Inverse of a Matrix

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 ]


           

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 A^T =        [   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. (A^T)^T = A                    Transpose of a transpose

2. (A+B)^T = A^T + B^T           Transpose of a Sum

3. (cA)^T = c(A^T)             Transpose of a Scalar Multiple

4. (AB)^T = B^TA^T             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 ]  

AB^T = [ 11    18    5  ]

           [ 57   76    25 ]

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 ]

Multiplication of Matrices

If A = [a_ij] is an m x n matrix and B = [b_ij] 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.

Addition of Matrices

If A = [ a_ij ] and B = [b_ij ] are matrices of size m x n, then their sum is the m x n matrix given by

A + B = [a_ij + b_ij ].

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.

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 (R_n) shown on the Amperes (I_n) 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

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.