`star` Applications of Determinants and Matrices

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

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

`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.

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

`\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) ]}`

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) ]}`

`\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.

`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.

`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."}`

`\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."}`

Q 3135101062

Solve the system of equations

2x + 5y = 1

3x + 2y = 7

Class 12 Chapter 4 Example 27

2x + 5y = 1

3x + 2y = 7

Class 12 Chapter 4 Example 27

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

3x – 2y + 3z = 8

2x + y – z = 1

4x – 3y + 2z = 4

Class 12 Chapter 4 Example 28

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

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

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