2P
- S = 2 x | (3
1) |
- | (2
1) | | | | (-2
0) | | (-1
-2) | |
| =
(6 2) | -
(2 1) |
+ |
(4 1) | | | (-4
0) | (-1
-2) | | (-3
20) | |
| (11)
S2 = | (2
1) (2 1) = (2 X 2 + 1 X 1 2 X 1 + 1 X -2) | | | (1
-2) (-1 -2) (-1 X 2 + -2 X 1 -1 X 1 + -2 X -2) | | | | | =
(3 0) | | | (0
3) | | | | | | Answer:
| (3
0) | | | (0
3) |
Now
that you are comfortable with multiplying two 2X2 matrices, we can proceed, note
the following: The
unit matrix with respect to multiplication is I
am sure that you can prove that given matrix | B
= | (3
6) | that
b X 1 = B. | | | (2
4) | | | |
| Given
the 2 X 2 matrix A | (a
b) | | | | | | (c
d) | | | | | | | | the
determinant of A, | (a
b) | is
ad - bc. | | | (c
d) | |
| Example:
If A is | (3
5), | then
the determinant of A = ad - bc = 3 X 4 - 5 X 2 = 2 | | | (2
4) | |
Given
the matrix A, then the inverse of
A is: 1 (d -b)
ad
-bc (-c a) | Example:
Given that A = | (3
5), | findthe
inverse of A, or A . | | | (2
4) | | | | | | | Using
the formula above, | A
1 = | 1/2
(4 -5) | | | | (-2
3) |
Simultaneous
equations, for example, 3X = 5Y = 19, may be expressed in matrix form as follows:
(3 5) (X) = (19)
(2
4) ( Y) (14) Or
alternatively, A x X = B Where
A is the coefficient matrix, X is the variable matrix and b the constant matrix.
Please
note the above well as they form the basis of finding the solution of simultaneous
equations using matrices. Homework
| 1.
Matrix C | (6
2) | is
a singular matrix. | | | (5
p) | |
Calculate
the value of P | 2.
Given that | -3x
+ 2y = - 11 | | | 5x
+ 4y = 33 |
Express
the simultaneous equations in the form C X = D Given
the 2 x 2 matrix C, find:
The
determinant of C
(11)
The inverse of C Clement
Radcliffe is the principal of Glenmuir High School in May Pen. | 
