Chapter  2    Review  of  Matrix  Algebra  

2-1 Matrix Algebra In this course, we need to solve system of linear equations in the form

nnnnnn

nn

nn

bxaxaxa

bxaxaxabxaxaxa

=+++

=+++

=+++

...............................................

......

2211

22222121

11212111

where x1, x2, …, xn are the unknowns.

(2-1)

Eqn. (2-1) can be written in a matrix form as

[ ]{ } { }bxA = (2-2)

where [A] is a (n x n) square matrix, {x} and {b} are (n x 1) vectors.

… (2-3)

Note:

Element located at ith row and jth column of matrix [A] is denoted by aij. For example, element at the 2nd row and 2nd column is a22.

[ ] { } { }

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nnnnnn

n

n

b

bb

x

xx

aaa

aaaaaa

:b and ,

:x ,

...::::

...

...

A 2

1

2

1

21

22221

11211

The square matrix [A] and the {x} and {b} vectors are given by,

2-2 Matrix Multiplication The product of matrix [A] of size (m x n) and matrix [B] of size (n x p) will results in matrix [C], with size (m x p).

Note: The (ij)th component of [C], i.e. cij, is obtained by taking the DOT product,

(2-4)

])[ ofcolumn th ( ])[ of rowth ( BjAicij ⋅= (2-5)

Example:

[ ] [ ] [ ] ) x ( ) x ( ) x (

pmpnnm

CBA =

2) x (2 2) x (3 )3 x 2(

710-157

3025

41

120312

⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎥⎦

⎤⎢⎣

⎡

−

2-3 Matrix Transposition If matrix [A] = [aij], then transpose of [A], denoted by [A]T, is given by [A]T = [aji]. Thus, the rows of [A] becomes the columns of [A]T.

Note: In general, if [A] is of dimension (m x n), then [A]T has the dimension of (n x m).

Example:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=

24326051

][A

Then, ⎥⎦

⎤⎢⎣

⎡

−

−=

23654201

][ TA

2-4 Transpose of a Product The transpose of a product of matrices is given by the product of the transposes of each matrices, in reverse order, i.e.

TTTT ABCCBA ][][][])][][([ = (2-6)

2-5 Determinant of a Matrix Consider a 2 x 2 square matrix [x],

[ ] ⎥⎦

⎤⎢⎣

⎡=

2221

1211

xxxx

x

(2-7)

The determinant of this matrix is give by, [ ] 12212211 xxxxxdet −=

EXERCISE Given that:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

321

}{ 432321654

][

1098765

][ 4321

][

ED

CA

1. Find the product for the following cases:

a) [A][C] b) [D]{E} c) [C]T[A][C]

2. Find the determinant of [A][A].

2-6 Solution of System of Linear Equations System of linear algebraic equations can be solved for the unknown using the following methods: a)  Cramer’s Rule b)  Inversion of Coefficient Matrix c)  Gaussian Elimination** d)  Gauss-Seidel Iteration

Example: Solve the following SLEs using Gaussian elimination.

(iii) 6 122(ii) 8324(i) 11312

321

321

321

−=−+−

=+−

=−+

xxxxxxxxx

Eliminate x1 from eq.(ii) and eq.(iii). Multiply eq.(ii) by 0.5 we get,

Subtract eq.(ii)* from eq.(i), we obtain

Add eq.(iii) with eq.(i), yields

(iii)6 122**(ii)75.420

(i)11312

321

321

321

−=−+−

=−+

=−+

xxxxxx

xxx

(iii)6 122*(ii)45.112

(i)11312

321

321

321

−=−+−

=+−

=−+

xxxxxxxxx

*iii)(5 430**(ii)75.420

(i)11312

321

321

321

=−+

=−+

=−+

xxxxxx

xxx

Eliminate x2 from eq.(iii)*. Multiply eq.(ii)** by 3 and eq.(iii)* by 2 we get

Subtract eq.(iii)** from eq.(ii)***, we obtain

***(iii)11 5.500**(ii)75.420

(i)11312

321

321

321

=−+

=−+

=−+

xxxxxx

xxx

From eq.(iii)*** we determine the value of x3, i.e.

25.5

113 −=

−=x

**(iii)10 860***(ii)215.1360

(i)11312

321

321

321

=−+

=−+

=−+

xxxxxx

xxx

Back substitute value of x3 into eq.(ii)** and solve for x2, we get

12

)2(5.472 −=

−+=x

Back substitute value of x2 and x3 into eq.(i) and solve for x1, we get

31 =x

EXERCISE Solve the following systems of linear equations by using the Gaussian elimination method.

6222832411312

3401242223

321

321

321

321

321

321

−=−+−

=+−

=−+

=++

=+−

=−+−

xxxxxxxxx

xxxxxxxxxa)

b)

Solve the following systems of linear equations by using the Gaussian elimination method.

92431354222212432

4321

4321

4321

4321

=+−+

−=−−+

−=−+−

=+−+

xxxxxxxxxxxxxxxx

ASSIGNMENT 1