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EECM 3714

Lecture 12: Unit 12

Matrix algebra
Renshaw, Ch
...
19
• Definitions, notation
• Matrix operations (Transposition – page 620; Addition; subtraction; Scalar multiplication; Matrix
multiplication - Page 621-3)
• Determinants

• Matrix inversion
• 2 by 2 inversion

• 3 by 3 inversion
• Solving systems of linear equations (Matrix Inversion and Cramer’s rule)

DEFINITIONS, NOTATION
• Matrix is a rectangular array of numbers/variables, e
...
(Page 578-579):

• 𝐴3×3

𝑎
= 𝑑
𝑔

𝑏
𝑒
ℎ

𝑐
𝑓 , 𝐵2×2 = 1 2
3 0
𝑖

• Order = dimensions of a matrix

• Order = number of rows (r) by number of columns (c)
• Usually denoted as m n, m = rows, n = columns

• An element is an entry in a matrix, denoted as 𝑎𝑖𝑗 , e
...
the element 𝑎23 = 𝑓 in matrix A, while
the element 𝑏22 = 0 in matrix B

SPECIAL MATRICES
• Square matrix: number of rows = number of
columns, i
...
𝑚 = 𝑛
1 3 5
• E
...
𝐶 = 7 6 4
0 23 1
• Null matrix: every element of matrix = 0 e
...

0 0
0=
0 0
• Identity matrix: diagonal elements are all 1; all
other elements are 0
1 0
• Note: must be a square matrix, e
...
𝐼 =
0 1

VECTORS AND SCALARS
• Scalar is a 1 × 1 matrix, i
...
a constant

• Row vector: matrix with only one row,
i
...
𝑚 = 1, e
...
𝑅 = 1 5 2
• Column vector = matrix with only one
2
column, i
...
𝑛 = 1, e
...
𝐷 = 4
1

EQUALITY OF TWO MATRICES
Two matrices A and B are equal if and only if
1) they have the same order and
2) if every element in A is equal to the corresponding element in B, e
...

𝐴=

2
1

5
2
,𝐵 =
2
1

5
⟹𝐴=𝐵
2

TRANSPOSITION
• Transposition involves interchanging the row and column entries of a matrix
• Notation: If A is a matrix with m rows and n columns, then its transpose, denoted by 𝐴𝑇 = 𝐴′ has
n rows and m columns
1 3 4
• Suppose that 𝐴 = 𝑎 𝑏 𝑐 and 𝐵 =
6 2 5
𝑎
1 6
• The transposes are then 𝐴′ = 𝑏 and 𝐵′ = 3 2
𝑐
4 5
• Note how the first row becomes the first column, the second row becomes the second column,
etc
...
e
...
e
...
)

0 9
6 7
3
;𝐵 =
;𝐶 =
1 50
0 4
10
• Note that A and B are conformable for addition/subtraction, while A and C; B and C are not
...

𝑎 𝑏
• Suppose that 𝐴 =

...

𝑐 𝑑
𝑘𝑎 𝑘𝑏
• 𝑘𝐴 =
𝑘𝑐 𝑘𝑑

MATRIX MULTIPLICATION, 1
• Before multiplying two matrices, first ensure that they are conformable for multiplication

• This involves checking if the number of columns of the first matrix = number of rows of second
matrix
• The order of the new matrix is given by the number of rows of the first matrix and the number of
columns of the second matrix

• Suppose A is a 2 × 3 matrix and B is a 3 × 2 matrix
...
Because B has 2 columns while A has 2 rows
...
621
• Multiply the first element of the first row with the first element of the first column; then multiply the second
element of the first row with the second element of the first column, etc
...


• Do the same with all row and column combinations (sum of the products of elements of row 1 and column 2
will yield the 1,2 element in the product matrix, etc
...
Find AB
3 4
2
A is a 2 by 2 matrix; B is a 2 by 1 column vector
...
The resulting matrix is a 2 by 1
column vector…
(1 × 1) + (2 × 2)
1 2 1
5
𝐴𝐵 =
=
=
(3 × 1) + (4 × 2)
3 4 2
11
2 4
1 3
If 𝐶 =
;𝐷 =
, find C-D
6 8
5 7
Note that C and D are conformable for subtraction (same order, 2 by 2)
2 4
1 3
1 1
𝐶−𝐷 =
−
=
6 8
5 7
1 1
Do examples 19
...


• Suppose that 𝐴 =
•

•
•
•
•

•

Examples
𝑎
• If 𝐸 = 𝑐
𝑒

𝑏
𝑑 , find E’
...
find 100𝑍
...


𝐴𝑛𝑠: 100𝑍 = 100 100

1000 ∴ 100𝑍 = 10000

100000

1 2 3
1
• If 𝑋 = 4 5 6 , 𝑌 = 2 , find XY
...
e
...
e
...

𝑎 𝑏
, 𝑡ℎ𝑒𝑛 𝐷𝑒𝑡 𝐴 = 𝑎𝑑 − 𝑏𝑐
𝑐 𝑑
• For a 3 × 3 matrix, we can use Sarrus’ rule (ONLY works for 3 by 3 matrices!!): If
• For a 2 × 2 matrix: If 𝐴 =

+
𝑐
𝑎
𝑓 , 𝐷𝑒𝑡 𝐴 =
𝑑
𝑖
𝑔
• Then, 𝐷𝑒𝑡 𝐴 = 𝑎𝑒𝑖 + 𝑏𝑓𝑔 + 𝑐𝑑ℎ − 𝑎𝑓ℎ − 𝑏𝑑𝑖 − 𝑐𝑒𝑔
𝑎
𝐴= 𝑑
𝑔

𝑏
𝑒
ℎ

+ +
𝑏 𝑐
𝑒 𝑓
ℎ 𝑖

− −
𝑎 𝑏
𝑑 𝑒
𝑔 ℎ

−
𝑐
𝑓
𝑖

EXAMPLES
2 4
Find the determinants of the following matrices • 𝐵 = 4 3
4 −7
2 1
• 𝐴=
1 3
+
• 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
2
𝐷𝑒𝑡 𝐵 =
4
𝐷𝑒𝑡 𝐴 = (4 × 3) − −7 × 1
...

𝐷𝑒𝑡 𝐴 = 19
...

𝑏

• Normally, we’d solve this as 𝑥 = 𝐴
...
But
...


• Therefore, we can solve for x as follows : 𝑥 = 𝐴−1 𝑏
• So we can solve for a system if the matrix A has an inverse

• Only square matrices have inverses; only non-singular matrices have inverses
...
Find the inverse of the matrix A as follows:
𝑐 𝑑
• First, find 𝐷𝑒𝑡(𝐴)
...

𝐷𝑒𝑡(𝐴) −𝑐
𝑎𝑑−𝑏𝑐
𝑎
−𝑐 𝑎
𝐴−1 =

𝑑
𝑎𝑑−𝑏𝑐
−𝑐
𝑎𝑑−𝑏𝑐

−𝑏
𝑎𝑑−𝑏𝑐
𝑎
𝑎𝑑−𝑏𝑐


...
6 & 19
...


INVERSE OF 3 BY 3 MATRIX, 1
• The inverse of A is then
𝐴−1 =

1
𝐴𝑑𝑗(𝐴)
...
8

− 𝑀21
𝑀22
− 𝑀23
𝐶31
𝐷𝑒𝑡(𝐴)
𝐶32
𝐷𝑒𝑡(𝐴)
𝐶33
𝐷𝑒𝑡(𝐴)

1
𝐷𝑒𝑡(𝐴)

𝑀31
− 𝑀32
𝑀33

𝐶11
𝐶12
𝐶13

𝐶21
𝐶22
𝐶23

𝐶31
𝐶32
...


EXAMPLE 2
2
• Find the inverse of 𝐵 = 4
2
• Solution:
• 𝐵−1 =

1
𝐴𝑑𝑗(𝐵)
𝐷𝑒𝑡(𝐵)

• Recall that 𝐷𝑒𝑡(𝐵) = 10
• First, find all of the minors:
• Then, find all of the cofactors:

4 1
3 7
1 3

SOLUTION, 2
MINORS

𝑀11 = 3 3 − 7 1 = 2; 𝑀12 = 4 3 − 7 2 = −2 𝑀13 = 4 1 − 3 2 = −2
𝑀21 = 4 3 − 1 1 = 11; 𝑀22 = 2 3 − 1 2 = 4;𝑀23 = 2 1 − 4 2 = −6
𝑀31 = 4 7 − 3 1 = 25;𝑀32 = 2 7 − 4 1 = 10; 𝑀33 = 2 3 − 4 4 = −10
...

6

𝑀33 = −10
...
2 −1
...
5
−10 = 0
...
4 −1
−10
−0
...
6 −1

MATRIX ALGEBRA & SYSTEMS OF EQUATIONS
• Matrix algebra is often used as a compact notation, especially for linear systems of equations
• We can then use matrix inversion and Cramer’s rule to solve for systems of equations using matrix
algebra, if they are written in the correct form
• Consider the following system of equations:
4𝑥 + 𝑦 − 5𝑧 = 8

−2𝑥 + 3𝑦 + 𝑧 = 12
3𝑥 − 𝑦 + 4𝑧 = 5
Note that here we have 3 equations wth 3 variables (unknowns)

• All unknowns on the LHS, all constants on RHS
• Note the order in which x, y, z appear in each equation in the system
...


MATRICES & SYSTEMS OF EQUATIONS
• The matrix algebra of the system of equations is as follows:
𝑥
8
4
1 −5
Ax = b, where 𝐴 = −2 3
1 , 𝑏 = 12 , 𝑥 = 𝑦
...

𝐴𝑥 = 𝑏 = −2 3
1
𝑧
5
3 −1 4
Note the result of multiplying A and x:
4
1
𝐴𝑥 = −2 3
3 −1

−5
1
4

𝑥
𝑦
𝑧

4𝑥 + 𝑦 − 5𝑧
= −2𝑥 + 3𝑦 + 𝑧
3𝑥 − 𝑦 + 4𝑧

EXAMPLE
• Consider the system of equations
𝑥 = 12 − 2𝑦
...

2𝑧 = −1 + 4𝑥 − 𝑦
...

−5𝑥 + 4𝑦 + 3𝑧 = 13
...


SOLUTION
• Therefore
1 2 0
𝐵 = −5 4 3 , 𝑑 =
−4 1 2
• Note that
𝑥 + 2𝑦 + 0𝑧
𝐵𝑑 = −5𝑥 + 4𝑦 + 3𝑧
−4𝑥 + 𝑦 + 2𝑧

𝑥
12
𝑦 , 𝑒 = 13
...


SOLVING SYSTEMS OF LINEAR EQUATIONS
• Suppose that we have a linear system of equations in the form 𝐴𝑥 = 𝑏, and suppose that we have
to solve for x
• Note: A is a n by n matrix of coefficients, x is a column vector containing the variables to be solved
and b is a column vector containing constants
• Two methods:
• Matrix inversion
• Cramer’s rule
• NB! Before using either method to solve the system, make sure that all variables that you have to
solve for are on the LHS in every equation, while everything else appears on the RHS!

SOLVING SYSTEMS OF EQUATIONS:
MATRIX INVERSION
• With matrix inversion, we can solve for a 𝑥 in 𝐴𝑥 = 𝑏 by finding the inverse of the matrix A,
• Then multiplying this with b, i
...

1

𝑥 = 𝐴−1 𝑏, where 𝐴−1 = 𝐷𝑒𝑡

𝐴

𝑎𝑑𝑗(𝐴)

CRAMER’S RULE
• Suppose that Ax = b
...
Instead of matrix inversion, we can also use
Cramer’s rule to solve for x
• According to Cramer’s rule, 𝑥𝑖 =

𝐷𝑒𝑡(𝐴𝑖 )
𝐷𝑒𝑡(𝐴)

• 𝐷𝑒𝑡(𝐴𝑖 ) is the determinant of the matrix 𝐴𝑖 , where 𝐴𝑖 is a matrix that is formed by replacing the ith column
of A with b
• 𝐴1 replace first column of A with b
• 𝐴2 replace second column of A with b
• 𝐴3 replace third column of A with b

• So, if there are two unknowns in two equations, you need three determinants (A; 𝐴1 ; 𝐴2 );
• If there are three equations in three unknowns, you need four determinants (A; 𝐴1 ; 𝐴2 ;𝐴3 )
• Do example 19
...
632

EXAMPLE 1
1
...
6𝑌, 𝐼 = 100 −
150𝑟, 𝐺 = 29
...
1𝑌 − 250𝑟
...

a) Use the goods and money market equilibrium conditions (𝑻𝑬 = 𝒚 and 𝑴𝒅 = 𝑴𝒔 ) write down the IS and LM
equations
...

c) Finally, solve for 𝒙, i
...


SOLUTION:
𝑥=

250×209
150×35
+
115
115
0
...
4×35
+ 115
115

𝑥=

500

...
06

a) In goods market equilibrium, 𝑌 = 𝑇𝐸; 𝑇𝐸 = 𝐶 + 𝐼 + 𝐺
• 𝐶 + 𝐼 + 𝐺 = 80 + 0
...
6𝑌 − 150𝑟
• 𝑌 = 209 + 0
...
4𝑌 + 150𝑟 = 209 (IS)
• In money market equilibrium, Ms = Md
• 275 = 240 + 0
...
1𝑌 − 250𝑟 = 35 (LM)
𝑌
209
0
...

35
0
...
4 −250 − 150 0
...
1

−150
=
0
...
1
0
...
4Τ
115

250Τ
115
0
...
1Τ
−0
...
06


...
4
• First, find determinant of A: 𝐴 =
0
...
4 −250 − 150 0
...


• 𝐷𝑒𝑡 𝐴1 = 209 −250 − 35 150 = −57500
• 𝐷𝑒𝑡 𝐴2 = 0
...
1 209 = −6
...

second column of 𝐴 by 𝑏 =
𝐷𝑒𝑡(𝐴)
−115
35
209
𝐴1 =
35

150

...
4 209
𝐴2 =

...
1 35

𝑟=

𝐷𝑒𝑡(𝐴2 )
𝐷𝑒𝑡(𝐴)

𝑥=

−6
...
06
...

0
...


Suppose that equilibrium conditions for three related products are

2𝑝1 = 77 − 4𝑝2 − 𝑝3
...

3𝑝3 = 48 − 2𝑝1 − 𝑝2
...
Solve for 𝑝1 , 𝑝2 and 𝑝3 using Cramer’s rule
...

4𝑝1 + 3𝑝2 + 7𝑝3 = 114
...

𝑝1
77
2 4 1
𝐴 = 4 3 7 , 𝑏 = 114 , 𝑝 = 𝑝2
𝑝3
2 1 3
48

SOLUTION
• Find the determinant of A:
• Note, A is 3 by 3, so use Sarrus’ rule
...

7
3

𝐷𝑒𝑡 𝐴1
= 77 3 3 + 4 7 48 + 1 114 1 − 1 3 48 − 4 114 3 − 77 7 1

𝐷𝑒𝑡 𝐴1 = 100
...

7
3

𝐷𝑒𝑡 𝐴2
= 2 114 3 + 77 7 2 + 1 4 48 − 1 114 2 − 77 4 3 − 2 7 48
𝐷𝑒𝑡 𝐴2 = 130
+ + +
2 4 77
𝐷𝑒𝑡 𝐴3 =
4 3 114
2 1 48
𝐷𝑒𝑡 𝐴3 = 2 3 48 + 4
2 114 1
𝐷𝑒𝑡 𝐴3 = 5

− −
2 4
4 3
2 1
114 2

−
77

...


𝑝2 =

𝐷𝑒𝑡(𝐴2 )
𝐷𝑒𝑡(𝐴)

=

130
10

= 13
...




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