Mathematics Applications of Determinants and Matrices, Solution of system of linear equations using inverse of a matrix For CBSE-NCERT
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star Applications of Determinants and Matrices
star Solution of system of linear equations using inverse of a matrix

### Applications of Determinants and Matrices

color{red}{"Consistent system :"} A system of equations is said to be consistent if its solution (one or more) exists.

color{red}{"Inconsistent system :"} A system of equations is said to be inconsistent if its solution does not exist.

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

\color{green} ✍️ Let us express the system of linear equations as matrix equations and solve them using inverse of the coefficient matrix.

Consider the system of equations

a_1 x + b_1 y + c_1 z = d_1

a_2 x + b_2 y + c_2 z = d_2

a_3 x + b_3 y + c_3 z = d_3

Let A = [ (a_1, b_1, c_1), ( a_2, b_2, c_2 ), (a_3, b_3, c_3) ] , X = [ (x), (y), (z) ] and B = [ (d_1), ( d_2),(d_3) ]

Then, the system of equations can be written as, AX = B, i.e.,

color{orange}{[ (a_1, b_1, c_1 ), ( a_2, b_2 , c_2 ), ( a_3, b_3 , c_3) ] [ (x), (y), (z) ] = [ (d_1), (d_2), (d_3) ]}

### Case I : System has unique solution (Consistent System)

\color{green} ✍️ If A is a nonsingular matrix, then its inverse exists. Now

AX = B

=> A^(–1) (AX) = A^(–1) B (premultiplying by A^(–1 ))

=> (A^(–1) A) X = A^(–1) B (by associative property)

=>  I X = A^(–1) B

=> X = A^(–1) B

\color{green} ✍️ This matrix equation provides color{green{"}unique solution" for the given system of equations as inverse of a matrix is unique.

\color{green} ✍️ This method of solving system of equations is known as Matrix Method.

### Case II : Inconsistent System

color{red}{"Inconsistent System" }

\color{green} ✍️ If A is a singular matrix, then |A| = 0.
\color{green} ✍️ In this case, we calculate (adj A) B.

color{green}{X = A^(-1)B = ((Adj A) B)/(|A|)

\color{green} ✍️ Now If (adj A) B ≠ O, (O being zero matrix),

\color{green} ✍️ Then solution does not exist and the system of equations is called inconsistent.

\color{green} ✍️ If (adj A) B = O, color{orange}{" then system may be either consistent or inconsistent according"}

color{orange}{"as the system have either infinitely many solutions or no solution."}

### Example

Q 3135101062

Solve the system of equations
2x + 5y = 1
3x + 2y = 7
Class 12 Chapter 4 Example 27
Solution:

The system of equations can be written in the form AX = B, where

A = [ (2,5), (3,2) ] ,X = [ (x), (y) ] and B = [ (1), (7) ]

Now,| A | = –11 ≠ 0 , Hence, A is nonsingular matrix and so has a unique solution.

Note that A^(-1) = - 1/11 [ (2,-5), ( -3,2) ]

Therefore X=A^(-1) B = - 1/11 [ (2,-5), ( -3,2) ] [ (1), (7) ]

i.e., [ (x), (y) ] = -1/11 [ (-33), (11) ] = [ (3), (-1) ]

Hence x=3 , y = -1
Q 3165101065

Solve the following system of equations by matrix method.
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Class 12 Chapter 4 Example 28
Solution:

The system of equations can be written in the form AX = B, where

A= [ (3,-2,3), ( 2,1,-1), ( 4, -3,2) ] ,X= [ (x), (y), (x) ]  and B = [ (8), (1), (4) ]

We see that

 | A | = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0

Hence, A is nonsingular and so its inverse exists. Now

A_11 = –1, A_12 = – 8, A_13 = –10

A_21 = –5, A_22 = – 6, A_23 = 1

A_31 = –1, A_32 = 9, A_33 = 7

Therefore A^(-1) = -1/17 [ (-1,-5 ,-1 ), ( -8,-6,9), ( -10,1,7) ]

So X= A^(-1) B = -1/17 [ (-1,-5 ,-1 ), ( -8,-6, 9 ), ( -10, 1,7) ] [ (8), (1), (4) ]

i.e.,  [ (x), (y),(x) ] = -1/17 [ (-17), (-34), (-51) ] = [ (1), (2), (3) ]

Hence x = 1, y = 2 and z = 3.
Q 3115201160

The sum of three numbers is 6. If we multiply third number by 3 and add
second number to it, we get 11. By adding first and third numbers, we get double of the
second number. Represent it algebraically and find the numbers using matrix method.
Class 12 Chapter 4 Example 29
Solution:

Let first, second and third numbers be denoted by x, y and z, respectively.
Then, according to given conditions, we have
x + y + z = 6
y + 3z = 11
x + z = 2y or x – 2y + z = 0
This system can be written as A X = B, where

A= [ ( 1,1,1), (0,1,3 ), (1,-2,1) ] ,X = [ (x), (y), (z) ] and B = [ (6), (11), (0) ]

Here A = 1 (1+ 6) – (0 – 3)+ (0 –1) = 9 ≠ 0 . Now we find adj A

A_11 = 1 (1 + 6) = 7, A_12 = – (0 – 3) = 3, A_13 = – 1

A_21 = – (1 + 2) = – 3, A_22 = 0, A_23 = – (– 2 – 1) = 3

A_31 = (3 – 1) = 2, A_32 = – (3 – 0) = – 3, A_33 = (1 – 0) = 1

Hence adj A = [ (7,-3,2), (3,0, -3), ( -1,3,1) ]

Thus A^(-1) = 1/(|A|) adj (A) =1/9 [ ( 7, -3,2 ),( 3,0,-3),(-1,3,1) ]

Since X = A^(-1) B

X = 1/9 [ (7,-3,2 ), ( 3,0,-3), ( -1,3,1) ] [ (6), (11), (0) ]

or  [ (x), (y), (x) ] =1/9 [ ( 42-33+0), (18+0+0), (-6+33+0) ] = 1/9 [ (9), (18), (27) ]= [ (1), (2), (3) ]

Thus x = 1, y = 2, z = 3