For 2 x 2 Matrices
A = [ 2 -3 ] where A = [ a b ]
[ 2 1 ] [ c d ]
Find the inverse of matrix A.
Solution:
A-1 = 1/(ad-bc) [ d -b ]
[ -c a ]
= 1/(2 + 6) [ 1 3 ]
[-2 2 ]
= 1/8 [ 1 3 ]
[-2 2 ]
A-1 = [ 1/8 3/8 ]
[-1/4 1/4 ]
Friday, September 12, 2014
Friday, September 5, 2014
The Transpose of a Matrix
The transpose of a matrix is formed by writing its columns as rows.
For Example:
[ 2 3 6 ] [ 2 4 5 ]
If A = [ 4 2 1 ] , then the transposed matrix will be AT = [ 3 2 9 ]
For Example:
[ 2 3 6 ] [ 2 4 5 ]
If A = [ 4 2 1 ] , then the transposed matrix will be AT = [ 3 2 9 ]
[ 5 9 8 ] [ 6 1 8 ]
Properties of Matrices
If A and B are matrices ( with sizes such that the given matrix operations are defined ) and c is a scalar, then the following properties are true.
1. (AT)T = A Transpose of a transpose
2. (A+B)T = AT + BT Transpose of a Sum
3. (cA)T =
c(AT) Transpose of
a Scalar Multiple
4. (AB)T = BTAT Transpose of a Product
Finding the Transpose of a Product
Show (AB)T.
[ 2 7 ]
A = [ 5 8 ] B = [ 2 2 4 ]
[ 1 3 ] [ 6 1 7]
Solution:
[ 2 7 ]
AB = [ 5 8 ] x [ 2 2 4 ]
[ 1 3 ] [ 6 1 7 ]
[ (2)(2) + (7)(6) (2)(2) + (7)(1) (2)(4) + (7)(7)]
AB = [ (5)(2) + (8)(6) (5)(2) + (8)(1) (5)(4) + (8)(7)]
[ (1)(2) + (3)(6) (1)(2) + (3)(1) (1)(4) + (3)(7)]
[ 4 + 42 4 + 7 8 + 49 ]
AB = [ 10 + 48 10 + 8 20 + 56 ]
[ 2 + 18 2 + 3 4 + 21 ]
[ 46 11 57 ]
AB = [ 58 18 76 ]
[ 20 5 25 ]
[ 46 58 20 ]
ABT = [ 11 18 5 ]
[ 57 76 25 ]
Friday, August 15, 2014
Subtraction of Matrices
The subtraction of matrices is basically like
matrix addition. You just subtract them instead of adding. For example:
[3 7
5] [5 2 9]
A = [2
0 6] B = [4 7 1]
[8 9
1] [3 2 6]
Where we have to solve A - B.
The solution is simply
[ 3 - 5 7 - 2
5 - 9 ]
[ 2 - 4 0 - 7
6 - 1 ]
[ 8 - 3 9 - 2
1 - 6 ]
[ -2
5 -4 ]
= [ -2 -7
5 ]
[ 5 7 -5 ]
Like
addition, we also cannot subtract matrices that have different sizes such as
[ 7 ]
A = [
2 ] and B = [ 3 4 6 ]
[ 4 ]
Friday, August 8, 2014
Multiplication of Matrices
If A = [aij]
is an m x n matrix and B = [bij] is an n x p matrix, then the
product AB is an m x p matrix.
Sample
Problem #1
[ 1 5 2 ] [ 3 ]
A= [ 2 9 5 ] B= [ 5 ]
[ 5 2 6 ] [ 1 ]
Matrix A is measured to be 3 x 3, while Matrix B is 3 x 1.
This means that we can multiply the the matrices with each other.
This Results to
[ (1)(3) + (5)(5) + (2)(1) ]
[ (2)(3) + (9)(5) + (5)(1) ]
[ (5)(3) + (2)(5) + (6)(1) ]
[ 3 + 25 + 2 ]
= [ 6 + 45 + 5 ]
[ 15 + 10 +6]
[ 30 ]
= [ 56 ]
[ 31 ]
Sample Problem #2
[ 2 7 5 ]
A = [ 1 3 6 ] B = [ 3 1 8 ]
[ 3 2 1 ]
Since Matrix A is measured to be 3 x 3, and Matrix B is 1 x 3, we cannot multiply the matrices with each other. Therefore, it is undefined.
Friday, August 1, 2014
Addition of Matrices
If A = [ aij ] and B = [bij ] are matrices of size m x n,
then their sum is the m x n matrix given by
A + B = [aij + bij ].
The sum of two matrices of
different sizes is undefined.
Examples
a. [ -2 3 ]
+ [ 4 2 ] = [
-2 + 4, 3 + 2 ] = [ 2 5
]
[ 5 8
] [ -1 3 ]
[ 5 + (-1), 8 + 3 ] [ 4 11
]
b. [ -1 6
4 ] + [ 3 1 6 ]
= [ 2 7 10 ]
[ 4 1
5 ] [ -8 5 3
] [ -4
6 8 ]
c. [ 1 ] [ 2
] [ 3 ]
[ 7 ] +
[ 3 ] = [10]
[ 5 ]
[ 4 ] [ 9 ]
d. The sum of
[ 5
1 ] [
1 2 6 ]
A = [ 2 3 ] and
B = [ 4 3
8 ]
[ 1
4 ] [ -1
2 0 ]
is undefined. Because the
matrices of both A and B have different sizes.
Friday, July 25, 2014
Network Analysis for Electric Circuits
Determine the currents I1,I2,
and I3 for the electrical network.
There are two junctions shown, but both of them are equal, so there is no need to name them as J1 and J2.
The answer is I2 = I1+I3, when we use Network
Analysis.
After that, we will use a matrix to solve for the Amperes (I). We will combine the resistances (Rn) shown on the Amperes (In) and make them equal to the amount of their voltage (v) for determining the entries for the matrix. The answers are:
I1 – I2 + I3 = 0 Volts
4I1
+ 3I2 = 3 Volts
1I3
+ 3I2 = 4 Volts
Then we use them as entries for the matrix.
[ 1 -1 1 0 ]
[ 4 3 0 3 ] -4(R1)
+ R2 ---> R2
[ 0 3 1 4 ]
[ 1 -1 1 0 ]
[ 0 7 -4 3 ] -2(R3)
+ R2 ---> R2
[ 0 3 1 4 ]
[ 1 -1 1 0 ]
[ 0 1 -6 -5 ] -3(R2)
+ R3 ---> R3
[ 0 3 1 4 ]
[ 1 -1 1 0 ]
[ 0 1 -6 -5 ] 1/19 (R3)
[ 0 0 19 19 ]
[ 1 -1 1 0 ]
[ 0 1 -6 -5 ] R1
+ R2 ---> R1
[ 0 0 1 1 ]
[ 1 0 -5 -5 ]
[ 0 1 -6 -5 ] 5(R3)
+ R1 ---> R1
[ 0 0 1 1 ]
[ 1 0 0 0 ]
[ 0 1 -6 -5 ] 6(R3)
+ R2 ---> R2
[ 0 0 1 1 ]
[ 1 0 0 0 ]
[ 0 1 0 1 ]
[ 0 0 1 1 ]
Now we know that:
I1
= 0 Amperes
I2
= 1 Ampere
and
I3
= 1 Ampere
Friday, July 18, 2014
Network Analysis
Below is an example of a network analysis.
In this example, we assume that the circles in the upper part of the picture are 1 and 2 from left to right, while the ones below are 3 and 4 from left to right. The answers are as follows:
J1
300 = x1+x2
J2
x1+x3 = 150+x4
J3
200+x2 = x3+x5
J4
x5+x4 = 350
J is known as the junction. The ones within the junctions are the numbers on the arrows going toward the junction, while the ones not within the junction in the equation are the numbers on the arrows going away from the junction. For example, in J1, 300 is the number on the arrow going toward the junction, while x1 and x2 are the numbers on the arrows going away from
the said junction. The equation can also be J1 x1+x2 = 300, since it is also produces the same result.
Friday, July 11, 2014
Translating Large x-Values Before Curve Fitting
Find a polynomial that fits the points
(2005, 1) (2006, 2) (2007,3) (2009,4) (2010,10)
Because the given x-values are large, use the translation z = x - 2007 to obtain
(-2, 1) (-1, 2) (0, 3) (2, 4) (3, 10)
This makes the polynomial functions as
p(2) = a0 + a1(2) + a2(2)2 + a3(2)3 + a4(2)4 = a0 + 2a1 + 4a2 + 8a3 + 16a4 = 4
p(3) = a0 + a1(3) + a2(3)2 + a3(3)3 + a4(3)4 = a0 + 3a1 + 9a2 + 27a3 + 81a4 = 10
Then use the Gauss-Jordan Elimination. At the end, the solution is a0=3, a1=29/60, a2=-67/120, a3=1/15. and a4=13/120. This means that p(x) = 3 + 29/60x - 67/120x2 + 1/15x3 + 13/120x4. But because we used z to translate the x-value, the p(x) is now written as 3 + 29/60(x-2007) - 67/120(x-2007)2 + 1/15(x-2007)3 + 13/120(x-2007)4
(2005, 1) (2006, 2) (2007,3) (2009,4) (2010,10)
Because the given x-values are large, use the translation z = x - 2007 to obtain
(-2, 1) (-1, 2) (0, 3) (2, 4) (3, 10)
This makes the polynomial functions as
p(-2) = a0 + a1(-2) + a2(-2)2 + a3(-2)3
+ a4(-2)4 = a0 - 2a1 + 4a2
– 8a3 + 16a4 = 1
p(-1) = a0 + a1(-1) + a2(-1)2 + a3(-1)3
+ a4(-1)4 = a0 - a1 + a2
– a3 + a4 = 2
p(0) = a0 + a1(0) + a2(0)2 + a3(0)3 + a4(0)4 = a0 = 3p(2) = a0 + a1(2) + a2(2)2 + a3(2)3 + a4(2)4 = a0 + 2a1 + 4a2 + 8a3 + 16a4 = 4
p(3) = a0 + a1(3) + a2(3)2 + a3(3)3 + a4(3)4 = a0 + 3a1 + 9a2 + 27a3 + 81a4 = 10
Then use the Gauss-Jordan Elimination. At the end, the solution is a0=3, a1=29/60, a2=-67/120, a3=1/15. and a4=13/120. This means that p(x) = 3 + 29/60x - 67/120x2 + 1/15x3 + 13/120x4. But because we used z to translate the x-value, the p(x) is now written as 3 + 29/60(x-2007) - 67/120(x-2007)2 + 1/15(x-2007)3 + 13/120(x-2007)4
| 1 | -2 | 4 | -8 | 16 | 1 |
| 1 | -1 | 1 | -1 | 1 | 2 |
| 1 | 0 | 0 | 0 | 0 | 3 |
| 1 | 2 | 4 | 8 | 16 | 4 |
| 1 | 3 | 9 | 27 | 81 | 10 |
-1(R1) + R2 ---> R2
-1(R1) + R3 ---> R3
-1(R1) + R4 ---> R4
-1(R1) + R5 ---> R5
-2(R2) + R3 ---> R3
-4(R2) + R4 ---> R4
-5(R2) + R5 ---> R5
1/2(R3) ---> R3
-12(R3) + R4 ---> R4
-20(R3) + R5 ---> R5
1/24(R4)
-60(R4) + R5 ---> R5
1/60(R5) ---> R5
1(R5) + R4 ---> R4
-7(R5) + R3 ---> R3
15(R5) + R2 ---> R2
-16(R5) + R1 ---> R1
3(R4) + R3 ---> R3
-7(R4) + R2 ---> R2
8(R4) + R1 ---> R1
3(R3) + R2 ---> R2
-4(R3) + R1 ---> R1
-2(R2) + R1 ---> R1
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 1 | 0 | 0 | 0 | 0 | 3 |
| 1 | 2 | 4 | 8 | 16 | 4 |
| 1 | 3 | 9 | 27 | 81 | 10 |
-1(R1) + R3 ---> R3
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 2 | -4 | 8 | -16 | 2 |
| 1 | 2 | 4 | 8 | 16 | 4 |
| 1 | 3 | 9 | 27 | 81 | 10 |
-1(R1) + R4 ---> R4
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 2 | -4 | 8 | -16 | 2 |
| 0 | 4 | 0 | 16 | 0 | 3 |
| 1 | 3 | 9 | 27 | 81 | 10 |
-1(R1) + R5 ---> R5
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 2 | -4 | 8 | -16 | 2 |
| 0 | 4 | 0 | 16 | 0 | 3 |
| 0 | 5 | 5 | 35 | 65 | 9 |
-2(R2) + R3 ---> R3
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 2 | -6 | 14 | 0 |
| 0 | 4 | 0 | 16 | 0 | 3 |
| 0 | 5 | 5 | 35 | 65 | 9 |
-4(R2) + R4 ---> R4
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 2 | -6 | 14 | 0 |
| 0 | 0 | 12 | -12 | 60 | -1 |
| 0 | 5 | 5 | 35 | 65 | 9 |
-5(R2) + R5 ---> R5
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 2 | -6 | 14 | 0 |
| 0 | 0 | 12 | -12 | 60 | -1 |
| 0 | 0 | 20 | 0 | 140 | 4 |
1/2(R3) ---> R3
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 1 | -3 | 7 | 0 |
| 0 | 0 | 12 | -12 | 60 | -1 |
| 0 | 0 | 20 | 0 | 140 | 4 |
-12(R3) + R4 ---> R4
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 1 | -3 | 7 | 0 |
| 0 | 0 | 0 | 24 | -24 | -1 |
| 0 | 0 | 20 | 0 | 140 | 4 |
-20(R3) + R5 ---> R5
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 1 | -3 | 7 | 0 |
| 0 | 0 | 0 | 24 | -24 | -1 |
| 0 | 0 | 0 | 60 | 0 | 4 |
1/24(R4)
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 1 | -3 | 7 | 0 |
| 0 | 0 | 0 | 1 | -1 | -1/24 |
| 0 | 0 | 0 | 60 | 0 | 4 |
-60(R4) + R5 ---> R5
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 1 | -3 | 7 | 0 |
| 0 | 0 | 0 | 1 | -1 | -1/24 |
| 0 | 0 | 0 | 0 | 60 | 13/2 |
1/60(R5) ---> R5
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 1 | -3 | 7 | 0 |
| 0 | 0 | 0 | 1 | -1 | -1/24 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
1(R5) + R4 ---> R4
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 1 | -3 | 7 | 0 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
-7(R5) + R3 ---> R3
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | -15 | 1 |
| 0 | 0 | 1 | -3 | 0 | -91/120 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
15(R5) + R2 ---> R2
| 1 | -2 | 4 | -8 | 16 | 1 |
| 0 | 1 | -3 | 7 | 0 | 21/8 |
| 0 | 0 | 1 | -3 | 0 | -91/120 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
-16(R5) + R1 ---> R1
| 1 | -2 | 4 | -8 | 0 | -11/15 |
| 0 | 1 | -3 | 7 | 0 | 21/8 |
| 0 | 0 | 1 | -3 | 0 | -91/120 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
3(R4) + R3 ---> R3
| 1 | -2 | 4 | -8 | 0 | -11/15 |
| 0 | 1 | -3 | 7 | 0 | 21/8 |
| 0 | 0 | 1 | 0 | 0 | -67/120 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
-7(R4) + R2 ---> R2
| 1 | -2 | 4 | -8 | 0 | -11/15 |
| 0 | 1 | -3 | 0 | 0 | 259/120 |
| 0 | 0 | 1 | 0 | 0 | -67/120 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
8(R4) + R1 ---> R1
| 1 | -2 | 4 | 0 | 0 | -1/5 |
| 0 | 1 | -3 | 0 | 0 | 259/120 |
| 0 | 0 | 1 | 0 | 0 | -67/120 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
3(R3) + R2 ---> R2
| 1 | -2 | 4 | 0 | 0 | -1/5 |
| 0 | 1 | 0 | 0 | 0 | 29/60 |
| 0 | 0 | 1 | 0 | 0 | -67/120 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
-4(R3) + R1 ---> R1
| 1 | -2 | 0 | 0 | 0 | 61/30 |
| 0 | 1 | 0 | 0 | 0 | 29/60 |
| 0 | 0 | 1 | 0 | 0 | -67/120 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
-2(R2) + R1 ---> R1
| 1 | 0 | 0 | 0 | 0 | 3 |
| 0 | 1 | 0 | 0 | 0 | 29/60 |
| 0 | 0 | 1 | 0 | 0 | -67/120 |
| 0 | 0 | 0 | 1 | 0 | 1/15 |
| 0 | 0 | 0 | 0 | 1 | 13/120 |
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