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**Unformatted text preview: ** Matrices: Definition & Equality of Matrices Addition and Subtraction of Matrices Multiplication of Matrices Identitiy Matrices Determinant of Matrices Inverse of Matrix Solving System Equations: Inverse of Matrices Method & Cramer’s Rule Method 1 A matrix is an ordered rectangular array of numbers.
A matrix with m rows and n columns has size m ☓ n.
The entry in the ith row and jth column is denoted
by aij. 2 Definition of Matrices
A rectangular array of numbers
written within brackets 1 3 7 A 5
0 4 2 row: 3 column
2 x 3 matrix 5 0
H 2 3 4 11 1 6 2 8 0 1 4 row: 3 column
4 x 3 matrix Definition of Matrices
n x n = Square Matrix of order n 0.5 0.2 1.0 0.0 0.3 0.5 0.7 0.0 0.2 Square Matrix of
order 3
3 x 3 matrix 3 2 1 0 Column Matrix
4 x 1 matrix 2 1
2 2
0 3 Row Matrix
1 x 4 matrix Definition of Matrices
Notation: aij; i = row and j = column 1 3 7 A 5
0 4 a12 = “a sub one two” a11 = 1 a12 = -3 a13 = 7 a21 = 5 a22 = 0 a23 = -4 Two matrices are equal if they have the same size and their corresponding
entries are equal. 6 Solve the following matrix equation for x, y, and z: x
3 1 4 z 1 2 y 1 2 2 1 2 Solution
Since the corresponding elements of the two
matrices must be equal, we find that x = 4, z
= 3, and y – 1 = 1, or y = 2. 7 If A and B are two matrices of the same size, then: 1. The sum A + B is the matrix obtained by adding the corresponding entries in the two matrices.
2. The difference A – B is the matrix obtained by
subtracting the corresponding entries in B from
those in A. 8 If A, B, and C are matrices of the same size, then
1. A + B = B + A Commutative law
2. (A + B) + C = A + (B + C) Associative law 9 The total output of Acrosonic for May is
Model A Model B Model C Model D Location I 320 280 460 280 Location II 480 360 580 0 Location III 540 420 200 880 The total output of Acrosonic for June is
Model A Model B Model C Model D Location I 210 180 330 180 Location II 400 300 450 40 Location III 420 280 180 740 Find the total output of the company for May and June. 10 Solution
Expressing the output for May and June as
matrices: The total output of Acrosonic for May is 320
A 480 540 280
360
420 460 280
580
0 200 880 The total output of Acrosonic for June is 210
B 400 420 180 330 180
300 450 40 280 180 740 11 Solution
The total output of the company for May and
June is given by the matrix 320
A B 480 540 280
360
420 530 880 960 460
660
700 460 280 210
580
0 400 200 880 420
790 460
1030
40 380 1620 180
300
280 330 180 450 40 180 740 12 The Acrosonic Company manufactures four different loudspeaker systems at three
separate locations.
The company’s May output is as follows:
Model A Model B Model C Model D
Location I 320 280 460 280 Location II 480 360 580 0 Location III 540 420 200 880 If we agree to preserve the relative location 460we
280 320 summarize
of each entry in
the280
table,
can
360 580
0
the set of data 480
as follows: 540 420 200 880 13 We have Acrosonic’s May output expressed as a matrix: 320 280 460 280 P 480 360 580
0 540 420 200 880 a. What is the size of the matrix P? Solution
Matrix P has three rows and four columns
and hence has size 3 ☓ 4.
14 We have Acrosonic’s May output expressed as a matrix: 320 280 460 280 P 480 360 580
0 540 420 200 880 b. Find a24 (the entry in row 2 and column 4 of the matrix P) and give an interpretation of
this number.
Solution
The required entry lies in row 2 and column
4, and is the number 0. This means that no
model D loudspeaker system was
manufactured at location II in May.
15 We have Acrosonic’s May output expressed as a matrix: 320 280 460 280 P 480 360 580
0 540 420 200 880 c. Find the sum of the entries that make up row 1 of P and interpret the result.
Solution
The required sum is given by
320 + 280 + 460 + 280 = 1340
which gives the total number of loudspeaker
systems manufactured at location I in May as
1340 units. 16 We have Acrosonic’s May output expressed as a matrix: 320 280 460 280 P 480 360 580
0 540 420 200 880 d. Find the sum of the entries that make up column 4 of P and interpret the result.
Solution
The required sum is given by
280 + 0 + 880 = 1160
giving the output of Model D loudspeaker
systems at all locations in May as 1160 units.
17 If A is an m ☓ n matrix with elements aij, then the transpose of A is the n
m matrix AT with elements aji. ☓ 18 Find the transpose of the matrix 1
A 4 7 2
5
8 3
6 9 Solution
The transpose of the matrix A is
1
AT 2 3 4
5
6 7
8 9 19 If A is a matrix and c is a real number, then the scalar product cA
is the matrix obtained by
multiplying each entry of A by c. 20 Given 3 4
A 1
2 and 3 2
B 1
2 find the matrix X that satisfies 2X + B = 3A
Solution
2 X B 3 A
2 X 3 A B 3
3 1 4 2 3 1 2
2 9 12 3 2 6 10 2
4 3
6 1
2 1 6 10 3 5 X 2 2 4 1 2 21 The management of Acrosonic has decided to increase its July production of loudspeaker
systems by 10%
(over June
output).
Find a matrix giving the targeted production
for July.
Solution
We have seen that
180 330 180total
output for 210 Acrosonic’s 400 300 450
June may be
represented
by the
B
40 matrix 420 280 180 740 22 The management of Acrosonic has decided to increase its July production of loudspeaker
systems by 10%
(over June
output).
Find a matrix giving the targeted production
for July.
Solution 210 180 330 180
The required matrix
is given by (1.1) B 1.1 400 300 450 40 420 280 180 740 231 440 462 198
330
308 363 198
495 44 198 814 23 If we have a row matrix of sizeA 1
☓[an,
a2
1 a3 an ] b1 b And a column matrix of size n ☓ 1, 2 B b3 bn Then we may define the matrix product of A and B, written AB, by AB [a1 a2 a3 b1 b 2 an ] b3 a1b1 a2b2 a3b3 anbn 24 bn Let A [1 2 3 5] and 2 3
B 0 1 Find the matrix product AB. Solution 2 3
AB [1 2 3 5] (1)(2) ( 2)(3) (3)(0) (5)( 1) 9 0 1
25 Note from the last example that for the multiplication to be feasible, the number of columns of the row matrix A
must be equal to the number of rows of the column matrix
B. From last example, note that the product matrix AB has size 1 ☓ 1. This has to do with the fact that we are multiplying a row
matrix with a column matrix. Size of AB
(1 ☓ 1) We can establish the dimensions of a product matrix schematically: Size of A (1 ☓ n) (n ☓ 1) Size of B Same 26 More generally, if A is a matrix of size m ☓ n
and B is a matrix of size n ☓ p, then the matrix product of A and B, AB, is defined and is a
matrix of size m ☓ p.
Schematically:
Size of AB
(m ☓ p) Size of A (m ☓ n) (n ☓ p) Size of B Same The number of columns of A must be the same as the number of rows of B for the
multiplication to be feasible. 27 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a
a
a
23 21 22 b11
B b21 b31 b12 b13
b22 b23
b32 b33 b14 b24 b34 From the schematic Size of A (2 ☓ 3) Size of AB
(2 ☓ 4) (3 ☓ 4) Size of B Same we see that the matrix product C = AB is feasible
(since the number of columns of A equals the number
of rows of B) and has size 2 ☓ 4.
28 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a
a
a
23 21 22 Thus, b11
B b21 b31 b12 b13
b22 b23
b32 b33 b14 b24 b34 c11 c12 c13 c14 C c
c
c
c 21 22 23 24 To see how to calculate the entries of C consider entry c11: c11 [a11 a12 b11 a13 ] b21 a11b11 a12b21 a13b31 b31 29 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a
a
a
23 21 22 Thus, b11
B b21 b31 b12 b13
b22 b23
b32 b33 b14 b24 b34 c11 c12 c13 c14 C c
c
c
c 21 22 23 24 Now consider calculating the entry c12: c12 [a11 a12 b12 a13 ] b22 a11b12 a12b22 a13b32 b32 30 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a
a
a
23 21 22 Thus, b11
B b21 b31 b12 b13
b22 b23
b32 b33 b14 b24 b34 c11 c12 c13 c14 C c
c
c
c 21 22 23 24 Now consider calculating the entry c21: c21 [a21 a22 b11 a23 ] b21 a21b11 a22b21 a23b31 b31 31 To see how to compute the product of a 2 ☓ 3 matrix A and a 3 ☓ 4 matrix B, suppose a11 a12 a13 A a
a
a
23 21 22 Thus, b11
B b21 b31 b12 b13
b22 b23
b32 b33 b14 b24 b34 c11 c12 c13 c14 C c
c
c
c 21 22 23 24 Other entries are computed in a similar manner. 32 Let 3 1 4
A 1
2
3 1 3 3
B 4 1 2 2 4 1 Compute AB. Solution
Since the number of columns of A is equal to
the number of rows of B, the matrix product
C = AB is defined.
The size of C is 2 ☓ 3. 33 3 1 4
A 1
2
3 Let 1 3 3
B 4 1 2 2 4 1 Compute AB. Solution
Thus, 1 3 3 c11 c12 c13 3 1 4 C AB 4 1 2 c
c
c 1
2
3 2 4 1
21
22
23 Calculate all entries for C: 1
c11 [3 1 4] 4 (3)(1) (1)(4) (4)(2) 15 2 34 3 1 4
A 1
2
3 Let 1 3 3
B 4 1 2 2 4 1 Compute AB. Solution
Thus, 1 3 3 15 c12 c13 3 1 4 C AB 4 1 2 c
c
c 1
2
3 2 4 1
21
22
23 Calculate all entries for C: 3
c12 [3 1 4] 1 (3)(3) (1)( 1) (4)(4) 24 4 35 3 1 4
A 1
2
3 Let 1 3 3
B 4 1 2 2 4 1 Compute AB. Solution
Thus, 1 3 3 15 24 c13 3 1 4 C AB 4 1 2 c
c
c 1
2
3 2 4 1
21
22
23 Calculate all entries for C: 3
c13 [3 1 4] 2 (3)( 3) (1)(2) (4)(1) 3 1 36 3 1 4
A 1
2
3 Let 1 3 3
B 4 1 2 2 4 1 Compute AB. Solution
Thus, 1 3 3 15 24 3 3 1 4 C AB 4 1 2 c
c
c 1
2
3 2 4 1
21
22
23 Calculate all entries for C: 1
c21 [ 1 2 3] 4 ( 1)(1) (2)(4) (3)(2) 13 2 37 3 1 4
A 1
2
3 Let 1 3 3
B 4 1 2 2 4 1 Compute AB. Solution
Thus, 1 3 3 15 24 3 3 1 4 C AB 4 1 2 13
c
c 1
2
3 2 4 1
22
23 Calculate all entries for C: 3
c22 [ 1 2 3] 1 ( 1)(3) (2)( 1) (3)(4) 7 4 38 3 1 4
A 1
2
3 Let 1 3 3
B 4 1 2 2 4 1 Compute AB. Solution
Thus, 1 3 3 15 24 3 3 1 4 C AB 4 1 2 13
7
c 1
2
3 2 4 1
23 Calculate all entries for C: 3
c23 [ 1 2 3] 2 ( 1)( 3) (2)(2) (3)(1) 10 1 39 Let 3 1 4
A 1
2
3 1 3 3
B 4 1 2 2 4 1 Compute AB. Solution
Thus, 1 3 3 3 1 4 15 24 3 C AB 4 1 2 1
2
3
13
7
10 2 4 1 40 If the products and sums are defined for the matrices A, B, and C, then
1. (AB)C = A(BC) Associative law
2. A(B + C) = AB + AC
Distributive law 41 Definition of Matrices
Diagonal Matrix
A square matrix where all elements are
zero except elements in the principal
diagonal a11 0 : 0 0
a22
:
0 .. 0 .. 0 :
: 0 anm Definition of Matrices
Identity Matrix
A diagonal matrix with all elements in the
principal are 1. Denoted by I 1 0 0 1 1 0 0 0 1 0 0 0 1 The identity matrix of size n is given by 1
0
I n 0 0
1 0 0
0 n rows 1 n columns 44 The identity matrix has the properties that In A = A for any n ☓ r matrix A. BIn = B for any s ☓ n matrix B. In particular, if A is a square matrix of size n, Ithen
n A AI n A 45 Let 1 3 1
A 4 3 2 1 0 1 Then 1 0 0 1 3 1 1 3 1 I 3 A 0 1 0 4 3 2 4 3 2 A 0 0 1 1 0 1 1 0 1 1 3 1 1 0 0 1 3 1 AI 3 4 3 2 0 1 0 4 3 2 A 1 0 1 0 0 1 1 0 1 So, I3A = AI3 = A. 46 Determinant of Matrices
Determinant of 2 x 2 Matrix
Second Order Determinant a11
A a21 a12 a11 a22 a21a12 a22 a
c b
ad bc d 1 2 Det ( A) ( 1)( 4) (3)(2) 4 6 2 3 4 Minor of Matrix
Determinant of matrix obtained by deleting the ith row and jth
row column of A
2 3 1 A 4 6 5 9 8 7 Minor of element A21 Minor of element A23 2 3 1
A 4 6 5
9 8 7
3 1 8 7 3(7) 8(1) 13 2 3 1
A 4 6 5
9 8 7
2 3 9 8 2(8) 9(3) 11 Cofactor Matrix
Cofactor
Cofactor
Theminor
minor of
of aaparticular
particular element
element
The
togetherwith
with its
itsassociated
associatedsign
sign
together a13 = + a32 = - Cofactor Matrix 2 3 1
A 4 6 5
9 8 7 Cofactor of A21 3 1 8 7 [3(7) 8(1)] 13 Cofactor of A23 2 3
9 8 [2(8) 9(3)] 11 Determinants of Matrices
Third-Order Determinant (I) –
Cofactor Method a22
A a11
a32 a11
A a21 a12
a22 a13
a23 a31 a32 a33 a23
a21 a12
a33
a31 a23
a21 a22 a13
a33
a31 a32 Determinants of Matrices
Third-Order Determinant (I) – Cofactor
Method 2 1 3
4 0 1
1 2 3 EXAMPLE
4
2 0 1
2 3 1 4 1
1 3 ( 3) 4 0
1 2 2[ 0(3) 2(1) ] 1[ 4(3) 1(1) ] 3[ 4(2) 1(0) ]
2( 2) 1(11) 3(8) 4 11 24 39 Determinants of Matrices
Third-Order Determinant (II) –
Expansion Method a11
a 21 a31
P4 a12
a 22
a32
P5 a13 a11 a12 a 23 a 21 a 22 a33 a31 a32 P6 P1 P2 P3 ( P1 P2 P3 ) ( P4 P5 P6 )
P1 P2 P3 P4 P5 P6
a11a22 a33 a12 a23a31 a13a21a32 a13a22 a31 a11a23a32 a12 a21a33 Determinants of Matrices
Third-Order Determinant (II) –
Expansion Method 2 1 3 EXAMPLE
5
2 1 3
4 0 1 2
4 1
0 1 2 1 2 3 4 0 1 1 2 3 [2(0)(3) 1(1)(1) ( 3)(4)(2)] [1(0)( 3) 2(1)(2) 3(4)(1)]
[0 1 24] [0 4 12] 23 16 39 Evaluate the determinant of the following:
a)Cofactor method EXAMPLE
6 b)Expansion method 6 1 2 A 5
3 7 4 2 1 2 A 5 4 6
3 2 1 7
1 Cofactor Method EXAMPLE
6
SOLUTION 2 3 7 2 1 6 5 7 4 1 ( 1) 5 3 4 2 2[ 3(1) ( 7)( 2) ] 6[ 5(1) ( 7)( 4) ] ( 1)[ 5( 2) (3)( 4) ]
2( 11) 6( 23) ( 1)( 2) 22 138 2
114 Expansion Method
2
6 1 2
6
5
3 7 5
3 4 2 1 4 2
[2(3)(1) 6( 7)( 4) ( 1)(5)( 2)] [ 1(3)( 4) 2( 7)( 2) 6(5)(1)]
[6 168 10] [12 28 30]
184 70
114 Matrix Equation
Linear equations can be represented by
using matrix multiplication x1 4 x2 2 x3 4
2 x1 3 x2 x3 3 A system of linear equations can be expressed in the form of an equation of
matrices. Consider
2 x the
4 y system
z 6 3x 6 y 5z 1
x 3y 7z 0 The coefficients on the left-hand side of the equation can be expressed as matrix A below,
the variables as matrix X, and the constants 4 of
1 the equation
6 2 side x
on right-hand
as matrix
A 3 6 5
X y B 1
B: 1 3 7 z 0 58 A system of linear equations can be expressed in the form of an equation of
matrices. Consider
2 x the
4 y system
z 6 3x 6 y 5z 1
x 3y 7z 0 The matrix representation of the system of linear equations
4
1 by 2is given x AX 6= B, or 3 6 5 y 1 1 3 7 z 0 59 A system of linear equations can be expressed in the form of an equation of matrices. Consider the
system
2x 4 y z 6 3x 6 y 5z 1 x 3y 7z 0
To confirm this, we can multiply the two matrices on the left-hand side of the equation, obtaining 2 x 4 y z 6 3x 6 y 5z 1 x 3 y 7 z 0
which, by matrix equality, is easily seen to be
equivalent to the given system of linear
equations. 60 Let A be a square matrix of size n.
A square matrix A–1 of size n such that A 1 A AA 1 I n is called the inverse of A.
Not every matrix has an inverse. A square matrix that has an inverse is said to be nonsingular. A square matrix that does not have
an inverse is said to be singular.
61 Inverse Matrix [2x2]
Step Details 1 Find the determinant of the matrix. 2 Find the adjoint matrix. 3 Find inverse matrix: 1
A Adjoin t
A
1 Inverse Matrix
a
A c b
d d ad bc
1
A c ad bc b ad bc 1 d
a A c ad bc Determinant b
a Adjoint Matrix A ad bc; provided A 0
• If A= 0, it is impossible to calculate the inverse, because we cannot divide by zero.
• If the matrix has zero determinant it is said to be non-singular (inverse
does not exist), otherwise it is said to be singular (inverse exist). Inverse Matrix [2x2] 3 1 A 4 2 EXAMPLE
7 A 3(2) 4(1) 6 4 2 d b
1
ad bc c a 1 1 1 2 1 1
2 A 2 4 3 2 11 2 Inverse Matrix [2x2] EXAMPLE
8 3 0
A 2
1 2 3
B 1
0 6 3
C 2
1 Given the above A, B and C matrix, find the inverse for
each matrix of the above. 3
A 2 0
1 Step 1: Find the determinant of the matrix. EXAMPLE
8
SOLUTION 3
A 2 0
(3)(1) ( 2)(0) 3 0 3 1 Step 2: Find the adjoint matrix. 3
Adjoin t of A 2 0 1 1 2 Step 3: Find inverse matrix
A 1 A 1 1
Adjoin t
A 1 1 3 2 0 1 3 3 2
3 0 1 0
3 1 2 0
3 2
B 1 3
0 Step 1: Find the determinant of the matrix. EXAMPLE
8
SOLUTION 2
B 1 3
( 2)(0) (1)(3) 0 3 3 0 Step 2: Find the adjoint matrix. 2 3
Adjoin t of B 1 0 0 3 1 2 Step 3: Find inverse matrix 1
B Adjoin t
B
1 1 1 0 3 0
B 1 2 1
2
3 3
3 1 6 3
C 2
1 EXAMPLE
8
SOLUTION Step 1: Find the determinant of the matrix. 6 3
C (6)(1) ( 2)(3) 6 6 0 2 1
Answer: Unable to find the inverse matrix as the
determinant is 0 1 2
The matrixA 3
4 1 2
A 3
has a matrix
1 2 2
1 as its inverse.
This can be demonstrated by multiplying
them:
1 1 0 1 2 2
AA I 3
1 3 4 2 2 0 1
1 1 1 2 1 0 2
A A 3 I 1 2 2 3 4 0 1
1 69 0 1
The matrixB 0
0 does not have an inverse.
a b
where c d 1
If B had an inverse given
B by a, b, c, and d are some appropriate numbers,
then by definition of an inverse we would have
BB–1 = I. 0 1 a b 1 0
That is 0 0 c d 0 1 c d 1 0 0 0 0 1 implying that 0 = 1, which is impossible! 70 Finding the Inverse of a Square Matrix
Inverse Matrix [3x3]
Step Details 1 Find the determinant of the matrix. 2 Find the minors of the matrix. 3 Find the cofactor of the matrix: 4 Find the adjoint matrix. 5 Find the inverse matrix. Inverse Matrix [3x3]
Find the inverse matrix of the following. EXAMPLE
9 2 1 3 A 0 3 1 1 0 2 Find the inverse matrix of the following. 2 1 3 A 0 3 1 1 0 2 EXAMPLE
9
SOLUTION Step 1: Find the determinant of the matrix. 2 1 3 2 1
A0 3 1 0 3 1 0 2 1 0 [( 2)(3)( 2) ( 1)(1)(1) (3)(0)(0)] [(1)(3)(3) (0)(1)( 2) (2)(0)( 1)]
[12 1 0] [9 0 0]
11 9
2 Step 2: Find the minor of the matrix. EXAMPLE
9 2 1 3
3 1
M 11 0 3 1 6 0 6
0 2
1 0 2
2 1 3
1 3
M 21 0 3 1 ( 2 0) 2
0 2
1 0 2
2 1 3
1 3
M 31 0 3 1 1 9 10
3 1
1 0 2 SOLUTION
M 12 2
0
1 M 22 2
0
1 M 32 2
0
1 1 3
0
3 1 1
0 2 1
(0 1) 1
2 1 3
2
3 1 1
0 2 3
4 3 1
2 1 3
2 3
3 1 (2 0) 2
0 1
0 2 M 13 2 1 3
0 3
0 3 1 0 3 3
1 0
1 0 2 M 23 2 1 3
2 1
0 3 1 (0 1) 1
1 0
1 0 2 M 33 2 1 3
2 1
0 3 1 6 0 6
0 3
1 0 2 The minor of the matrix 1 3 6 1 2 1 10 2 6 Step 3: Find the cofactor of the matrix. EXAMPLE
9
SOLUTION 1 3 6 1 3 6 1 2
1 1
2 1 10 2 10 2 6 6 The minor of the matrix Adjoint Matrix a11 a12 A a21 a22
a 31 a32 a13 a23 a33 11 12 13 21 22 23 32
33 31
Matrix of Cofactor
T 11 12 13 11 21 31 adj A 21 22 23 12 22 32 32
33 23
33 31 13 Step 4: Find the adjoint matrix. EXAMPLE
9
SOLUTION 6 Cofactor matrix 2 10 1
1 2 6 Adj A 2 10 T 1
1 2 3 1
6 Step 5: Find the inverse matrix. 3 1
6 6 1 3 2
1
1 A 1 2 10 3 6 1
1
A 1
1 2 1
2 2 3 1 6 3 2 10 2
6 1
adj A
A 1 5 1 1
2 1
3 2 If AX = B is a linear system of n equations in n unknowns and if A–1
exists, then
X = A–1B
is the unique solution of the system. 78 Solving Equation
(Inverse Matrix [2x2])
Step Details 1 Convert to matrix equation. 2 Find the determinant of the matrix. 3 Find the adjoint matrix. 4 Find inverse matrix: 5 Solve the equation. 1
A Adjoin t
A
1 Using Inverse to Solve a System
x1 2 x2 5
3 x1 7 x2 18 EXAMPLE
Step 1: Convert to matrix equation. 10 1 2 x1 5 3 7 x 18 2 A X B Ax = B Step 2: Find the determinant of the matrix. A EXAMPLE
10 1 2
3 7 (1)(7) (3)(2) 7 6 1 Step 3: Find the adjoint matrix. 1 2 7 2
Adjoin t of A 3
7 3
1 Step 4: Find inverse matrix A 1 EXAMPLE
10 1
Adjoin t
A 1 7 2 7 2
A 1 3 1 3 1 1 Step 5: Solve the matrix equation. x=A B
-1 1
3 2 x1 5 7 x 2 18 A X B x1 7 2 5 35 36 1 x 3 1 18 15 18 3 2 Therefore x1 = -1 and x2 = 3 Ax = B Solving Equation
(Inverse Matrix [3x3])
Step Details 1 Convert to matrix equation. 2 Find the determinant of the matrix. 3 Find the minors of the matrix. 4 Find the cofactor of the matrix: 5 Find the adjoint matrix. 6 Find the inverse matrix. 7 Solve the equation. Solving Equation
(Inverse Matrix [3x3])
x y z 6
x y z 2 EXAMPLE
12 2 x y 3 z 6
Step 1: Convert to matrix equation. 1 1 1 x 6 1 1 1 y 2 2 1 3 z 6
A X B Ax = B Step 2: Find the determinant of the matrix. EXAMPLE
12 1 1 1 1 1
A1 1 1 1 1
2 1 3 2 1
[(1)( 1)(3) (1)(1)(2) (1)(1)( 1)] [(2)( 1)(1) ( 1)(1)(1) (3)(1)(1)]
[ 3 2 1] [ 2 1 3] 2 0 2 Step 3: Find the minor of the matrix. EXAMPL
E
12 1 1 1
1 1
M 11 1 1 1 3 ( 1) 2
1 3
2 1 3
1 1 M 13 1 1
1
2
1 1 1
M 21 1 1 1 3 ( 1) 4
1 3
2 1 3 M 23 1
2 1 1 1
1 1
M 31 1 1 1 1 ( 1) 2
1 1
2 1 3 1
1
2 M 12 1
1
2 1
1
1 1
1
1 2
3 1
3 2 1
3 M 22 1
1
2 1
1
1 1
1
1 2
3 1
3 2 1
3 M 32 1
1
2 1
1
1 1
1 1
1 1 1 0
1 1
3 M 33 1 1
1
1 1
2
1 3
1 1 1
1 1
2
1 3 1 1 ( 2) 1
1
1 1 2 3
1 1 1
1 1
1 1 1 1 2
1 1
1 3 The minor of the matrix 2 1 1 4 1 3 2 0 2 Step 4: Find the cofactor of the matrix. EXAMPLE
12 . 2 1 1 2 1 1 3 4 1 3 4 1 2 0 2 2
0 2 Step 5: Find the adjoint matrix.
T 2 1 1 2 4 2 adj A 4 1
3 1 1
0 2 1 0 2
3 2 EXAMPLE
12 2 4 2 adj A 1 1
0 1
3 2 1
A adj A
A
1 Step 6: Find the inverse matrix. 2 2 4 2 1 1 1
A 0 1 1 1 1
2 2 2 3 2 1 3 1
2
2 1
0 1 Step 7: Solve the equation. X = A-1 B EXAMPLE
12 2 1 6 6 4 ( 6) 4 x 1 y 1 2 1 0
2 3 1 0 2
2 3 z 1 2 2 1 6 3 3 6 0 Therefore x = 4, y = 2 and z = 0 Solve the system of linear equations 2x y z 1
3x 2 y z 2
2 x y 2 z 1
a)Write the system of equations in the form AX =B
b)Solve the matrices equation using inverse of
matrices method 90 Solve the system of linear equations 2x y z 1
3x 2 y z 2
2 x y 2 z 1 Solution
Write the system of equations in the form AX
= B where
2
A 3 2 1
2
1 1
1 2 x
X y z 1
B 2 1 91 Solve the system of linear equations 2x y z ...

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- Fall '19
- rita atake