Solution Of Simultaneous Linear Equations

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = 30 – 28 = 2

So, the above system has a unique solution, given by

X = A ^{– 1}B

Let C_ij be the cofactor of a_ijin A, then

C₁₁ = (– 1)^{1 + 1} 6 = 6

C₁₂ = (– 1)^{1 + 2} 4 = – 4

C₂₁ = (– 1)^{2 + 1} 7 = – 7

C₂₂ = (– 1)^{2 + 2} 5 = 5

Also, adj A =

=

A ^{– 1} =

A ^{– 1} =

Now, X = A ^{– 1}B

Hence, X = Y =

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = 10 – 6 = 4

So, the above system has a unique solution, given by

X = A ^{– 1}B

Let C_ij be the cofactor of a_ijin A, then

C₁₁ = (– 1)^{1 + 1} 2 = 2

C₁₂ = (– 1)^{1 + 2} 3 = – 3

C₂₁ = (– 1)^{2 + 1} 2 = – 2

C₂₂ = (– 1)^{2 + 2} 2 = 5

Also, adj A =

=

A ^{– 1} =

A ^{– 1} =

Now, X = A ^{– 1}B

Hence, X = Y = 4

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = – 3 – 4 = – 7

So, the above system has a unique solution, given by

X = A ^{– 1}B

Let C_ij be the cofactor of a_ijin A, then

C₁₁ = (– 1)^{1 + 1} – 1 = – 1

C₁₂ = (– 1)^{1 + 2} 1 = – 1

C₂₁ = (– 1)^{2 + 1} 4 = – 4

C₂₂ = (– 1)^{2 + 2} 3 = 3

Also, adj A =

=

A ^{– 1} =

A ^{– 1} =

Now, X = A ^{– 1}B

Hence, X = 1 Y = – 2

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = – 3 – 3 = – 6

So, the above system has a unique solution, given by

X = A ^{– 1}B

Let C_ij be the cofactor of a_ijin A, then

C₁₁ = (– 1)^{1 + 1} – 1 = – 1

C₁₂ = (– 1)^{1 + 2} 3 = – 3

C₂₁ = (– 1)^{2 + 1} 1 = – 1

C₂₂ = (– 1)^{2 + 2} 3 = 3

Also, adj A =

=

A ^{– 1} =

A ^{– 1} =

Now, X = A ^{– 1}B

Hence, X = 7 Y = – 2

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = 6 – 7 = – 1

So, the above system has a unique solution, given by

X = A ^{– 1}B

Let C_ij be the cofactor of a_ijin A, then

C₁₁ = (– 1)^{1 + 1} 2 = 2

C₁₂ = (– 1)^{1 + 2} 1 = – 1

C₂₁ = (– 1)^{2 + 1} 7 = – 7

C₂₂ = (– 1)^{2 + 2} 3 = 3

Also, adj A =

=

A ^{– 1} =

A ^{– 1} =

Now, X = A ^{– 1}B

Hence, X = – 15 Y = 7

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = 9 – 5 = 4

So, the above system has a unique solution, given by

X = A ^{– 1}B

Let C_ij be the cofactor of a_ijin A, then

C₁₁ = (– 1)^{1 + 1} 3 = 3

C₁₂ = (– 1)^{1 + 2} 5 = – 5

C₂₁ = (– 1)^{2 + 1} 1 = – 1

C₂₂ = (– 1)^{2 + 2} 3 = 3

Also, adj A =

=

A ^{– 1} =

A ^{– 1} =

Now, X = A ^{– 1}B

Hence, X = Y =

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 1

= (– 20) – 1(– 17) – 1(11)

= – 20 + 17 + 11 = 8

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} – 21 + 1 = – 20

C₂₁ = (– 1)^{2 + 1} – 7 – 1 = 8

C₃₁ = (– 1)^{3 + 1} 1 + 3 = 4

C₁₂ = (– 1)^{1 + 2} – 14 – 3 = 17

C₂₂ = (– 1)^{2 + 1} – 7 + 3 = – 4

C₃₂ = (– 1)^{3 + 1} 1 + 2 = – 3

C₁₃ = (– 1)^{1 + 2} – 2 – 9 = – 11

C₂₃ = (– 1)^{2 + 1} – 1 – 3 = 4

C₃₃ = (– 1)^{3 + 1} 3 – 2 = 1

adj A =

=

Now, X = A ^{– 1}B =

X =

X =

Hence, X = 3,Y = 1 and Z = 1

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 1

= (3 – 1) – 1(– 6 – 2) + 1(2 + 2)

= 2 + 8 + 4

= 14

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 3 – 1 = 2

C₂₁ = (– 1)^{2 + 1} – 3 – 1 = 4

C₃₁ = (– 1)^{3 + 1} 1 + 1 = 2

C₁₂ = (– 1)^{1 + 2} – 6 – 2 = 8

C₂₂ = (– 1)^{2 + 1} – 3 – 2 = – 5

C₃₂ = (– 1)^{3 + 1} 1 – 2 = 1

C₁₃ = (– 1)^{1 + 2} 2 + 2 = 4

C₂₃ = (– 1)^{2 + 1} 1 – 2 = 1

C₃₃ = (– 1)^{3 + 1} – 1 – 2 = – 3

adj A =

=

Now, X = A ^{– 1}B =

X =

X =

Hence, X = ,Y = and Z =

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 6

= 6(225 + 360) + 12(60 + 40) + 25(72 – 30)

= 3510 + 1200 + 1050

= 5760

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} (225 + 360) = 585

C₂₁ = (– 1)^{2 + 1} (– 180 – 450) = 630

C₃₁ = (– 1)^{3 + 1} (240 – 375) = – 135

C₁₂ = (– 1)^{1 + 2} (60 + 40) = – 100

C₂₂ = (– 1)^{2 + 1} (90 – 50) = 40

C₃₂ = (– 1)^{3 + 1} (– 120 – 100) = 220

C₁₃ = (– 1)^{1 + 2} (72 – 30) = 42

C₂₃ = (– 1)^{2 + 1}(108 + 24) = – 132

C₃₃ = (– 1)^{3 + 1} (90 + 48) = 138

adj A =

=

Now, X = A ^{– 1}B =

X =

X =

Hence, X = ,Y = and Z =

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 3

= 3(3 – 6) – 4(– 6 – 3) + 7(4 + 1)

= – 9 + 36 + 35

= 62

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 3 – 6 = – 3

C₂₁ = (– 1)^{2 + 1} – 12 – 14 = 26

C₃₁ = (– 1)^{3 + 1}12 + 7 = 19

C₁₂ = (– 1)^{1 + 2} – 6 – 3 = 9

C₂₂ = (– 1)^{2 + 1} – 3 – 7 = – 10

C₃₂ = (– 1)^{3 + 1} 9 – 14 = 5

C₁₃ = (– 1)^{1 + 2} 4 + 1 = 5

C₂₃ = (– 1)^{2 + 1} 6 – 4 = – 2

C₃₃ = (– 1)^{3 + 1} – 3 – 8 = – 11

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = 1,Y = 1 and Z = 1

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 5

= 5(4 – 6) – 3(8 – 3) + 1(4 – 2)

= – 10 – 15 + 3

= – 22

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} (4 – 6) = – 2

C₂₁ = (– 1)^{2 + 1}(12 – 2) = – 10

C₃₁ = (– 1)^{3 + 1}(9 – 1) = 8

C₁₂ = (– 1)^{1 + 2} (8 – 3) = – 5

C₂₂ = (– 1)^{2 + 1} 20 – 1 = 19

C₃₂ = (– 1)^{3 + 1} 15 – 2 = – 13

C₁₃ = (– 1)^{1 + 2} (4 – 2) = 2

C₂₃ = (– 1)^{2 + 1} 10 – 3 = – 7

C₃₃ = (– 1)^{3 + 1} 5 – 6 = – 1

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = 1,Y = 2 and Z = 5

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 3

= 3(12 – 6) – 4(0 + 3) + 2(0 – 2)

= 18 – 12 – 4

= 2

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} (12 – 6) = 6

C₂₁ = (– 1)^{2 + 1}(24 + 4) = – 28

C₃₁ = (– 1)^{3 + 1}(– 12 – 4) = – 16

C₁₂ = (– 1)^{1 + 2} (0 + 3) = – 3

C₂₂ = (– 1)^{2 + 1} 18 – 2 = 16

C₃₂ = (– 1)^{3 + 1} – 9 – 0 = 9

C₁₃ = (– 1)^{1 + 2} (0 – 2) = – 2

C₂₃ = (– 1)^{2 + 1} (– 6 – 4) = 10

C₃₃ = (– 1)^{3 + 1} 6 – 0 = 6

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = – 2,Y = 3 and Z = 1

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 2

= 2(– 6 + 1) – 1(– 2 + 3) + 1(1 – 9)

= – 10 – 1 – 8

= – 19

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} – 6 + 1 = – 5

C₂₁ = (– 1)^{2 + 1}(24 + 4) = – 28

C₃₁ = (– 1)^{3 + 1} – 1 – 3 = – 4

C₁₂ = (– 1)^{1 + 2} – 2 + 3 = – 1

C₂₂ = (– 1)^{2 + 1} – 4 – 3 = – 7

C₃₂ = (– 1)^{3 + 1} – 2 – 1 = 3

C₁₃ = (– 1)^{1 + 2}1 – 9 = – 8

C₂₃ = (– 1)^{2 + 1}2 – 3 = – 1

C₃₃ = (– 1)^{3 + 1} 6 – 1 = 5

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = 1,Y = 1 and Z = – 1

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 2

= 2(0 – 1) – 6(3 + 2)

= – 2 – 30

= – 32

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 0 – 1 = – 1

C₂₁ = (– 1)^{2 + 1}6 + 0 = – 6

C₃₁ = (– 1)^{3 + 1} – 6 = – 6

C₁₂ = (– 1)^{1 + 2} 3 + 2 = 5

C₂₂ = (– 1)^{2 + 1} 2 – 0 = 2

C₃₂ = (– 1)^{3 + 1} – 2 – 0 = 2

C₁₃ = (– 1)^{1 + 2} – 3 – 0 = – 3

C₂₃ = (– 1)^{2 + 1} – 2 – 12 = 14

C₃₃ = (– 1)^{3 + 1} 0 – 18 = – 18

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = – 2,Y = 1 and Z = 2

The given system can be written in matrix form as:

AX = B

Now, |A| = 0

= 0 + 4 – 1

= 3

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 1 – 0 = 1

C₂₁ = (– 1)^{2 + 1}1 – 2 = 1

C₃₁ = (– 1)^{3 + 1}0 + 1 = 1

C₁₂ = (– 1)^{1 + 2} – 2 – 0 = 2

C₂₂ = (– 1)^{2 + 1} – 1 – 0 = – 1

C₃₂ = (– 1)^{3 + 1} 0 – 2 = 2

C₁₃ = (– 1)^{1 + 2} 4 – 0 = 4

C₂₃ = (– 1)^{2 + 1} 2 – 0 = – 2

C₃₃ = (– 1)^{3 + 1} – 1 + 2 = 1

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = 1,Y = 2 and Z = 3

The given system can be written in matrix form as:

AX = B

Now, |A| = 8

= 8(– 1) – 4(1) + 3(3)

= – 8 – 4 + 9

= – 3

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 1 – 2 = – 1

C₂₁ = (– 1)^{2 + 1} 4 – 6 = 2

C₃₁ = (– 1)^{3 + 1} 4 – 3 = 1

C₁₂ = (– 1)^{1 + 2} 2 – 1 = – 1

C₂₂ = (– 1)^{2 + 1} 8 – 3 = 5

C₃₂ = (– 1)^{3 + 1} 8 – 6 = – 2

C₁₃ = (– 1)^{1 + 2} 4 – 1 = 3

C₂₃ = (– 1)^{2 + 1} 16 – 4 = – 12

C₃₃ = (– 1)^{3 + 1} 8 – 8 = 0

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = 1,Y = 1 and Z = 2

The given system can be written in matrix form as:

A X = B

Now, |A| = 1

= 1(– 2) – 1(1 – 6) + 1(1)

= – 2 + 5 + 1

= 4

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 0 – 2 = – 2

C₂₁ = (– 1)^{2 + 1} 1 – 1 = 0

C₃₁ = (– 1)^{3 + 1} 2 – 0 = 2

C₁₂ = (– 1)^{1 + 2} 1 – 6 = 5

C₂₂ = (– 1)^{2 + 1} 1 – 3 = – 2

C₃₂ = (– 1)^{3 + 1} 2 – 1 = – 1

C₁₃ = (– 1)^{1 + 2} 1 – 0 = 1

C₂₃ = (– 1)^{2 + 1} 1 – 3 = 2

C₃₃ = (– 1)^{3 + 1} 0 – 1 = – 1

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = 3,Y = 1 and Z = 2

The given system can be written in matrix form as:

AX = B

Now,

|A| = 2(75) – 3(– 110) + 10(72)

= 150 + 330 + 720

= 1200

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 120 – 45 = 75

C₂₁ = (– 1)^{2 + 1} – 60 – 90 = 150

C₃₁ = (– 1)^{3 + 1} 15 + 60 = 75

C₁₂ = (– 1)^{1 + 2} – 80 – 30 = 110

C₂₂ = (– 1)^{2 + 1} – 40 – 60 = – 100

C₃₂ = (– 1)^{3 + 1} 10 – 40 = 30

C₁₃ = (– 1)^{1 + 2} 36 + 36 = 72

C₂₃ = (– 1)^{2 + 1} 18 – 18 = 0

C₃₃ = (– 1)^{3 + 1} – 12 – 12 = – 24

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

Hence, X = 2,Y = 3 and Z = 5

The given system can be written in matrix form as:

A X = B

Now,

|A| = 1(12 – 5) + 1(9 + 10) + 2(– 3 – 8)

= 7 + 19 – 22

= 4

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 12 – 5 = 7

C₂₁ = (– 1)^{2 + 1} – 3 + 2 = 1

C₃₁ = (– 1)^{3 + 1} 5 – 8 = – 3

C₁₂ = (– 1)^{1 + 2} 9 + 10 = – 19

C₂₂ = (– 1)^{2 + 1} 3 – 4 = – 1

C₃₂ = (– 1)^{3 + 1} – 5 – 6 = 11

C₁₃ = (– 1)^{1 + 2} – 3 – 8 = – 11

C₂₃ = (– 1)^{2 + 1} – 1 + 2 = – 1

C₃₃ = (– 1)^{3 + 1} 4 + 3 = 7

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

Hence, X = 2,Y = 1 and Z = 3

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = 36 – 36 = 0

So, A is singular, Now X will be consistence if (Adj A)xB = 0

C₁₁ = (– 1)^{1 + 1} 6 = 6

C₁₂ = (– 1)^{1 + 2} 9 = – 9

C₂₁ = (– 1)^{2 + 1} 4 = – 4

C₂₂ = (– 1)^{2 + 2} 6 = 6

Also, adj A =

=

(Adj A).B =

=

Thus, AX = B will be infinite solution,

Let y = k

Hence, 6x = 2 – 4k or 9x = 3 – 6k

X =

Hence, X = , Y = k

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = 18 – 18 = 0

So, A is singular,Now X will be consistence if (Adj A)xB = 0

C₁₁ = (– 1)^{1 + 1} 9 = 9

C₁₂ = (– 1)^{1 + 2} 6 = – 6

C₂₁ = (– 1)^{2 + 1} 3 = – 3

C₂₂ = (– 1)^{2 + 2} 2 = 2

Also, adj A =

=

(Adj A).B =

=

Thus, AX = B will be infinite solution,

Let y = k

Hence,

2x = 5 – 3k or X =

x = 15 – 9k or X =

Hence, X = , Y = k

This can be written as:

|A| = 5(260 – 4) – 3(30 – 14) + 7(6 – 182)

= 5(256) – 3(16) + 7(176)

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:

(Adj A)x B≠0 or (Adj A)x B = 0

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 260 – 4 = 256

C₂₁ = (– 1)^{2 + 1} 30 – 14 = – 16

C₃₁ = (– 1)^{3 + 1} 6 – 182 = – 176

C₁₂ = (– 1)^{1 + 2} 30 – 14 = – 16

C₂₂ = (– 1)^{2 + 1} 50 – 49 = 1

C₃₂ = (– 1)^{3 + 1} 10 – 21 = 11

C₁₃ = (– 1)^{1 + 2} 6 – 182 = – 176

C₂₃ = (– 1)^{2 + 1} 10 – 21 = 11

C₃₃ = (– 1)^{3 + 1} 130 – 9 = 121

adj A =

=

Adj A x B =

Now, AX = B has infinite many solution

Let z = k

Then, 5x + 3y = 4 – 7k

3x + 26y = 9 – 2k

This can be written as:

|A| = 121

Adj A =

Now, X = A ^{– 1}B =

=

=

There values of x,y,z satisfy the third equation

This can be written as:

|A| = 1(2) – 1(4) + 1(2)

= 2 – 4 + 2

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:

(Adj A)x B≠0 or (Adj A)x B = 0

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 14 – 12 = 2

C₂₁ = (– 1)^{2 + 1} 7 – 4 = – 3

C₃₁ = (– 1)^{3 + 1} 3 – 2 = 1

C₁₂ = (– 1)^{1 + 2} 7 – 3 = – 4

C₂₂ = (– 1)^{2 + 1} 7 – 1 = 6

C₃₂ = (– 1)^{3 + 1} 3 – 1 = 2

C₁₃ = (– 1)^{1 + 2} 4 – 2 = 2

C₂₃ = (– 1)^{2 + 1} 4 – 1 = – 3

C₃₃ = (– 1)^{3 + 1} 2 – 1 = 1

adj A =

=

Adj A x B =

Now, AX = B has infinite many solution

Let z = k

Then, x + y = 6 – k

X + 2y = 14 – 3k

This can be written as:

|A| = 1

Adj A =

Now, X = A ^{– 1}B =

=

=

There values of x,y,z satisfy the third equation

Hence, x = k – 2, y = 8 – 2k, z = k

|A| = 2(14) – 2(14) – 2(0)

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:

(Adj A)x B≠0 or (Adj A)x B = 0

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1}8 + 6 = 14

C₂₁ = (– 1)^{2 + 1} 4 + 12 = – 16

C₃₁ = (– 1)^{3 + 1} – 2 + 8 = 6

C₁₂ = (– 1)^{1 + 2} 8 + 6 = – 14

C₂₂ = (– 1)^{2 + 1} 4 + 12 = 16

C₃₂ = (– 1)^{3 + 1} – 2 + 8 = – 6

C₁₃ = (– 1)^{1 + 2} 24 – 24 = 0

C₂₃ = (– 1)^{2 + 1} 12 – 12 = 0

C₃₃ = (– 1)^{3 + 1} 8 – 8 = 0

adj A =

=

Adj A x B =

Now, AX = B has infinite many solution

Let z = k

Then, 2x + 2y = 1 + 2k

4x + 4y = 2 + k

This can be written as:

Hence, |A| = 0 z = 0

Hence, The given equation doesn’t satisfy .

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = 30 – 30 = 0

So, A is singular,Now X will be consistence if (Adj A)xB = 0

C₁₁ = (– 1)^{1 + 1} 15 = 15

C₁₂ = (– 1)^{1 + 2} 6 = – 6

C₂₁ = (– 1)^{2 + 1} 5 = – 5

C₂₂ = (– 1)^{2 + 2} 2 = 2

Also, adj A =

=

(Adj A).B =

=

= ≠

Hence, The given system is inconsistent.

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = 18 – 18 = 0

So, A is singular,Now X will be consistence if (Adj A)xB = 0

C₁₁ = (– 1)^{1 + 1} 9 = 9

C₁₂ = (– 1)^{1 + 2} 6 = – 6

C₂₁ = (– 1)^{2 + 1} 3 = – 3

C₂₂ = (– 1)^{2 + 2} 2 = 2

Also, adj A =

=

(Adj A).B =

=

= ≠0

Hence, The given system is inconsistent.

The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = – 12 + 12 = 0

So, A is singular,Now X will be consistence if (Adj A)xB = 0

C₁₁ = (– 1)^{1 + 1} – 3 = – 3

C₁₂ = (– 1)^{1 + 2} 6 = – 6

C₂₁ = (– 1)^{2 + 1} – 2 = 2

C₂₂ = (– 1)^{2 + 2} 4 = 4

Also, adj A =

=

(Adj A).B =

=

Hence, The given system is inconsistent.

|A| = 4(– 36) + 5(36) – 2(18)

|A| = 0

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} – 32 – 4 = – 36

C₂₁ = (– 1)^{2 + 1} – 40 + 4 = – 36

C₃₁ = (– 1)^{3 + 1} – 10 – 8 = – 18

C₁₂ = (– 1)^{1 + 2} 40 – 4 = – 36

C₂₂ = (– 1)^{2 + 1} 32 + 4 = 36

C₃₂ = (– 1)^{3 + 1} 8 + 10 = – 18

C₁₃ = (– 1)^{1 + 2} 10 + 8 = 18

C₂₃ = (– 1)^{2 + 1} 8 + 10 = – 18

C₃₃ = (– 1)^{3 + 1} – 16 + 25 = 9

adj A =

=

Adj A x B =

=

Hence, The above system is inconsistent.

|A| = 3(– 5) + 1(3) – 2(– 6)

|A| = 0

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 0 – 5 = – 5

C₂₁ = (– 1)^{2 + 1} 0 – 10 = 10

C₃₁ = (– 1)^{3 + 1} 1 + 4 = 5

C₁₂ = (– 1)^{1 + 2} 0 + 3 = – 3

C₂₂ = (– 1)^{2 + 1} 0 + 6 = 6

C₃₂ = (– 1)^{3 + 1} – 3 – 0 = 3

C₁₃ = (– 1)^{1 + 2} 0 – 6 = – 6

C₂₃ = (– 1)^{2 + 1} – 15 + 3 = 12

C₃₃ = (– 1)^{3 + 1} 6 – 0 = 6

adj A =

=

Adj A x B =

=

Hence, The above system is inconsistent.

|A| = 1(– 3) – 1(3) – 2(– 3) = – 3 – 3 + 6

|A| = 0

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} – 2 – 1 = – 3

C₂₁ = (– 1)^{2 + 1} 1 + 2 = – 3

C₃₁ = (– 1)^{3 + 1} 1 – 4 = – 3

C₁₂ = (– 1)^{1 + 2} 1 + 2 = – 3

C₂₂ = (– 1)^{2 + 1} 1 – 4 = – 3

C₃₂ = (– 1)^{3 + 1} 1 + 2 = – 3

C₁₃ = (– 1)^{1 + 2} 1 – 4 = – 3

C₂₃ = (– 1)^{2 + 1} 1 + 2 = – 3

C₃₃ = (– 1)^{3 + 1} – 2 – 1 = – 3

adj A =

=

Adj A x B =

=

Hence, The above system is inconsistent.

A = B =

AB =

AB =

Now, we can see that it is AB = 6I. where I is the unit Matrix

Or, A ^{– 1} =

Now the given equation can be written as:

A X B

Or, X = A ^{– 1}B

=

=

X =

Hence, x = 2,y = – 1 and z = 4

A =

|A| = 2(0) + 3(– 2) + 5(1)

= – 1

Now, the cofactors of A

C₁₁ = (– 1)^{1 + 1} – 4 + 4 = 0

C₂₁ = (– 1)^{2 + 1} 6 – 5 = – 1

C₃₁ = (– 1)^{3 + 1} 12 – 10 = 2

C₁₂ = (– 1)^{1 + 2} – 6 + 4 = 2

C₂₂ = (– 1)^{2 + 1} – 4 – 5 = – 9

C₃₂ = (– 1)^{3 + 1} – 8 – 15 = 23

C₁₃ = (– 1)^{1 + 2} 3 – 2 = 1

C₂₃ = (– 1)^{2 + 1} 2 + 3 = – 5

C₃₃ = (– 1)^{3 + 1} 4 + 9 = 13

adj A =

A ^{– 1 =}

A ^{– 1} =

A ^{– 1} =

Now the given equation can be written as:

A X B

Or, X = A ^{– 1}B

=

X =

Hence, x = 1,y = 2 and z = 3

A =

|A| = 1(1 + 3) + 2(– 1 + 2) + 5(3 + 2)

= 4 + 2 + 25

= 27

Now, the cofactors of A

C₁₁ = (– 1)^{1 + 1} 1 + 3 = 4

C₂₁ = (– 1)^{2 + 1} – 2 – 15 = 17

C₃₁ = (– 1)^{3 + 1} – 2 + 5 = 3

C₁₂ = (– 1)^{1 + 2} – 1 + 2 = – 1

C₂₂ = (– 1)^{2 + 1} – 1 – 10 = – 11

C₃₂ = (– 1)^{3 + 1} – 1 – 5 = 6

C₁₃ = (– 1)^{1 + 2} 3 + 2 = 5

C₂₃ = (– 1)^{2 + 1} 3 – 4 = 1

C₃₃ = (– 1)^{3 + 1} – 1 – 2 = – 3

adj A =

A ^{– 1 =}

A ^{– 1} =

Now the given equation can be written as:

A X B

Or, X = A ^{– 1}B

=

X =

X =

Hence, x = – 1,y = – 2 and z = 3

A =

|A| = 1(1 + 6) + 2(2 – 0) + 0

= 7 + 4

= 11

Now, the cofactors of A

C₁₁ = (– 1)^{1 + 1} 1 + 6 = 7

C₂₁ = (– 1)^{2 + 1} – 2 – 0 = 2

C₃₁ = (– 1)^{3 + 1} – 6 – 0 = – 6

C₁₂ = (– 1)^{1 + 2} 2 – 0 = – 2

C₂₂ = (– 1)^{2 + 1} 1 – 0 = 1

C₃₂ = (– 1)^{3 + 1} 3 – 0 = – 3

C₁₃ = (– 1)^{1 + 2} – 4 – 0 = – 4

C₂₃ = (– 1)^{2 + 1} – 2 – 0 = 2

C₃₃ = (– 1)^{3 + 1} 1 + 4 = 5

adj A =

A ^{– 1 =}

A ^{– 1} =

Now the given equation can be written as:

A X B

Or, X = A ^{– 1}B

=

X =

X =

Hence, x = 4,y = – 3 and z = 1

A =

|A| = 3(3 – 0) + 4(2 – 5) + 2(0 – 3)

= 9 – 12 – 6

= – 9

Now, the cofactors of A

C₁₁ = (– 1)^{1 + 1} 3 – 0 = 3

C₂₁ = (– 1)^{2 + 1} – 4 – 0 = 4

C₃₁ = (– 1)^{3 + 1} – 20 – 6 = – 26

C₁₂ = (– 1)^{1 + 2} 2 – 5 = 3

C₂₂ = (– 1)^{2 + 1} 3 – 2 = 1

C₃₂ = (– 1)^{3 + 1} 15 – 4 = – 11

C₁₃ = (– 1)^{1 + 2} 0 – 3 = – 3

C₂₃ = (– 1)^{2 + 1} 0 + 4 = – 4

C₃₃ = (– 1)^{3 + 1} 9 + 8 = 17

adj A =

A ^{– 1 =}

A ^{– 1} =

Now the given equation can be written as:

A X B

Or, X = A ^{– 1}B

=

X =

X =

Hence, x = 3,y = 2 and z = – 1

A = B =

AB =

AB =

Now, we can see that it is AB = 11I. where I is the unit Matrix

Or, A ^{– 1} =

Now the given equation can be written as:

A X B

Or, X = A ^{– 1}B

=

=

X =

Hence, x = 4,y = – 3 and z = 1

A =

|A| = 1(– 1 – 1) – 2(– 2 – 0) + 0

= – 2 + 4

= 2

Now, the cofactors of A

C₁₁ = (– 1)^{1 + 1} – 1 – 1 = – 2

C₂₁ = (– 1)^{2 + 1} 2 – 0 = 2

C₃₁ = (– 1)^{3 + 1} – 2 – 0 = – 2

C₁₂ = (– 1)^{1 + 2} 2 – 0 = – 2

C₂₂ = (– 1)^{2 + 1} 1 – 0 = 1

C₃₂ = (– 1)^{3 + 1} – 1 – 0 = 1

C₁₃ = (– 1)^{1 + 2} – 2 – 0 = – 2

C₂₃ = (– 1)^{2 + 1} – 1 – 0 = 1

C₃₃ = (– 1)^{3 + 1} – 1 + 4 = 3

adj A =

A ^{– 1 =}

A ^{– 1} =

Now the given equation can be written as:

A X B

Or, X = A ^{– 1}B

=

X =

Hence, x = – 3,y = 2.5 and z = – 4.5

B = A =

BA =

BA =

Now, we can see that it is BA = 6I. where I is the unit Matrix

Or, B ^{– 1} =

Now the given equation can be written as:

A X B

Or, X = B ^{– 1}A

=

=

X =

Hence, x = – 8,y = – 15 and z =

Let the numbers are x, y, z

X + y + z = 2 ……(i)

Also, 2y + (x + z) + 1

X + 2y + z = 1 ……(ii)

Again,

x + z + 5(x) = 6

5x + y + z = 6 …… (iii)

A X = B

|A| = 1(1) – 1(– 4) + 1(– 9)

= 1 + 4 – 9

= – 4

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (2 – 1) = 1

C₁₂ = (– 1)^{1 + 2} (1 – 5) = 4

C₁₃ = (– 1)^{1 + 3} (1 – 10) = – 9

C₂₁ = (– 1)^{2 + 1} (1 – 1) = 0

C₂₂ = (– 1)^{2 + 2} (1 – 5) = – 4

C₂₃ = (– 1)^{2 + 3} (1 – 5) = 4

C₃₁ = (– 1)^{3 + 1} (1 – 2) = – 1

C₃₂ = (– 1)^{3 + 2} (1 – 1) = 0

C₃₃ = (– 1)^{3 + 3} (2 – 1) = 1

X = A ^{– 1} B =

Adj A =

X =

X =

=

Hence,

Let the numbers are x, y,z

x + y + z = 10,000 ……(i)

Also,

0.1x + 0.12y + 0.15z = 1310 …… (ii)

Again,

0.1x + 0.12y – 0.15z = – 190 …… (iii)

A X = B

|A| = 1(– 0.036) – 1(– 0.03) + 1(0)

= – 0.006

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} – 0.036

C₁₂ = 0.27

C₁₃ = 0

C₂₁ = 0.27

C₂₂ = – 0.25

C₂₃ = – 0.02

C₃₁ = 0.03

C₃₂ = – 0.05

C₃₃ = 0.02

X = A ^{– 1} B =

Adj A =

X =

X =

=

Hence, x = Rs 2000, y = Rs 3000 and z = Rs 5000

Let the numbers are x, y, z

x + y + z = 45 ……(i)

Also,

Z – x = 8 ……(ii)

Again,

x + z = 2y …… (iii)

A X = B

|A| = 1(2) – 1(– 2) + 1(2)

= 6

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (0 + 2) = 2

C₁₂ = (– 1)^{1 + 2} (– 1 – 1) = 2

C₁₃ = (– 1)^{1 + 3} (2 – 0) = 2

C₂₁ = (– 1)^{2 + 1} (1 + 2) = – 3

C₂₂ = (– 1)^{2 + 2} (1 – 1) = 0

C₂₃ = (– 1)^{2 + 3} (– 2 – 1 ) = 3

C₃₁ = (– 1)^{3 + 1} (1 – 0) = 1

C₃₂ = (– 1)^{3 + 2} (1 + 1) = – 2

C₃₃ = (– 1)^{3 + 3} (0 + 1) = 1

X = A ^{– 1} B =

Adj A =

X =

X =

=

Hence, x = 11, y = 15 and z = 19

Let the numbers are x, y, z

3x + 5y – 4 z = 6000 …… (i)

Also,

2x – 3y + z = 5000 …… (ii)

Again,

– x + 4y + 6z = 13000 ……(iii)

A X = B

|A| = 3(– 18 – 4) – 2(30 + 16) – 1(5 – 12)

= 3(– 22) – 2(46) + 7

= – 66 – 92 + 7

= – 151

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (– 18 – 4) = – 22

C₁₂ = (– 1)^{1 + 2} (12 + 1) = – 13

C₁₃ = (– 1)^{1 + 3} (8 – 3) = 5

C₂₁ = (– 1)^{2 + 1} (30 + 16) = – 46

C₂₂ = (– 1)^{2 + 2} (18 – 4) = 14

C₂₃ = (– 1)^{2 + 3} (12 + 5 ) = – 17

C₃₁ = (– 1)^{3 + 1} (5 – 12) = – 7

C₃₂ = (– 1)^{3 + 2} (3 + 8) = – 11

C₃₃ = (– 1)^{3 + 3} (– 9 – 10) = – 19

Adj A =

X = A ^{– 1} B =

X =

X =

=

Hence, x = 3000, y = 1000 and z = 2000

Let the numbers are x, y, z

3x + 5y – 4 z = 6000 …… (i)

Also,

2x – 3y + z = 5000 ……(ii)

Again,

– x + 4y + 6z = 13000 …… (iii)

A X = B

|A| = 1(3 + 6) – 1(2 – 3) + 1(– 4 – 3)

= 1(9) – 1(– 1) – 7

= 9 + 1 – 7

= 3

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (3 + 6) = 9

C₁₂ = (– 1)^{1 + 2} (2 – 3) = 1

C₁₃ = (– 1)^{1 + 3} (– 4 – 3) = – 7

C₂₁ = (– 1)^{2 + 1} (1 + 2) = – 3

C₂₂ = (– 1)^{2 + 2} (1 – 1) = 0

C₂₃ = (– 1)^{2 + 3} (– 2 – 1 ) = 3

C₃₁ = (– 1)^{3 + 1} (3 – 3) = 0

C₃₂ = (– 1)^{3 + 2} (3 – 2) = – 1

C₃₃ = (– 1)^{3 + 3} (3 – 2) = 1

Adj A =

X = A ^{– 1} B =

X =

X =

X =

=

Hence, x = 3, y = 4 and z = 5

Let the numbers are x, y, z be the cash awards for Honesty, Regularity and Hard Work

x + y + z = 6000 …… (i)

Also,

x + 3z = 11000 …… (ii)

Again,

x – 2y + z = 0 …… (iii)

A X = B

|A| = 1(0 + 6) – 1(1 – 3) + 1(– 2 – 0)

= 1(6) – 1(– 2) – 2

= 6 + 2 – 2

= 6

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (0 + 6) = 6

C₁₂ = (– 1)^{1 + 2} (1 – 3) = 2

C₁₃ = (– 1)^{1 + 3} (– 2 – 0) = – 2

C₂₁ = (– 1)^{2 + 1} (1 + 2) = – 3

C₂₂ = (– 1)^{2 + 2} (1 – 1) = 0

C₂₃ = (– 1)^{2 + 3} (– 2 – 1 ) = 3

C₃₁ = (– 1)^{3 + 1} (3 – 0) = 3

C₃₂ = (– 1)^{3 + 2} (3 – 1) = – 2

C₃₃ = (– 1)^{3 + 3} (0 – 1) = – 1

Adj A =

X = A ^{– 1} B =

X =

X =

=

Hence, x = 500, y = 2000 and z = 3500

Let the numbers are x, y, z be the cash awards for Resourcefulness, Competence, and Determination respectively

4x + 3y + 2z = 37000 …… (i)

Also,

5x + 3y + 4z = 47000 …… (ii)

Again,

x + y + z = 12000 …… (iii)

A X = B

|A| = 4(3 – 4) – 3(5 – 4) + 2(5 – 3)

= 4(– 1) – 3(1) + 2(2)

= – 4 – 3 + 4

= – 3

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (3 – 4) = – 1

C₁₂ = (– 1)^{1 + 2} (5 – 4) = – 1

C₁₃ = (– 1)^{1 + 3} (5 – 3) = 2

C₂₁ = (– 1)^{2 + 1} (3 – 2) = – 1

C₂₂ = (– 1)^{2 + 2} (4 – 2) = 2

C₂₃ = (– 1)^{2 + 3} (4 – 3 ) = – 1

C₃₁ = (– 1)^{3 + 1} (12 – 6) = 6

C₃₂ = (– 1)^{3 + 2} (16 – 10) = – 6

C₃₃ = (– 1)^{3 + 3} (12 – 15) = – 3

Adj A =

X = A ^{– 1} B =

X =

X =

=

Hence, x = 4000, y = 5000 and z = 3000

Thus, The value X, Y, Z describes the amount of prizes per person for Resourcefulness, Competence and Determination.

Let the numbers are x, y,z be the prize amount per person for adaptability, carefulness and calmness respectively

As per the given data we get,

2x + 4y + 3z = 29000

5x + 2y + 3z = 30500

X + y + z = 9500

These three equations can be written as

A X = B

|A| = 2(2 – 3) – 4(5 – 3) + 3(5 – 2)

= 2(– 1) – 4(2) + 3(3)

= – 2 – 8 + 9

= – 1

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (2 – 3) = – 1

C₁₂ = (– 1)^{1 + 2} (5 – 3) = – 2

C₁₃ = (– 1)^{1 + 3} (5 – 2) = 3

C₂₁ = (– 1)^{2 + 1} (4 – 3) = – 1

C₂₂ = (– 1)^{2 + 2} (2 – 3) = – 1

C₂₃ = (– 1)^{2 + 3} (2 – 4 ) = – 2

C₃₁ = (– 1)^{3 + 1} (12 – 6) = 6

C₃₂ = (– 1)^{3 + 2} (6 – 15) = – 9

C₃₃ = (– 1)^{3 + 3} (4 – 20) = – 16

Adj A =

X = A ^{– 1} B =

X =

X =

X =

=

Hence, x = 2500, y = 3000 and z = 4000

Let the numbers are x, y,z be the prize amount per person for sincerity, truthfulness and helpfulness respectively

As per the given data we get,

3x + 2y + z = 1600

4x + y + 3z = 2300

x + y + z = 900

These three equations can be written as

A X = B

|A| = 3(1 – 3) – 2(4 – 3) + 1(4 – 1)

= 3(– 2) – 2(1) + 1(3)

= – 6 – 2 + 3

= – 5

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (1 – 3) = – 2

C₁₂ = (– 1)^{1 + 2} (4 – 3) = – 1

C₁₃ = (– 1)^{1 + 3} (4 – 1) = 3

C₂₁ = (– 1)^{2 + 1} (2 – 1) = – 1

C₂₂ = (– 1)^{2 + 2} (3 – 1) = 2

C₂₃ = (– 1)^{2 + 3} (3 – 2 ) = – 1

C₃₁ = (– 1)^{3 + 1} (6 – 1) = 5

C₃₂ = (– 1)^{3 + 2} (9 – 4) = – 5

C₃₃ = (– 1)^{3 + 3} (3 – 8) = – 5

Adj A =

X = A ^{– 1} B =

X =

X =

X =

=

Hence, x = 200, y = 300 and z = 400

Excellence in extra curricular activities should be another value considered for an award.

x,y and z be the prize amount per student for Discipline, Politeness and Punctuality respectively.

3x + 2y + z = 1000

4x + y + 3z = 1500

x + y + z = 600

These three equations can be written as

A X = B

|A| = 3(1 – 3) – 2(4 – 3) + 1(4 – 1)

= 3(– 2) – 2(1) + 1(3)

= – 6 – 2 + 3

= – 5

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (1 – 3) = – 2

C₁₂ = (– 1)^{1 + 2} (4 – 3) = – 1

C₁₃ = (– 1)^{1 + 3} (4 – 1) = 3

C₂₁ = (– 1)^{2 + 1} (2 – 1) = – 1

C₂₂ = (– 1)^{2 + 2} (3 – 1) = 2

C₂₃ = (– 1)^{2 + 3} (3 – 2 ) = – 1

C₃₁ = (– 1)^{3 + 1} (6 – 1) = 5

C₃₂ = (– 1)^{3 + 2} (9 – 4) = – 5

C₃₃ = (– 1)^{3 + 3} (3 – 8) = – 5

Adj A =

X = A ^{– 1} B =

X =

X =

X =

=

Hence, x = 100, y = 200 and z = 300

x, y and z be the prize amount per student for Discipline, Politeness and Punctuality respectively.

3x + 2y + z = 2200

4x + y + 3z = 3100

x + y + z = 1200

These three equations can be written as

A X = B

|A| = 3(1 – 3) – 2(4 – 3) + 1(4 – 1)

= 3(– 2) – 2(1) + 1(3)

= – 6 – 2 + 3

= – 5

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (1 – 3) = – 2

C₁₂ = (– 1)^{1 + 2} (4 – 3) = – 1

C₁₃ = (– 1)^{1 + 3} (4 – 1) = 3

C₂₁ = (– 1)^{2 + 1} (2 – 1) = – 1

C₂₂ = (– 1)^{2 + 2} (3 – 1) = 2

C₂₃ = (– 1)^{2 + 3} (3 – 2 ) = – 1

C₃₁ = (– 1)^{3 + 1} (6 – 1) = 5

C₃₂ = (– 1)^{3 + 2} (9 – 4) = – 5

C₃₃ = (– 1)^{3 + 3} (3 – 8) = – 5

Adj A =

X = A ^{– 1} B =

X =

X =

X =

=

Hence, x = 300, y = 400 and z = 500

Let the deposited be x, y and z respectively.

As per the Data we get,

x + y + z = 7000

5%x + 8%y + 8.5%z = 550

i.e 5x + 8y + 8.5z = 55000

x – y = 0

These three equations can be written as

A X = B

|A| = 1(0 + 8.5) – 1(0 – 8.5) + 1(– 5 – 8)

= 1(8.5) – 1(– 8.5) + 1(– 13)

= 8.5 + 8.5 – 13

= 4

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (0 + 8.5) = 8.5

C₁₂ = (– 1)^{1 + 2} (0 – 8.5) = 8.5

C₁₃ = (– 1)^{1 + 3} (– 5 – 8) = – 13

C₂₁ = (– 1)^{2 + 1} (0 + 1) = – 1

C₂₂ = (– 1)^{2 + 2} (0 – 1) = – 1

C₂₃ = (– 1)^{2 + 3} (– 1 – 1 ) = 2

C₃₁ = (– 1)^{3 + 1} (8.5 – 8) = 0.5

C₃₂ = (– 1)^{3 + 2} (8.5 – 5) = – 3.5

C₃₃ = (– 1)^{3 + 3} (8 – 5) = 3

Adj A =

X = A ^{– 1} B =

X =

X =

X =

=

Hence, x = 1125, y = 1125 and z = 4750

Let the varieties of pen A, B and C be x, y and z respectively.

As per the Data we get,

x + y + z = 21

4x + 3y + 2z = 60

6x + 2y + 3z = 70

These three equations can be written as

A X = B

|A| = 1(9 – 4) – 1(12 – 12) + 1(8 – 18)

= 1(5) – 1(0) + 1(– 10)

= 5 – 0 – 10

= – 5

Hence, the unique solution given by x = A ^{– 1}B

C_{11 =} (– 1)^{1 + 1} (9 – 4) = 5

C₁₂ = (– 1)^{1 + 2} (12 – 12) = 0

C₁₃ = (– 1)^{1 + 3} (8 – 18) = – 10

C₂₁ = (– 1)^{2 + 1} (3 – 2) = – 1

C₂₂ = (– 1)^{2 + 2} (3 – 6) = – 3

C₂₃ = (– 1)^{2 + 3} (2 – 6 ) = 4

C₃₁ = (– 1)^{3 + 1} (2 – 3) = – 1

C₃₂ = (– 1)^{3 + 2} (2 – 4) = 2

C₃₃ = (– 1)^{3 + 3} (3 – 4) = – 1

Adj A =

X = A ^{– 1} B =

X =

X =

=

=

Hence, A = Rs 5, B = Rs 8 and C = Rs 8

2x – y + z = 0

3x + 2y – z = 0

X + 4y + 3z = 0

The system can be written as

A X = 0

Now, |A| = 2(6 + 4) + 1(9 + 1) + 1(12 – 2)

|A| = 2(10) + 10 + 10

|A| = 40 ≠ 0

Since, |A|≠ 0, hence x = y = z = 0 is the only solution of this homogeneous equation.

The system can be written as

A X = 0

Now, |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)

|A| = – 20 – 24 + 44

|A| = 0

Hence, the system has infinite solutions

Let z = k

2x – y = – 2k

5x + 3y = k

A X = B

|A| = 6 + 5 = 11≠0 So, A ^{– 1} exist

Now adj A = =

X = A ^{– 1} B =

X =

Hence, x = , y = and z = k

The system can be written as

A X = 0

Now, |A| = 3(12 – 21) + 1(16 – 15) + 2(28 – 15)

|A| = – 27 + 1 + 26

|A| = 0

Hence, the system has infinite solutions

Let z = k

3x – y = – 2k

4x + 3y = – 3k

A X = B

|A| = 9 + 4 = 13 ≠0 So, A ^{– 1} exist

Now adj A = =

X = A ^{– 1} B =

X =

Hence, x = , y = and z = k

The system can be written as

A X = 0

Now, |A| = 1(– 2 – 2) – 1(2 + 6) – 6(1 – 3)

|A| = – 4 – 8 + 12

|A| = 0

Hence, the system has infinite solutions

Let z = k

X + y = 6k

x – y = – 2k

A X = B

|A| = – 1 – 1 = – 2 ≠0 So, A ^{– 1} exist

Now adj A = =

X = A ^{– 1} B =

X =

X =

Hence, x = 2k, y = 4k and z = k

The system can be written as

A X = 0

Now, |A| = 1(6) – 1(9) + 1(3) = 0

|A| = 6 – 9 + 3

|A| = 0

Hence, the system has infinite solutions

Let z = k

X + y = – k

x – y = 5k

A X = B

|A| = – 1 – 1 = – 2 ≠0 So, A ^{– 1} exist

Now adj A = =

X = A ^{– 1} B =

X =

X =

Hence, x = 2k, y = – 3k and z = k

The system can be written as

A X = 0

Now, |A| = 1(4) – 1(– 8) – 1(12) = 0

|A| = 4 + 8 – 12

|A| = 0

Hence, the system has infinite solutions

Let z = k

X + y = k

x – 2y = – k

A X = B

|A| = – 2 – 1 = – 3 ≠0 So, A ^{– 1} exist

Now adj A = =

X = A ^{– 1} B =

X =

X =

Hence, x = , y = and z = k

The system can be written as

A X = 0

Now, |A| = 3(1 + 2) – 1(1 – 1) – 2(– 2 – 1) = 0

|A| = 9 – 0 + 6

|A| = 15

Hence, the given system has only trivial solution given by x = y = z = 0

The system can be written as

A X = 0

Now, |A| = 2(– 3 + 2) – 3(3 + 6) – 1(1 + 3) = 0

|A| = – 2 – 27 – 4

|A| = – 33

Hence, the given system has only trivial solution given by x = y = z = 0

The given system of equations is

x + y + z = 2

3x – y + 2z = 6

3x + y + z = -18

The matrix equation corresponding to this system is :-

Let A =

|A| = = 1(-1-2) -1(3-6) + 1(3+3)

= -3+3+6 = 6

i.e., |A| 0

A unique solution of the system exists.

The given system of equations is :-

x– 3y +2z = 1

x + 4y – 3z = 5

As there are three variables x,y,z and we have only two equations so it is impossible to find the solution. Thus, no solution exists for this system of equations.

Given that

X = , A = and B =

Also AX = B and we have to find the value of X ,

Pre-multiplying A^-1 both sides we get,

A^-1AX = A^-1B

IX = A^-1B ( A^-1A = I )

X = A^-1B ( IX = X) ……(i)

Now,

|A| = = 1(0-2) +1(2-3) +2(4-0) = -2-1+8 = 5

And adjA =

A^-1 = adjA =

From (i),

= = =

=

On comparing both sides we get,

The given system of equations is :-

2x + y – z = 7

x – 3y + 2z = 1

x + 4y – 3z = 5

The matrix equation corresponding to the above system is

Let A = and B =

|A| = = 2(9-8) -1(-3-2) -1(4+3) = 2+5-7 = 0

and adjA =

(adjA)B = = 0

So, No solution exists.

The system of linear equations:

x + y + z = 2

2x + y – z = 3

3x + 2y + kz = 4

The matrix equation corresponding to the above system is

=

Let A =

|A| = = 1(k+2) -1(2k+3) +1(4-3) = k+2-2k-3+1 = -k

For the system to have a unique solution , it is necessary that |A| 0

-k 0

Thus,

The given system of linear equations is :-

a₁x + b₁y + c₁z = 0

a₂x + b₂y + c₂z = 0

a₃x + b₃y + c₃z = 0

The matrix equation corresponding to the above system is :-

According to the question,

= 0

This matrix is a singular matrix. So, the system has infinitely many solutions including the trivial solution.

The given system of linear equations is :-

+ - = 1

- + = 1

- + + = 1

Let = p , = q , = r

The equation becomes,

+ - = 1

- + = 1

- + + = 1

The matrix equation corresponding to the above system of equation is

Let A =

|A| =

Taking common from , from and from we get,

|A| = = 0

A unique solution exists.

For the system of equations:

x + 2y + 3z = 1

2x + y + 3z = 2

5x + 5y + 9z = 4

The matrix equation corresponding to the above system is

Let A =

|A| = = 1(9-8) -2(18-15) +3(10-5) = 1-6+15 = 10

i.e., |A| 0

Hence, a unique solution exists of the above system.

The given system of linear equation :-

x + y + z = λ

5x – y + μz = 10

2x + 3y – z = 6

The matrix equation corresponding to the above system is :-

Let A =

|A| = = 1(1-3 -1(-5-2) +1(15+2)

= 1-3 +5 +2 +17 = 23-

For the existence of the unique solution, the value of |A| must be equal to 0.

Hence the existence of the unique solution merely depends on the value of

The given system of linear equations is :-

x + y + z = 5

x + 2y + 3z = 9

x + 3y + λz = μ

The matrix equation corresponding to the above system is

Let A =

|A| = = 1(2

= 2-9 –+3+1 =

For the existence of a unique solution |A| 0

0

The given matrix equation is :

The first matrix is an identity matrix which is denoted by I .

Let P = and B =

Thus the equation becomes

IP = B

As we know that any matrix when multiplied with the identity matrix will be the matrix itself. But here the order of I is 3while the order of P is 3, so the order of the result matrix is 3

IP = P =

So, we have

P = B

On comparing both sides we get,

The given matrix equation is :-

Let A = and B =

So, we can write the equation as

AP = B

Pre-multiplying A^-1 both sides we get,

A^-1AP = A^-1B

IP = A^-1B ( A^-1A = I )

P = A^-1B ( IP = P ) …….(i)

Now,

|A| = = 1

and

adj A =

A^-1 = adj A

A^-1 =

From (i) ,

=

=

Comparing both sides we get,

The given matrix equation is :-

or, =

Comparing both sides we get,

The given matrix equation is :-

=

Multiplying the two matrices in the LHS we get,

=

Comparing both sides we get,

3x-4y = 10 ………….(i)

9x+y = 2 ………..(ii)

Multiplying eq.(ii) by 4 and then adding both the equations we get,

Putting this value of x in eq.(ii) we get,

+ y = 2

or, y = 2 –

The given matrix equation is :-

Multiplying the two matrices in the LHS we get,

Comparing both sides we get,

It is given that

AX = B

Where

A = and B =

And we have to find the value of n.

AX =

or, AX =

Thus,

On comparing both sides we get,

2n+4 = 8 ……………...(i)

4n+3 = 11 ……………...(ii)

From (i),

2n = 4

n = 2

and from (ii) ,

4n = 8

n = 2

Hence