Notes On Application of Determinants and Matrices - CBSE Class 12 Maths

Solution of a system of linear equations using inverse of a matrix

Consider the system of equations

a1x + a2y + a3z = k1

b1x + b2y + b3z = k2

c1x + c2y + c3z = k3

In matrix form, the system of equations can be expressed as

a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3   x y z   = k 1 k 2 k 3         …(i)

Let A =  a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3   ; X =  x y z   and B =  k 1 k 2 k 3

From (i), we have

AX = B


Case (i): A is a non-singular matrix. ⇒ A-1 exists.

We have AX = B

⇒ A-1 (AX) = A-1B

⇒ (A-1 A)X = A-1B (by associative property)

(I)X = A-1B ('.' A-1A = I , inverse property)

X = A-1B ⇒ X = A-1B ('.' IX = X , identity property)

This method of solving a system of linear equations is known as the matrix method.


Case (ii): A is a singular matrix.

⇒ |A| = 0

We have

X = A-1B ⇒ X = (adj A / Det A)B


Sub-case (i): (Adj A)B is a non-zero matrix.

In this case, the system of linear equations is inconsistent.

In other words, the system of linear equations has no solution.


Sub-case (ii): (Adj A)B is a zero matrix.

In this case, the system of linear equations has infinitely many solutions or no solution at all.

Summary

Solution of a system of linear equations using inverse of a matrix

Consider the system of equations

a1x + a2y + a3z = k1

b1x + b2y + b3z = k2

c1x + c2y + c3z = k3

In matrix form, the system of equations can be expressed as

a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3   x y z   = k 1 k 2 k 3         …(i)

Let A =  a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3   ; X =  x y z   and B =  k 1 k 2 k 3

From (i), we have

AX = B


Case (i): A is a non-singular matrix. ⇒ A-1 exists.

We have AX = B

⇒ A-1 (AX) = A-1B

⇒ (A-1 A)X = A-1B (by associative property)

(I)X = A-1B ('.' A-1A = I , inverse property)

X = A-1B ⇒ X = A-1B ('.' IX = X , identity property)

This method of solving a system of linear equations is known as the matrix method.


Case (ii): A is a singular matrix.

⇒ |A| = 0

We have

X = A-1B ⇒ X = (adj A / Det A)B


Sub-case (i): (Adj A)B is a non-zero matrix.

In this case, the system of linear equations is inconsistent.

In other words, the system of linear equations has no solution.


Sub-case (ii): (Adj A)B is a zero matrix.

In this case, the system of linear equations has infinitely many solutions or no solution at all.

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