Applications Of Matrices And Determinants

In this chapter, we present some applications of matrices in solving system of linear equations. To be specific, we study four methods, namely

i)Matrix inversion method,

(ii) Cramer’s rule

(iii) Gaussian elimination method, and

(iv) Rank method.

Before knowing these methods, we introduce the following:

(i) Inverse of a non-singular square Inverse of a Matrix

The multiplicative inverse of a square matrix is called its inverse matrix. If a matrix AA has an inverse, then AA is said to be nonsingular or invertible. A singular matrix does not have an inverse.

To find the inverse of a square matrix AA , you need to find a matrix A−1A−1 such that the product of AA and A−1A−1 is the identity matrix.

In other words, for every square matrix AA which is nonsingular there exist an inverse matrix, with the property that, AA−1=A−1A=IAA−1=A−1A=I , where II is the identity matrix of the appropriate size.

Method 1:

Let AA be an n×n matrix.

Write the doubly augmented matrix [A | In] .

Apply elementary row operations to write the matrix in reduced row-echelon form.

Decide whether the matrix A is invertible (nonsingular).

If A can be reduced to the identity matrix In  , then A−1 is the matrix on the right of the transformed augmented matrix.

If A cannot be reduced to the identity matrix, then A is singular.

Method 2:

You may use the following formula when finding the inverse of n×n matrix.

If A is non-singular matrix, there exists an inverse which is given by , where | A || A | is the determinant of the matrix.

The Classical Adjoin of a Square Matrix :

Let A = [ a _ij ] be a square matrix. The transpose of the matrix whose ( i, j) entry is the a _ij cofactor is called the classical adjoint of A:

Example 1:

Find the adjoint of the matrix

Solution:

The first step is to evaluate the cofactor of every entry: 

Therefore, 

Why form the adjoint matrix?

First, verify the following calculation where the matrix A above is multiplied by its adjoint:

----------------------------------(*)

Now, since a Laplace expansion by the first column of A gives

equation (*) becomes

This result gives the following equation for the inverse of A:

Example 2:

Determine the inverse of the following matrix by first computing its adjoin:

Solution:

First, evaluate the cofactor of each entry in A: 

These computations imply that 

Now, since Laplace expansion along the first row gives 

the inverse of A is

which may be verified by checking that AA ⁻¹ = A ⁻¹ A = I.

Example 3:

If A is an invertible n by n matrix, compute the determinant of Adj A in terms of det A.

Solution:

Because A is invertible, the equation A ⁻¹ = Adj A/det A implies 

Recall that if B is n x n and k is a scalar, then det( kB) = k ⁿ det B. Applying this formula with k = det A and B = A ⁻¹ gives 

Thus,

Example 4:

Show that the adjoin of the adjoin of A is guaranteed to equal A if A is an invertible 2 by 2 matrix, but not if A is an invertible square matrix of higher order.

Solution:

First, the equation A · Adj A = (det A) I can be rewritten

which implies

Next, the equation A · Adj A = (det A) I also implies

This expression, along with the result of Example 3, transforms (*) into 

where n is the size of the square matrix A. If n = 2, then (det A) ⁿ⁻² = (det A) ⁰ = 1—since det A ≠ 0—which implies Adj (Adj A) = A, as desired. However, if n > 2, then (det A) ⁿ⁻² will not equal 1 for every nonzero value of det A, so Adj (Adj A) will not necessarily equal A. Yet this proof does show that whatever the size of the matrix, Adj (Adj A) will equal A if det A = 1.

The relation between adjoint and inverse of a matrix :

To find the inverse of a matrix A, i.e A-1 we shall first define the adjoint of a matrix. Let A be an n x n matrix. The (i,j) cofactor of A is defined to be

A_ij = (-1)^ij det(M_ij),

where M_ij is the (i,j)^th minor matrix obtained from A after removing the ith row and jth column. Let’s consider the n x n matrix A = (Aij) and define the n x n matrix Adj(A) = A^T. The matrix Adj(A) is called the adjoint of matrix A. When A is invertible, then its inverse can be obtained by the formula given below.

The inverse is defined only for non-singular square matrices. The following relationship holds between a matrix and its inverse:

AA^-1 = A^-1A = I, where I is the identity matrix

Theorems On Ad joint And Inverse Of A Matrix

Theorem 1

If A be any given square matrix of order n, then A adj(A) = adj(A) A = |A|I, where I is the identitiy matrix of order n.

Proof:

Let

Since the sum of the product of elements of a row (or a column) with corresponding cofactors is equal to |A| and otherwise zero, we have

Similarly, we can show that adj(A) A = |A| I

Hence, A adj(A) = adj(A) A

Theorem 2

If A and B are non-singular matrices of the same order, then AB and BA are also non-singular matrices of the same order.

Theorem 3

The determinant of the product matrices is equal to the product of their respective determinants, that is, |AB| = |A||B|, where A and B are square matrices of the same order.

Remark: 

We know that

Writing determinants of matrices on both sides, we have

i.e,

i.e, |adj(A)| |A| = |A|³

or |adj(A)| = |A|²

In general, if A is a quare matrix of order n, then |adj(A)| = |A|^n-1.

Theorem 4

A square matrix A is invertible if and only if A is a non-singular matrix.

Proof:

Let A be an invertible matrix of order n and I be the identity matrix of the same order. Then there exists a square matrix B of order n such that AB = BA = I. Now, AB = I.

So |A| |B| = |I| = 1 (since |I| = 1 and |AB| = |A| |B|). This gives |A| to be a non-zero value. Hence A is a non-singular matrix. Conversely, let A be a non-singular matrix, then |A| is non-zero. Now A adj(A) = adj(A) A = |A| I (Theorem 1). Or,

or AB = BA = I, where

Thus A is invertible and

Creating the Adjugate Matrix to Find the Inverse Matrix:

Properties of Inverse Matrices:
If A is nonsingular, then so is A ^-1 and (A ^-1) ^-1 = A
If A and B are nonsingular matrices, then AB is nonsingular and (AB) ^-1 = B^-1 A ^-1
If A is nonsingular then (A^T)^-1 = (A^-1)^T
If A and B are matrices with AB=I_n then A and B are inverses of each other.

Notice that the fourth property implies that if AB = I then BA = I

Let A, A₁ and A₂ be n×n matrices, the following statements are true:
1. If A^-1 = B, then A (col k of B) = e_k
2. If A has an inverse matrix, then there is only one inverse matrix.
3. If A₁ and A₂ have inverses, then A₁ A₂ has an inverse and (A₁ A₂)^-1 = A₁^-1 A₂^-1
4. If A has an inverse, then x = A^-1d is the solution of Ax = d and this is the only solution.
5. The following are equivalent:
(i)  A has an inverse.
(ii)  det (A) is not zero.
(iii)  Ax = 0 implies x = 0.

If c is any non-zero scalar then cA is invertible and (cA)^-1 = A^-1/c.
For n = 0, 1, 2…, Aⁿ is invertible and (Aⁿ)^-1 = A^-n = (A^-1)ⁿ.
If A is a square matrix and n > 0 then:
A^-n = (A^-1)ⁿ

Example 1:

Compute A^-3 for the matrix:

Solution:

 First of all, we need to find the inverse of the given matrix. The method to find the inverse is only applicable for 2 × 2 matrices.

Steps are as follows:
[1] Interchange leading diagonal elements:
-7 → 2
2 → -7


[2] Change signs of the other 2 elements:
-3 → 3
4 → -4


[3] Find the determinant |A|


[4] Multiply result of [2] by 1/|A|

Now:

Example 2:

Given the matrix A verify that the indicated matrix is in fact the inverse.
Solution:

To verify that we do in fact have the inverse we’ll need to check the condition
AA^-1 = A^-1A = I

Now we check whether AA^-1 = A^-1A = I:

Hence the required is verified.

Example 3:

Let A be the 2 × 2 matrix,

Show that A has no inverse.
Solution:

An inverse for A must be a 2 × 2 matrix.
Let such that AB = BA = I. If such a matrix B exists, it must satisfy the following equation:

The preceding equation requires that:
a + 2c = 1 and 3a + 6c = 0
which is clearly impossible, so we can conclude that A has no inverse.

Left cancellation property  & the right cancellation property :

In mathematics, the notion of cancellative is a generalization of the notion of invertible.

An element a in a magma (M, ∗) has the left cancellation property (or is left-cancellative) if for all b and c in M, a ∗ b = a ∗ c always implies that b = c.

An element a in a magma (M, ∗) has the right cancellation property (or is right-cancellative) if for all b and c in M, b ∗ a = c ∗ a always implies that b = c.

An element a in a magma (M, ∗) has the two-sided cancellation property (or is cancellative) if it is both left- and right-cancellative.

A magma (M, ∗) has the left cancellation property (or is left-cancellative) if all a in the magma are left cancellative, and similar definitions apply for the right cancellative or two-sided cancellative properties.

A left-invertible element is left-cancellative, and analogously for right and two-sided.

Application of matrices to geometry :

There is a special type of non-singular matrices which are widely used in applications of matrices to geometry. For simplicity, we consider two-dimensional analytical geometry.

Let  O be the origin, and  x 'O x  and  y 'Oy be the  x -axis and y -axis. Let P be a point in the plane whose coordinates are (x, y) with respect to the coordinate system. Suppose that we rotate the x -axis and   y -axis about the origin, through an angle θ as shown in the figure. Let X 'OX and Y 'OY be the new X -axis and new Y -axis. Let  ( X ,Y )  be the new set of  coordinates of P with respect to the new coordinate system.  Referring  to  Fig.1.1, we get

x = OL = ON − LN = X cos θ – QT =  X cos θ − Y sin θ ,

y = PL = PT + TL = QN + PT =  X sin θ + Y cos θ .

These equations provide transformation of one coordinate system into another coordinate system.

The above two equations can be written in the matrix form

Hence, we get the transformation X = x cosθ - y sinθ , Y = x sinθ + y cosθ .

This transformation is used in Computer Graphics and determined by the matrix 

We note that the matrix W satisfies a special property W ^-1 = W^T ; that is, WW ^T = W^TW =I

Definition 1.3

A square matrix A is called orthogonal if AA^T = A^TA = I.

Note

A is orthogonal if and only if A is non-singular and A⁻¹ = A^T .

Example 1.11

Prove that  is orthogonal.

Solution

Similarly, we get A^TA = I₂ . Hence AA^T = A^TA = I₂ ⇒ A is orthogonal.

Application of matrices to Cryptography :

One of the important applications of inverse of a non-singular square matrix is in cryptography. Cryptography is an art of communication between two people by keeping the information not known to others. It is based upon two factors, namely encryption and decryption. Encryption means the process of transformation of an information (plain form) into an unreadable form (coded form). On the other hand, Decryption means the transformation of the coded message back into original form. Encryption and decryption require a secret technique which is known only to the sender and the receiver.

This secret is called a key. One way of generating a key is by using a non-singular matrix to encrypt a message by the sender. The receiver decodes (decrypts) the message to retrieve the original message by using the inverse of the matrix. The matrix used for encryption is called encryption matrix (encoding matrix) and that used for decoding is called decryption matrix (decoding matrix).

We explain the process of encryption and decryption by means of an example.

Suppose that the sender and receiver consider messages in alphabets A − Z  only, both assign the numbers 1-26 to the letters A − Z respectively, and the number 0 to a blank space. For simplicity, the sender employs a key as post-multiplication by a non-singular matrix of order 3 of his own choice. The receiver uses post-multiplication by the inverse of the matrix which has been chosen by the sender.

Let the encoding matrix be

Let the message to be sent by the sender be “WELCOME”.

Since the key is taken as the operation of post-multiplication by a square matrix of order 3, the message is cut into pieces (WEL), (COM), (E), each of length 3, and converted into a sequence of row matrices of numbers:

[23 5 12],[3 15 13],[5 0 0].

Note that, we have included two zeros in the last row matrix. The reason is to get a row matrix with 5 as the first entry.

Next, we encode the message by post-multiplying each row matrix as given below:

So the encoded message is [45 − 28 −23] [46 -18 3] [ 5  −5 5]

The receiver will decode the message by the reverse key, post-multiplying by the inverse of A.

So the decoding matrix is

The receiver decodes the coded message as follows:

So, the sequence of decoded row matrices is [23 5 12] ,  [3 15 13], [5 0 0].

Thus, the receiver reads the message as “WELCOME”

Elementary Transformations of a Matrix

A matrix can be transformed to another matrix by certain operations called elementary row operations and elementary column operations.

Elementary row and column operations

Elementary row (column) operations on a matrix are as follows:

The interchanging of any two rows (columns) of the matrix

Replacing a row (column) of the matrix by a non-zero scalar multiple of the row (column) by a non-zero scalar.

Replacing a row (column) of the matrix by a sum of the row (column) with a non-zero scalar multiple of another row (column) of the matrix.

Elementary row operations and elementary column operations on a matrix are known as elementary transformations.

We use the following notations for elementary row transformations:

i.   Interchanging of i^th and j^th rows is denoted by R_i ↔ R_j .

ii. The multiplication of each element of i^th row by a non-zero constant λ is denoted by R_i → λ R_i .

iii. Addition to i^th row, a non-zero constant λ multiple of j^th row is denoted by R_i → R_i + λ R_j . Similar notations are used for elementary column transformations.

Definition 1.4

Two matrices A and B of same order are said to be equivalent to one another if one can be obtained from the other by the applications of elementary transformations. Symbolically, we write A ~ B to mean that the matrix A is equivalent to the matrix B .

For instance, let us consider a matrix 

After performing the elementary row operation R₂ → R₂ + R₁ on A , we get a matrix B in which the second row is the sum of the second row in A and the first row in A .

Thus, we get 

The above elementary row transformation is also represented as follows:

Note

An elementary transformation transforms a given matrix into another matrix which need not be equal to the given matrix.

 Row-Echelon form

Using the row elementary operations, we can transform a given non-zero matrix to a simplified form called a Row-echelon form. In a row-echelon form, we may have rows all of whose entries are zero. Such rows are called zero rows. A non-zero row is one in which at least one of the entries is not zero. For instance, in the matrix,

,

R₁ and R₂ are non-zero rows and R₃ is a zero row

Definition 1.5

A non-zero matrix E is said to be in a row-echelon form if:

i.                     All zero rows of E occur below every non-zero row of E.

ii.                     The first non-zero element in any row i of E occurs in the j^th column of E , then all other entries in the j^th column of E below the first non-zero element of row i are zeros.

iii.                     The first non-zero entry in the ith row of E lies to the left of the first non-zero entry in (i +1)^th row of E .

The following matrices are in row-echelon form:

Consider the matrix in (i). Go up row by row from the last row of the matrix. The third row is a zero row. The first non-zero entry in the second row occurs in the third column and it lies to the right of the first non-zero entry in the first row which occurs in the second column. So the matrix is in row- echelon form.

Consider the matrix in (ii). Go up row by row from the last row of the matrix. All the rows are non-zero rows. The first non-zero entry in the third row occurs in the fourth column and it occurs to the right of the first non-zero entry in the second row which occurs in the third column. The first non-zero entry in the second row occurs in the third column and it occurs to the right of the first non-zero entry in the first row which occurs in the first column. So the matrix is in row-echelon form.

The following matrices are not in row-echelon form:

Consider the matrix in (i). In this matrix, the first non-zero entry in the third row occurs in the second column and it is on the left of the first non-zero entry in the second row which occurs in the third column. So the matrix is not in row-echelon form.

Consider the matrix in (ii). In this matrix, the first non-zero entry in the second row occurs in the first column and it is on the left of the first non-zero entry in the first row which occurs in the second column. So the matrix is not in row-echelon form.

Method to reduce a matrix  [a_ij]_m ×n to a row-echelon form.

Step 1

Inspect the first row. If the first row is a zero row, then the row is interchanged with a non-zero row below the first row. If a₁₁ is not equal to 0, then go to step 2. Otherwise, interchange the first row R₁ with any other row below the first row which has a non-zero element in the first column; if no row below the first row has non-zero entry in the first column, then consider a₁₂ . If a₁₂ is not equal to 0, then go to step 2. Otherwise, interchange the first row R₁ with any other row below the first row which has a non-zero element in the second column; if no row below the first row has non-zero entry in the second column, then consider a₁₃. Proceed in the same way till we get a non-zero entry in the first row. This is called pivoting and the first non-zero element in the first row is called the pivot of the first row.

Step 2

Use the first row and elementary row operations to transform all elements under the pivot to become zeros.

Step 3

Consider the next row as first row and perform steps 1 and 2 with the rows below this row only.

Repeat the step until all rows are exhausted.

Example 1.13

Reduce the matrix  to a row-echelon form.

Solution

Note

This is also a row-echelon form of the given matrix.

So, a row-echelon form of a matrix is not necessarily unique.

Example 1.14

Reduce the matrix  to a row-echelon form.

Solution

Rank of a Matrix

To define the rank of a matrix, we have to know about sub-matrices and minors of a matrix.

Let A be a given matrix. A matrix obtained by deleting some rows and some columns of A is called a sub-matrix of A. A matrix is a sub-matrix of itself because it is obtained by leaving zero number of rows and zero number of columns.

Recall that the determinant of a square sub-matrix of a matrix is called a minor of the matrix.

 Definition 1.6

The rank of a matrix A is defined as the order of a highest order non-vanishing minor of the matrix A. It is denoted by the symbol ρ (A). The rank of a zero matrix is defined to be 0.

Note

If a matrix contains at-least one non-zero element, then ρ ( A) ≥ 1.

The rank of the identity matrix I_n is n.

If the rank of a matrix A is r, then there exists at-least one minor of A of order r which does not vanish and every minor of A of order r +1 and higher order (if any) vanishes.

If A is an m × n matrix, then ρ (A) ≤ min{m, n} = minimum of m, n.

A square matrix A of order n has inverse if and only if ρ ( A) = n.

Example 1.15

Find the rank of each of the following matrices:

Solution

(i) Let A =. Then A is a matrix of order 3× 3. So ρ(A) ≤ min {3, 3}  = 3. The highest  order of minors of A is 3 . There is only one third order minor of A .

 It is  =  3 (6− 6) − 2 (6−6) +  5 (3 − 3) = 0. So, ρ(A) < 3.

 Next consider the second-order minors of A .

We find that the second order minor  = 3 − 2 = 1 ≠ 0 . So ρ(A) = 2 .

(ii) Let A = . Then A is a matrix of order 3×4 . So ρ(A) ≤ min {3, 4}  = 3.

The highest order of minors of A is 3 . We search for a non-zero third-order minor of A . But

we find that all of them vanish. In fact, we have

 So, ρ(A) < 3. Next, we search for a non-zero second-order minor of A .

We find that  = -4+9 =5 ≠ 0 . So, ρ(A) = 2 .

Remark

Finding the rank of a matrix by searching a highest order non-vanishing minor is quite tedious when the order of the matrix is quite large. There is another easy method for finding the rank of a matrix even if the order of the matrix is quite high. This method is by computing the rank of an equivalent row-echelon form of the matrix. If a matrix is in row-echelon form, then all entries below the leading diagonal (it is the line joining the positions of the diagonal elements a₁₁ , a₂₂ , a₃₃ ,. of the matrix) are zeros. So, checking whether a minor is zero or not, is quite simple.

Example 1.16

Find the rank of the following matrices which are in row-echelon form :

Solution

(i) Let A = . Then A is a matrix of order 3 3 × and ρ(A) ≤ 3

 The third order minor |A| =   = (2) (3)( 1) = 6 ≠ 0 . So, ρ(A) = 3 .

Note

that there are three non-zero rows.

(ii) Let A = . Then A is a matrix of order 3× 3 and ρ(A) ≤ 3.

The only third order minor is |A| =  = (-2) (5) (0) = 0 . So ρ(A) ≤ 2 .

There are several second order minors. We find that there is a second order minor, for  example,  = (-2)(5) = -10 ≠ 0 . So, ρ(A) = 2.

Note

that there are two non-zero rows. The third row is a zero row.

(iii) Let A = . Then A is a matrix of order 4 × 3 and ρ(A) ≤ 3.

The last two rows are zero rows. There are several second order minors. We find that there  is a second order minor, for example,  = (6) (2) = 12 ≠ 0 .

So, ρ(A) = 2.

Note

that there are two non-zero rows. The third and fourth rows are zero rows.

We observe from the above example that the rank of a matrix in row echelon form is equal  to the number of non-zero rows in it. We state this observation as a theorem without proof.

Theorem 1.11

The rank of a matrix in row echelon form is the number of non-zero rows in it.

The rank of a matrix which is not in a row-echelon form, can be found by applying the following result which is stated without proof.

Theorem 1.12

The rank of a non-zero matrix is equal to the number of non-zero rows in a row-echelon form of the matrix.

Example 1.17

Find the rank of the matrix  by reducing it to a row-echelon form.

Solution

Let A = . Applying elementary row operations, we get

The last equivalent matrix is in row-echelon form. It has two non-zero rows. So, ρ (A)= 2.

Example 1.18

Find the rank of the matrix  by reducing it to a row-echelon form.

Solution

Let A be the matrix. Performing elementary row operations, we get

The last equivalent matrix is in row-echelon form. It has three non-zero rows. So, ρ(A) = 3 .

Elementary row operations on a matrix can be performed by pre-multiplying the given matrix by a special class of matrices called elementary matrices.

Definition 1.7

An elementary matrix is defined as a matrix which is obtained from an identity matrix by applying only one elementary transformation.

Remark

If we are dealing with matrices with three rows, then all elementary matrices are square matrices of order 3 which are obtained by carrying out only one elementary row operations on the unit matrix I₃. Every elementary row operation that is carried out on a given matrix A can be obtained by pre-multiplying A with elementary matrix. Similarly, every elementary column operation that is carried out on a given matrix A can be obtained by post-multiplying Awith an elementary matrix. In the present chapter, we use elementary row operations only.

For instance, let us consider the matrix A = 

Suppose that we do the transformation R₂  → R₂ + λR₃ on A, where λ ≠ 0 is a constant. Then, we get

The matrix   is an elementary matrix, since we have 

Pre-multiplying A by   , we get

From (1) and (2), we get 

So, the effect of applying the elementary transformation R₂ → R₂ + λR₃ on A is the same as that of pre-multiplying the matrix A with the elementary matrix 

Similarly, we can show that

(i) the effect of applying the elementary transformation R₂↔ R₃ on A is the same as that of  pre-multiplying the matrix A with the elementary matrix 

(ii) the effect of applying the elementary transformation R₂ → R₂λ on A is the same as that of  pre-multiplying the matrix A with the elementary matrix 

We state the following result without proof.

Theorem 1.13

Every non-singular matrix can be transformed to an identity matrix, by a sequence of elementary row operations.

As an illustration of the above theorem, let us consider the matrix A = 

Then, |A| =  12+ 3 = 15 ≠  0. So, A is non-singular. Let us transform A into I₂ by a sequence of  elementary row operations. First, we search for a row operation to make a₁₁ of A as 1. The elementary row operation needed for this is R₁ → (1/2) R₁. The corresponding elementary matrix is 

Next, let us make all elements below a₁₁ of E₁A as 0. There is only one element a₂₁.

The elementary row operation needed for this is R₂  → R₂  + (−3) R₁ .

The corresponding elementary matrix is E₂ =

Next, let us make a₂₂ of E₂(E₂A) as 1. The elementary row operation needed for this is  

The corresponding elementary matrix is E₃ = 

Then, we get E₃(E₂(E₁A)) = 

Finally, let us find an elementary row operation to make a₁₂ of E₃(E₂(E₁A)) as 0. The elementary  row operation needed for this is R₁ → R₁ + (1/2) R₂. The corresponding elementary matrix is 

We write the above sequence of elementary transformations in the following manner:

Example 1.19

Show that the matrix  is non-singular and reduce it to the identity matrix by elementary row transformations.

Solution

Let A = .Then, |A| = 3 (0+2 ) – 1(2+5) + 4(4-0) = 6-7+16 ≠ 0. So, A is non-singular. Keeping the identity matrix as our goal, we perform the row operations sequentially on A as follows:

Gauss-Jordan Method

Let A be a non-singular square matrix of order n . Let B be the inverse of A.

Then, we have AB = BA = I_n . By the property of I_n , we have A = I_n A = AI_n .

Consider the equation A = I_n A          …………………………………………..(1)

Since A is non-singular, pre-multiplying by a sequence of elementary matrices (row operations) on both sides of (1), A on the left-hand-side of (1) is transformed to the identity matrix I_n and the same sequence of elementary matrices (row operations) transforms I_n of the right-hand-side of (1) to a matrix B. So, equation (1) transforms to I_n = BA. Hence, the inverse of A is B. That is, A⁻¹ = B.

Note

If E₁ , E₂ ,…, E_k are elementary matrices (row operations) such that (E_k … E₂ E₁ ) A = I_n ,  then A⁻¹ = E_k … E₂ E₁.

Transforming a non-singular matrix A to the form I_n by applying elementary row operations, is called Gauss-Jordan method. The steps in finding A⁻¹ by Gauss-Jordan method are given below:

Step 1

Augment the identity matrix I_n on the right-side of A to get the matrix [ A | I_n ] .

Step 2

Obtain elementary matrices (row operations)  E₁ , E₂ ,…, E_k such that  (E_k … E₂ E₁ ) A = I_n .

Apply  E₁ , E₂ ,…, E_k on [ A | I_n ] . Then  [(E_k …… E₂ E₁ ) A | (E_k ….. E₂ E₁ ) I_n]. That is, [I_n | A⁻¹ ].

Example 1.20

Find the inverse of the non-singular matrix A = , by Gauss-Jordan method.

Solution

Applying Gauss-Jordan method, we get

Example 1.21

Find the inverse of A =  by Gauss-Jordan method.

Solution

Applying Gauss-Jordan method, we get

Applications of Matrices: Solving System of Linear Equations

One of the important applications of matrices and determinants is solving of system of linear equations. Systems of linear equations arise as mathematical models of several phenomena occurring in biology, chemistry, commerce, economics, physics and engineering. For instance, analysis of circuit theory, analysis of input-output models, and analysis of chemical reactions require solutions of systems of linear equations. 

System of Linear Equations in Matrix Form

A system of m linear equations in n unknowns is of the following form:

a₁₁x₁ + a₁₂x₂ + a₁₃x₃ + ……… + a_1nx_n + = b₁

a₂₁x₁ + a₂₂x₂ + a₂₃x₃ + ……… + a_2nx_n + = b₂

a₃₁x₁ + a₃₂x₂ + a₃x₃ + ……… + a_3nx_n + = b₃

…..    ….   …..   …..   …..    ...

A_m1x₁ + a_m2x₂ + a_m3x₃ + ……… + a_mnx_n + = b_m

where the coefficients  a_ij , i = 1, 2, …. , m; j = 1, 2,….., n  and b_k , k = 1, 2,….., m   are constants. If all  the b_k 's are zeros, then the above system is called a homogeneous system of linear equations. On the other hand, if at least one of the b_k 's is non-zero, then the above system is called a non-homogeneous system of linear equations. If there exist values α₁ , α₂ , ….. , α_n for x₁, x₂ , …. , x_n respectively which satisfy every  equation of (1), then the ordered  n − tuple (α₁ ,  α₂ , …. , α_n ) is called a solution of (1). The above system (1) can be put in a matrix form as follows:

Let A =  be   the m x n matrix formed by the coefficients of x₁, x₂ , x₃,…. , x_n . The first row of A is formed by the coefficients of x₁, x₂ , x₃,…. , x_n in the same order in which they occur in the first equation. Likewise, the other rows of A are formed. The first column is formed by the coefficients of x₁ in the m equations in the same order. The other columns are formed in a similar way.

Let  X =  be the n x1 order column matrix formed by the unknowns x₁, x₂ , x₃,…. , x_n

Let   B  =  be the m x 1 order column matrix formed by the right-hand side constants b₁, b₂ , b₃ , …. , b_m .

Then we get

Then AX = B is a matrix equation involving matrices and it is called the matrix form of the system of linear equations (1). The matrix A is called the coefficient matrix of the system and the matrix

is called the augmented matrix of the system. We denote the augmented matrix by [ A | B ].

As an example, the matrix form of the system of linear equations

2x + 3y - 5z + 7 = 0, 7 y + 2z - 3x = 17, 6x - 3y - 8z + 24 = 0 is 

Solution to a System of Linear equations

The meaning of solution to a system of linear equations can be understood by considering the following cases :

Case (i)

Consider the system of linear equations

2x - y = 5 , ... (1)

x + 3y = 6 . ... (2)

These two equations represent a pair of straight lines in two dimensional analytical geometry (see the Fig. 1.2). Using (1), we get

x  =  [5 + y] / 2     (3)

Substituting (3) in (2) and simplifying, we get y = 1.

 Substituting         y = 1 in (1) and simplifying, we get  x = 3 .

Both equations (1) and (2) are satisfied by x = 3 and y = 1.

That is, a solution of (1) is also a solution of (2).

So, we say that the system is consistent and has unique solution (3,1) . The point (3,1) is the point of intersection of the two lines 2x - y = 5 and x + 3y = 6 .

Case (ii)

Consider the system of linear equations

3x + 2 y = 5 , ... (1)

6x + 4 y = 10 ... (2)

 Using equation (1), we get

x = [5 - 2 y] / 3 ... (3)

Substituting (3) in (2) and simplifying, we get 0 = 0.

This informs us that equation (2) is an elementary transformation of equation (1). In fact, by dividing equation (2) by 2, we get equation (1). It is not possible to find uniquely x and y with just a single equation (1).

So we are forced to assume the value of one unknown, say y = t , where t is any real number.

Then, x = [5 - 2t] /3. The two equations (1) and (2) represent one and only one straight line (coincident lines) in two dimensional analytical geometry (see Fig. 1.3) . In other words, the system is consistent (a solution of (1) is also a solution of (2)) and has infinitely many solutions, as t can assume any real value.

Case (iii)

Consider the system of linear equations

4x + y = 6 , ... (1)

8x + 2 y = 18 . ... (2)

Using equation (1), we get

x = [6 – y] / 4 ... (3)

Substituting (3) in (2) and simplifying, we get 12 = 18 .

This is a contradicting result, which informs us that equation (2) is inconsistent with equation (1). So, a solution of (1) is not a solution of (2).

In other words, the system is inconsistent and has no solution. We note that the two equations represent two parallel straight lines (non-coincident) in two dimensional analytical geometry (see Fig. 1.4). We know that two non-coincident parallel lines never meet in real points.

Note

(1) Interchanging any two equations of a system of linear equations does not alter the solution of the system.

(2) Replacing an equation of a system of linear equations by a non-zero constant multiple of itself does not alter the solution of the system.

(3) Replacing an equation of a system of linear equations by addition of itself with a non-zero multiple of any other equation of the system does not alter the solution of the system.

Matrix Inversion Method

This method can be applied only when the coefficient matrix is a square matrix and non-singular.

Consider the matrix equation

     AX   = B ,      … (1)

where A is a square matrix and non-singular. Since A is non-singular, A⁻¹ exists and A⁻¹ A = AA⁻¹ = I. Pre-multiplying both sides of (1) by A⁻¹, we get A⁻¹ ( AX ) = A⁻¹B. That is, ( A⁻¹ A) X = A⁻¹B. Hence, we get X = A⁻¹B.

Example 1.22

Solve the following system of linear equations, using matrix inversion method:

5x + 2 y = 3, 3x + 2 y = 5 .

Solution

The matrix form of the system is AX = B , where 

We find |A| =   = 10 - 6= 4 ≠ 0. So, A⁻¹  exists and A⁻¹ = 

Then, applying the formula X = A^{−1B ,} we get

So the solution is (x = −1, y = 4).

Example 1.23

Solve the following system of equations, using matrix inversion method:

2x₁ + 3x₂ + 3x₃ = 5,

x₁ – 2x₂ + x₃ = -4,

3x₁ – x₂ – 2x₃ = 3

Solution

The matrix form of the system is AX = B,where

So, the solution is ( x₁ = 1, x₂ = 2, x₃ = −1) .

Example 1.24

If , find the products AB and BA and hence solve the system of equations x − y + z = 4, x – 2y – 2z = 9, 2x + y +3z =1.

Solution

Writing the given system of equations in matrix form, we get

Hence, the solution is (x = 3, y = - 2, z = −1).

CRAMER’S RULE

This rule can be applied only when the coefficient matrix is a square matrix and non-singular. It is explained by considering the following system of equations:

where the coefficient matrix  is non-singular. Then 

Let us put Δ = . Then, we have

Note

Replacing the first column elements a₁₁ , a₂₁ , a₃₁ of Δ with b₁ , b₂ , b₃ respectively, we get Δ₁. Replacing the second column elements a₁₂ , a₂₂ , a₃₂ of Δ with b₁ , b₂ , b₃ respectively, we get Δ₂ . Replacing the third column elements a₁₃ , a₂₃ , a₃₃ of Δ with b₁ , b₂ , b₃ respectively, we get Δ₃.

If Δ = 0, Cramer’s rule cannot be applied.

Example 1.25

Solve, by Cramer’s rule, the system of equations

x₁ − x₂ = 3, 2x₁ + 3x₂ + 4x₃ = 17, x₂ + 2x₃ = 7.

Solution

First we evaluate the determinants

So, the solution is (x₁ = 2, x₂ = - 1, x₃ = 4).

Example 1.26

In a T20 match, Chennai Super Kings needed just 6 runs to win with 1 ball left to go in the last over. The last ball was bowled and the batsman at the crease hit it high up. The ball traversed along a path in a vertical plane and the equation of the path is y = ax² + bx + c with respect to a xy -coordinate system in the  vertical  plane  and  the  ball  traversed   through   the   points  (10,8), (20,16), (30,18) , can you conclude that Chennai Super Kings won the match?

Justify your answer. (All distances are measured in metres and the meeting point of the plane of the path with the farthest boundary line is (70, 0).)

Solution

The path y = ax2 + bx + c  passes through the points (10,8), (20,16), (40, 22) . So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c = 16,1600a + 40b + c = 22. To apply Cramer’s rule, we find

When x = 70, we get y = 6. So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before the boundary line to jump and catch the ball. Hence the ball went for a super six and the Chennai Super Kings won the match.

Gaussian Elimination Method

This method can be applied even if the coefficient matrix is singular matrix and rectangular matrix. It is essentially the method of substitution which we have already seen. In this method, we transform the augmented matrix of the system of linear equations into row-echelon form and then by back-substitution, we get the solution.

Example 1.27

Solve the following system of linear equations, by Gaussian elimination method :

4x + 3y + 6z = 25, x + 5 y + 7z = 13, 2x + 9 y + z = 1.

Solution

Transforming the augmented matrix to echelon form, we get

The equivalent system is written by using the echelon form:

 x + 5y  + 7z = 13 , … (1)

 17y + 22z = 27 , … (2)

 199z = 398 . … (3)

Substituting z = 2, y = -1  in (1), we get x = 13 - 5  × (−1 ) − 7 × 2 = 4 .

So, the solution is ( x =4, y = - 1, z = 2 ).

Note. The above method of going from the last equation to the first equation is called the method of back substitution.

Example 1.28

The  upward  speed  v(t) of a rocket  at time t is approximated by v(t) = at² + bt + c, 0 ≤  t ≤ 100 where a, b, and c are constants. It has been found that the speed at times t = 3, t = 6 , and t = 9 seconds are respectively, 64, 133, and 208 miles per second respectively. Find the speed at time  t = 15 seconds. (Use Gaussian elimination method.)

Solution

Since v(3) =64, v(6) = 133 and v(9) = 208 , we get the following system of linear equations

 9a +3b + c = 64 ,

 36a + 6b + c = 133,

 81a + 9b + c = 208 .

We solve the above system of linear equations by Gaussian elimination method.

Reducing the augmented matrix to an equivalent row-echelon form by using elementary row  operations, we get

Writing the equivalent equations from the row-echelon matrix, we get

9a + 3b + c =  64, 2b + c = 41, c= 1.

By back substitution, we get 

So, we get v (t) = 1/3 t²  + 20t + 1.

Hence, v(15) = 1/3 (225) + 20(15) + 1 = 75 + 300 + 1 = 376

Solution To A System Of Linear Equations

Example 1.22

Solve the following system of linear equations, using matrix inversion method:

5x + 2 y = 3, 3x + 2 y = 5 .

Solution

The matrix form of the system is AX = B , where 

We find |A| =   = 10 - 6= 4 ≠ 0. So, A−1  exists and A−1 = 

Then, applying the formula X = A−1B , we get

So the solution is (x = −1, y = 4).

Example 1.23

Solve the following system of equations, using matrix inversion method:

2x1 + 3x2 + 3x3 = 5,

x1 – 2x2 + x3 = -4,

3x1 – x2 – 2x3 = 3

Solution

The matrix form of the system is AX = B,where

So, the solution is ( x1 = 1, x2 = 2, x3 = −1) .

Example 1.24

If , find the products AB and BA and hence solve the system of equations x − y + z = 4, x – 2y – 2z = 9, 2x + y +3z =1.

Solution

Writing the given system of equations in matrix form, we get

Hence, the solution is (x = 3, y = - 2, z = −1).

(Ii) Cramer’s Rule

Example 1.25

Solve, by Cramer’s rule, the system of equations

x1 − x2 = 3, 2x1 + 3x2 + 4x3 = 17, x2 + 2x3 = 7.

Solution

First we evaluate the determinants

So, the solution is (x1 = 2, x2 = - 1, x3 = 4).

Example 1.26

In a T20 match, Chennai Super Kings needed just 6 runs to win with 1 ball left to go in the last over. The last ball was bowled and the batsman at the crease hit it high up. The ball traversed along a path in a vertical plane and the equation of the path is y = ax2 + bx + c with respect to a xy -coordinate system in the  vertical  plane  and  the  ball  traversed   through   the   points  (10,8), (20,16), (30,18) , can you conclude that Chennai Super Kings won the match

Justify your answer. (All distances are measured in metres and the meeting point of the plane of the path with the farthest boundary line is (70, 0).)

Solution

The path y = ax2 + bx + c  passes through the points (10,8), (20,16), (40, 22) . So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c= 16,1600a + 40b + c = 22. To apply Cramer’s rule, we find

When x = 70, we get y = 6. 

So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before theboundary line to jump and catch the ball.

Hence the ball went for a super six and the Chennai Super Kings won the match.

(iii) Gaussian Elimination Method

Example 1.27

Solve the following system of linear equations, by Gaussian elimination method :

4x + 3y + 6z = 25, x + 5 y + 7z = 13, 2x + 9 y + z = 1.

Solution

Transforming the augmented matrix to echelon form, we get

The equivalent system is written by using the echelon form:

 x + 5y  + 7z = 13 , … (1)

 17y + 22z = 27 , … (2)

 199z = 398 . … (3)

Substituting z = 2, y = -1  in (1), we get x = 13 - 5  × (−1 ) − 7 × 2 = 4 .

So, the solution is ( x =4, y = - 1, z = 2 ).

Note. The above method of going from the last equation to the first equation is called the method of back substitution.

Example 1.28

The  upward  speed  v(t) of a rocket  at time t is approximated by v(t) = at2 + bt + c, 0 ≤  t ≤ 100 where a, b, and c are constants. It has been found that the speed at times t = 3, t = 6 , and t = 9 seconds are respectively, 64, 133, and 208 miles per second respectively. Find the speed at time  t = 15 seconds. (Use Gaussian elimination method.)

Solution

Since v(3) =64, v(6) = 133 and v(9) = 208 , we get the following system of linear equations

 9a +3b + c = 64 ,

 36a + 6b + c = 133,

 81a + 9b + c = 208 .

We solve the above system of linear equations by Gaussian elimination method.

Reducing the augmented matrix to an equivalent row-echelon form by using elementary row  operations, we get

Writing the equivalent equations from the row-echelon matrix, we get

9a + 3b + c =  64, 2b + c = 41, c= 1.

By back substitution, we get 

So, we get v (t) = 1/3 t2  + 20t + 1.

Hence, v(15) = 1/3 (225) + 20(15) + 1 = 75 + 300 + 1 = 376

Applications Of Matrices: Consistency Of System Of Linear Equations By Rank Method

In second previous section, we have already defined consistency of a system of linear equation. In this section, we investigate it by using rank method. We state the following theorem without proof:

Theorem 1.14 (Rouché - Capelli Theorem)

A system of linear equations, written in the matrix form as AX = B, is consistent if and only if the rank of the coefficient matrix is equal to the rank of the augmented matrix; that is, ρ ( A) = ρ ([ A | B]).

We apply the theorem in the following examples.

Non-homogeneous Linear Equations

Example 1.29

Test for consistency of the following system of linear equations and if possible solve:

x + 2 y − z = 3, 3x − y + 2z = 1, x − 2 y + 3z = 3, x − y + z +1 = 0 .

Solution

Here the number of unknowns is 3.

The matrix form of the system is AX = B, where

Applying Gaussian elimination method on [ A | B], we get

There are three non-zero rows in the row-echelon form of [A | B].So, ρ ([A | B] ) . = 3

So, the row-echelon form of A is  . There are three non-zero rows in it. So ρ(A) = 3.

Hence, ρ(A) = ρ([ A | B ]) = 3.

From the echelon form, we write the equivalent system of equations

x + 2y –z =3, 7y-5z = 8, z=4, 0=0.

The last equation 0 = 0 is meaningful. By the method of back substitution, we get

 z = 4

7y - 20 = 8 ⇒ y = 4 ,

 x = 3 - 8 + 4 ⇒ x = −1.

So, the solution is (x = −1, y = 4,  z = 4) .(Note that A is not a square matrix.)

Here the given system is consistent and the solution is unique.

Example 1.30

Test for consistency of the following system of linear equations and if possible solve:

4x − 2 y + 6z = 8, x + y − 3z = −1, 15x − 3y + 9z = 21.

Solution

Here the number of unknowns is 3.

The matrix form of the system is AX = B, where

Applying elementary row operations on the augmented matrix[ A | B], we get

So, ρ ( A) = ρ ([ A | B]) = 2 < 3. From the echelon form, we get the equivalent equations

x + y - 3z =  -1, y - 3z = -2 , 0 = 0 .

The equivalent system has two non-trivial equations and three unknowns. So, one of the unknowns should be fixed at our choice in order to get two equations for the other two unknowns. We fix z arbitrarily as a real number t , and we get y = 3t - 2, x = -1- (3t - 2) + 3t = 1. So, the solution is ( x =1, y = 3t - 2, z = t ), where t is real . The above solution set is a one-parameter family of solutions.

Here, the given system is consistent and has infinitely many solutions which form a one parameter family of solutions.

Note

In the above example, the square matrix A is singular and so matrix inversion method cannot be applied to solve the system of equations. However,Gaussian elimination method is applicable and we are able to decide whether the system is consistent or not. The next example also confirms the supremacy of Gaussian elimination method over other methods.

Example 1.31

Test for consistency of the following system of linear equations and if possible solve:

x − y + z = −9, 2x − 2 y + 2z = −18, 3x − 3y + 3z + 27 = 0.

Solution

Here the number of unknowns is 3.

The matrix form of the system is AX = B, where

Applying elementary row operations on the augmented matrix[ A | B], we get

So, ρ ( A) = ρ ([ A | B]) = 1 < 3.

From the echelon form, we get the equivalent equations x - y + z = -9, 0 = 0, 0 = 0.

The equivalent system has one non-trivial equation and three unknowns.

Taking y = s, z = t arbitrarily, we get x - s + t = -9; or x = -9 + s - t.

So, the solution is ( x = -9 + s - t, y = s, z = t ), where s and t are parameters.

The above solution set is a two-parameter family of solutions.

Here, the given system of equations is consistent and has infinitely many solutions which form a two parameter family of solutions.

Example 1.32

Test the consistency of the following system of linear equations

x − y + z = −9, 2x − y + z = 4, 3x − y + z = 6, 4x − y + 2z = 7.

Solution

Here the number of unknowns is 3.

The matrix form of the system of equations is AX = B, where

Applying elementary row operations on the augmented matrix [A|B], we get

So, ρ (A) = 3 and ρ ([ A | B]) = 4. Hence ρ ( A) ≠ ρ ([ A | B]).

If we write the equivalent system of equations using the echelon form, we get

 x - y + z = -9, y - z = 22, z = -23, 0 = -11.

The last equation is a contradiction.

So the given system of equations is inconsistent and has no solution. By Rouché - Capelli theorem, we have the following rule:

Example 1.33

Find the condition on a, b and c so that the following system of linear equations has one parameter family of solutions: x + y + z = a, x + 2 y + 3z = b, 3x + 5 y + 7z = c.

Solution

Here the number of unknowns is 3.

The matrix form of the system is AX = B, where A = 

Applying elementary row operations on the augmented matrix [ A | B], we get

In order that the system should have one parameter family of solutions, we must have ρ ( A) = ρ ([ A, B]) = 2. So, the third row in the echelon form should be a zero row.

So, c - 2b - a = 0 ⇒ c = a + 2b.

Example 1.34

Investigate for what values of λ and μ the system of linear equations

x + 2 y + z = 7, x + y + λ z = μ, x + 3y − 5z = 5 has

(i) no solution (ii) a unique solution (iii) an infinite number of solutions.

Solution

Here the number of unknowns is 3.

The matrix form of the system is AX = B, where A = 

Applying elementary row operations on the augmented matrix [ A | B], we get

(i) If  λ = 7 and μ ≠ 9 , then ρ (A) = 2 and ρ ([ A | B]) = 3. So ρ ( A) ≠ ρ ([ A | B]). Hence the given system is inconsistent and has no solution.

(ii)    If  λ ≠ 7 and m is any real number, then ρ (A) = 3 and ρ ([ A | B]) = 3.

So ρ (A) = ρ ([ A | B]) = 3 = Number of unknowns. Hence the given system is consistent and has a unique solution.

(iii) If  λ = 7 and μ = 9, then ρ(A) = 2 and ρ ([ A | B]) = 2.

So, ρ(A) = ρ ([ A | B]) = 2 < Number of unknowns. Hence the given system is consistent and has infinite number of solutions.