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Matrices
Revision: Substitution
1. Solve for one variable in one of the
equations.
2. Substitute this expression into the other
equation to get one equation with one
unknown.
3. Back substitute the value found in step 2
into the expression from step 1.
4. Check.
Solve:
 2x  y  1

3 x  4 y  14
Step 1: Solve equation 1
for y.
Step 2: Substitute this
expression into eqn 2 for y.
Step 3: Solve for x.
Step 4: Substitute this
value for x into eqn 1
and find the
corresponding y value.
Step 5: Check in both
equations.
y  1 2 x
3 x  41  2x   14
3 x  4  8 x  14
 5 x  10
x  2
y  1  2 2
y 5
Elimination
1. Adjust the coefficients.
2. Add the equations to eliminate one of the
variable. Then solve for the remaining
variable.
3. Back substitute the value found in step 2
into one of the original equations to solve
for the other variable.
4. Check.
Solve:
3 x  2y  14

 x  2y  2
Step1: The coefficients on
the y variables are opposites.
No multiplication is needed.
Step 2: Add the two equations
together.
Step 3: Solve for x.
Step 4: Substitute this value for
x into either eqn and find the
corresponding y value.
Step 5: Check in both
equations.
3 x  2y  14
x  2y  2
4 x  16
x4
34   2y
12  2y
2y
y
 14
 14
2
1
Definition:
• In mathematics, a matrix (plural matrices) is a
rectangular array of numbers, symbols, or
expressions, arranged in rows and columns.
• In simple terms, a table of numbers.
Example:
• John and Bill went to the tuck shop. John
bought a pie and cake and Bill bought 2 pies, a
drink and a cake.
John and Bill went to the tuck shop. John bought a pie
and cake and Bill bought 2 pies, a drink and a cake.
John
Bill
Matrix
Pies
1
2
Drinks
0
1
é 1 0 1 ù
ê
ú
êë 2 1 1 úû
Cakes
1
1
Basics for matrices
Order of a matrix
Order = ( rows ´ columns)
2´ 3 means 2 rows and 3 columns
é 1 0 1 ù
ê
ú
êë 2 1 1 úû
Equality:
• Two matrices are equal only if they have the
same order and their corresponding elements
are equal.
é 4 3 ù é 4 3 ù
ê
ú=ê
ú
ë 2 1 û ë 2 1 û
but
é 1 2 ù é 4 3 ù
ê
ú¹ê
ú
ë 3 4 û ë 2 1 û
Addition
• Two matrices must be of the same order
before they can be added.
é 4 3 ù é -4 2 ù é
ê
ú+ ê
ú=ê
ë 2 1 û ë 3 1 û ë
é
=ê
ë
4 - 4 3+ 2 ù
ú
2 + 3 1+1 û
ù
0 5
ú
5 2 û
Subtraction
• The negative matrix is obtained by taking the
opposite value of each element in the matrix.
é
A =ê
ë
é
-A = ê
ë
4 -3 ù
ú
2 1 û
ù
-4 3
ú
-2 -1 û
Subtraction is the
same as adding the
negative matrix.
Example
é 2 2 ù
é 1 2 ù
A =ê
ú
ú B =ê
ë 1 -3 û
ë 3 4 û
é 1- 2 2 - 2 ù é -1 0 ù
A- B =ê
ú
ú=ê
ë 3-1 4 + 3 û ë 2 7 û
Scalar Multiplication
• Multiplying by a number.
• Each element is multiplied by the number.
é 1 2 ù é 3 6 ù
3ê
ú=ê
ú
ë 3 4 û ë 9 12 û
• Also
3( A + B) = 3A + 3B
Zero Matrix
• Every element in the matrix is zero.
• This is the identity for addition.
• Example:
é 0 0 ù
ê
ú,
ë 0 0 û
é 0 0 0 ù
ê
ú
êë 0 0 0 úû
Identity or unit matrix
• The elements of the leading diagonal are all 1
and all other elements are zero.
• Example:
é 1 0 0 ù
é 1 0 ù ê
ú
ê
ú, ê 0 1 0 ú
ë 0 1 û ê 0 0 1 ú
ë
û
Matrix Multiplication
• Two matrices can be multiplied if the number
of columns of the first matrix equals the
number of rows of the second matrix.
• Example:
é 1 2 ùé 5 ù
ê
úê
ú
ë 3 4 ûë 3 û
2 ´ 2 2 ´1
Matrix Multiplication
é 1 2 ùé 5 ù
ê
úê
ú
ë 3 4 ûë 3 û
2 ´ 2 2 ´1
• This forms a 2 x 1 matrix
Matrix Multiplication
é 1 2 ùé 5 ù é 1´ 5 + 2 ´ 3
ê
úê
ú=ê
ë 3 4 ûë 3 û ë 3´ 5 + 4 ´ 3
2 ´ 2 2 ´1
• This forms a 2 x 1 matrix
ù é 11 ù
ú=ê
ú
û ë 27 û
Remember
• A linear equation with 2 variables is a line
• E.g.
2x + 3y =12
2x + 3y =12
A linear equation with 3 variables is a
plane.
Solving problems using matrices
• Scenario 1
• Two lines intersect at one point. There is a
unique solution.
• This means the system is independent and
consistent with a unique solution.
Solving problems using matrices
• Two small jugs and one large jug can hold 8
cups of water. One large jug minus one small
jug constitutes 2 cups of water. How many
cups of water can each jug hold?
Step 1: write down the equations
• Two small jugs and one large jug can hold 8
cups of water. One large jug minus one small
jug constitutes 2 cups of water. How many
cups of water can each jug hold?
• Let x = number of cups of water that the large
jug holds
• Let y = number of cups of water that the small
jug holds
Step 1: write down the equations
• Two small jugs and one large jug can hold 8
cups of water. One large jug minus one small
jug constitutes 2 cups of water. How many
cups of water can each jug hold?
x + 2y = 8
• Let x = number of cups of water that the small
jug holds
• Let y = number of cups of water that the large
jug holds
x-y=2
Step 2: Create the matrix
x + 2y = 8
x-y=2
• Two small jugs and one large jug can hold 8
cups of water. One large jug minus one small
jug constitutes 2 cups of water. How many
cups of water can each jug hold?
é 1 2
ê
êë 1 -1
ù
8
ú
2 úû
• Let x = number of cups of water that the small
jug holds
• Let y = number of cups of water that the large
jug holds
This is called the augmented matrix of
the system
x + 2y = 8
x-y=2
• Two small jugs and one large jug can hold 8
cups of water. One large jug minus one small
jug constitutes 2 cups of water. How many
cups of water can each jug hold?
é 1 2
ê
êë 1 -1
ù
8
ú
2 úû
• Let x = number of cups of water that the small
jug holds
• Let y = number of cups of water that the large
jug holds
Step 2: Keep equation (1) and
eliminate ‘x’ from (2)
(1)
( 2)
é 1 2
ê
êë 1 -1
ù
8
ú
2 úû
é 1 2
ê
êë 0 -3
ù
1
(
)
8
ú
-6 úû ( 2 ) - (1)
Step 3: Divide (2) by -3
é 1 2
ê
êë 0 -3
é 1 2
ê
êë 0 1
ù
1
(
)
8
ú
-6 úû ( 2 ) - (1)
ù
1
(
)
8
ú
2 úû ( 2 ) / - 3
The matrix is now in
row echelon form
é 1 2
ê
êë 0 1
ù
8
ú
2 úû
Using back substitution to solve:
é 1 2
ê
êë 0 1
y=2
x + 2y = 8
x=2
ù
8
ú
2 úû
We could have done 1 more step
é 1 2
ê
êë 0 1
ù
8
ú
2 úû
é 1 0
ê
êë 0 1
ù
2 (1) - 2 ( 2 )
ú
2 úû
( 2)
x=2
y=2
This is called
reduced row echelon form
é 1 0
ê
êë 0 1
2
2
ù
ú
úû
We have a unique solution (2, 2)
• This means the system is independent and
consistent with a unique solution.
• Geometrically: The lines intersect at a unique
point because the gradients are not the same.
Example 2:
• Solve the following equation:
3x - 4y = 2
x+y = 3
Write in matrix form:
3x - 4y = 2
x+y = 3
• Solve the following equation:
é 1 1
ê
êë 3 -4
ù
3
ú
2 úû
é 1 the
ù
(1)equation:
• Solve
following
1
3
ê
ú
êë 3 -4 2 úû ( 2 )
é 1 1
ê
êë 0 -7
é 1 1
ê
êë 0 1
3
-7
ù
(1)
ú
úû ( 2 ) - 3 (1)
ù
(1)
3
ú
1 úû ( 2 ) / - 7
• Solve the following equation:
é 1 1
ê
êë 0 1
ù
3 (1)
ú
1 úû ( 2 )
é 1 0
ê
êë 0 1
ù
2 (1) - ( 2 )
ú
1 úû ( 2 )
y =1
x=2
Solution is (1, 2)
• This means the system is independent and
consistent with a unique solution.
Scenario 2
• The two lines don’t intersect because the lines
are parallel.
• The system is “independent and inconsistent
and has no solution.”
Example:
• Solve the following
x + 2y = 3
2x + 4y = 9
Write down the augmented matrix
x + 2y = 3
2x + 4y = 9
é 1 2
ê
êë 2 4
ù
3
ú
9 úû
é 1 2
ê
êë 2 4
ù
1
(
)
3
ú
9 úû ( 2 )
é 1 2
ê
êë 0 0
ù
1
(
)
3
ú
3 úû ( 2 ) - 2 (1)
The system is “independent and
inconsistent and has no solution.”
é 1 2
ê
êë 2 4
ù
1
(
)
3
ú
9 úû ( 2 )
é 1 2
ê
êë 0 0
ù
1
(
)
3
ú
3 úû ( 2 ) - 2 (1)
The lines are parallel and hence there
is no solution
Scenario 3
Identical lines which intersect at an infinite
number of points.
The system is “dependent and consistent and
has an infinite number of solutions.”
Example:
• Solve the following
x + 2y = 3
2x + 4y = 6
Write down the augmented matrix
x + 2y = 3
2x + 4y = 6
é 1 2
ê
êë 2 4
ù
3
ú
6 úû
é 1 2
ê
êë 2 4
ù
1
(
)
3
ú
6 úû ( 2 )
é 1 2
ê
êë 0 0
ù
1
(
)
3
ú
0 úû ( 2 ) - 2 (1)
The system is “dependent and
consistent and has infinite solutions.”
é 1 2
ê
êë 2 4
ù
1
(
)
3
ú
6 úû ( 2 )
é 1 2
ê
êë 0 0
ù
1
(
)
3
ú
0 úû ( 2 ) - 2 (1)
The lines are identical and hence there is an infinite
number of solutions