Matrices III

  • Inverse of a Matrix

Unlike numbers, the division operation is not defined for matrices. A similar (but not same) operation is to find the inverse of a matrix. Note that only square matrices can have inverses. When a square matrix A is non-singular (|A| \ne 0), its inverse exists (= A^{-1}) and is unique.

Consider the number 1. When we multiply any number by 1, we get the same number and when we divide a number by 1, we get its reciprocal.

Similarly, we have identity matrix I. When we multiply any matrix by I (with appropriate order), we get the same matrix. When we multiply a matrix by its inverse, we get the identity matrix. Thus,

{4 \times \frac {1}{4} = 1 \ and \ A \times A^{-1} = I}

For a matrix A = \begin {bmatrix} 1 & 2 \\ 3 & 4 \end {bmatrix}, we have A^{-1} = \begin {bmatrix} -2 & 1 \\ \frac 3 2 & \frac {-1} 2 \end {bmatrix}. If we do A \times A^{-1}, we get \begin {bmatrix} 1 & 0 \\ 0 & 1 \end {bmatrix}, which is a 2 \times 2 identity matrix.

  • Finding the Inverse Using Row Transformations

We know

{A \times A^{-1} = I}

By some means, if we reduce A to I, LHS will become I \times A^{-1}, which is A^{-1}. The matrix on the RHS will have to undergo the same set of transformations in same order, as we did on A to get I.

The identity matrix on the right hand side will now get transformed to A^{-1}. Thus,

{I \times A^{-1} = A^{-1}}

To find inverse of a 2 \times 2 matrix using row transformations:

1) Make sure that the determinant of the matrix is non-zero.

2) Make a_{11}=1 by using transformation of the kind c \times R_{1}

3) Make all other elements in the 1st column 0 by using transformations R_2 + c R_1

4) Make a_{22}=1 by using transformation of the kind c \times R_{2}

5) Make all other elements in the 2nd column 0 by using transformations R_1 + c R_2

To find inverse of a 3 \times 3 matrix using row transformations:

1) Make sure that the determinant of the matrix is non-zero.

2) Make a_{11}=1 by using transformation of the kind c \times R_{1}

3) Make all other elements in the 1st column 0 by using transformations R_2 + c R_1, R_3 + c R_1

4) Make  a_{22}=1 by using transformation of the kind c \times R_{2}

5) Make all other elements in the 2nd column 0 by using transformations R_1 + c R_2, R_3 + c R_2

6) Make a_{33}=1 by using transformation of the kind c \times R_{3}

7) Make all other elements in the 3rd column 0 by using transformations R_1 + c R_3, R_2 + c R_3

Notice the order by which we are transforming the given matrix into identity matrix. By performing the same sequence of transformations on identity matrix on the other side, we get the inverse.

  • Finding the Inverse Using Column Transformations

To find inverse of a 2 \times 2 matrix using column transformations:

1) Make sure that the determinant of the matrix is non-zero.

2) Make a_{11}=1 by using transformation of the kind c \times C_{1}

3) Make all other elements in the 1st row 0 by using transformations C_2 + c C_1

4) Make a_{22}=1 by using transformation of the kind c \times C_{2}

5) Make all other elements in the 2nd row 0 by using transformations C_1 + c C_2

To find inverse of a 3 \times 3 matrix using column transformations:

1) Make sure that the determinant of the matrix is non-zero.

2) Make a_{11}=1 by using transformation of the kind c \times C_{1}

3) Make all other elements in the 1st row 0 by using transformations C_2 + c C_1, C_3 + c C_1

4) Make a_{22}=1 by using transformation of the kind c \times C_{2}

5) Make all other elements in the 2nd row 0 by using transformations C_1 + c C_2, C_3 + c C_2

6) Make a_{33}=1 by using transformation of the kind c \times C_{3}

7) Make all other elements in the 3rd row 0 by using transformations C_1 + c C_3, C_2 + c C_3

  • Finding the Inverse by Adjoint Method

A^{-1} is given by

{A^{-1} = \frac {1}{|A|} \ adj \ A }

The adjoint of A is a matrix obtained as follows:

1) Obtain |A|.

2) Obtain the matrix of minors. The minor of an element a_{ij} is obtained by eliminating ith row and jth column and finding the determinant of the newly formed matrix. Knowing the concept of minor is essential. We will define the rank of matrix using this.

3) Obtain the matrix of cofactors. The cofactor of an element a_{ij} is

{(-1)^{i+j} \times minor \ of \ a_{ij}}

4) Take the transpose of matrix of cofactors. Transpose of a matrix is obtained by interchanging the rows and columns. This will be the adjoint. Thus,

{A^{-1} = \frac {1}{|A|} \ adj \ A}

There is another method to find the inverse, which is using matrices in normal form. We will see this method in }.

  • Solving a System of Simultaneous Equations Using Matrices

Consider the following system of equations:

{x+y+z = 6, \ 2y+5z = -4, \ 2x + 5y -z = 27}

We want to find the values of x,y and z, which will satisfy all 3 equations simultaneously. This system of equations can be written in matrix form as

{\begin {bmatrix} 1 & 1 & 1 \\ 0 & 2 & 5 \\ 2 & 5 & -1 \end {bmatrix} \begin {bmatrix} x \\ y \\ z \end {bmatrix} = \begin {bmatrix} 6 \\ -4 \\ 27 \end {bmatrix}}

Let us call the 3 \times 3 matrix A, the matrix on the RHS B and the matrix of unknowns X. Thus, we have

{AX = B}

There are 2 methods of solving this:

  • Method 1 : Method of Inversion

In the method of inversion, we find the inverse of the matrix of the coefficients. \\

We know that A \times A^{-1} = I. On pre-multiplying both sides of AX = B by A^{-1}, we get A^{-1} \times A B = A^{-1}B, i.e. I X = A^{-1}B i.e. X = A^{-1}B.

So, all we have to do is first find the inverse of A and then pre-multiply B by $A^{-1}$.

In this example,

{A^{-1} = \frac {1}{-21} \begin {bmatrix} -27 & 6 & 3 \\ 10 & -3 & -5 \\ -4 & -3 & 2 \end {bmatrix}}

So,

{X = \begin {bmatrix} x \\ y \\ z \end {bmatrix} = \frac {1}{-21} \begin {bmatrix} -27 & 6 & 3 \\ 10 & -3 & -5 \\ -4 & -3 & 2 \end {bmatrix} \times \begin {bmatrix} 6 \\ -4 \\ 27 \end {bmatrix}= \begin {bmatrix} 5 \\ 3 \\-2 \end {bmatrix}}

Thus, x=5, y=3 and z= -2.

  • Method 2 : Method of Reduction

In this method, we reduce A to an upper-triangular matrix by performing row operations. Same operations are to be performed on matrix B in same sequence. Ultimately, we are left with following matrix:

{\begin {bmatrix} . & . & . \\ 0 & . & . \\ 0 & 0 & . \end {bmatrix} \begin {bmatrix} x \\ y \\z \end {bmatrix} = \begin {bmatrix} . \\ . \\ . \end {bmatrix}}

So, we obtain z from the last row, then we get y from the 2nd row and then, x from the 1st row.

[Note – Try it and see whether you get the same values o $x,y$ and $z$ as by inversion method.]

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