NCERT solutions for Class 12 Maths Chapter 4 - Determinants

Question 1: Evaluate the following determinant- |24−5−1|

Answer:

The determinant is evaluated as follows

| 2 4 − 5 − 1 | = 2 ( − 1 ) − 4 ( − 5 ) = − 2 + 20 = 18

Question 2: Evaluate the following determinant-

(i) |cos⁡θ−sin⁡θsin⁡θcos⁡θ|

(ii) |x2−x+1x−1x+1x+1|

Answer:

(i) The given two determinants are calculated as follows

100 | cos ⁡ θ − sin ⁡ θ sin ⁡ θ cos ⁡ θ | = c o s θ ( cos ⁡ θ ) − ( − sin ⁡ θ ) sin ⁡ θ = cos 2 ⁡ θ + sin 2 ⁡ θ = 1

(ii) We have determinant | x 2 − x + 1 x − 1 x + 1 x + 1 |

So, 100 | x 2 − x + 1 x − 1 x + 1 x + 1 | = ( x 2 − x + 1 ) ( x + 1 ) − ( x − 1 ) ( x + 1 )

⇒ ( x + 1 ) ( x 2 − x + 1 − x + 1 ) = ( x + 1 ) ( x 2 − 2 x + 2 )

⇒ x 3 − 2 x 2 + 2 x + x 2 − 2 x + 2

⇒ x 3 − x 2 + 2

Question 3: If A=[1242] , then show that |2A|=4|A|

Answer:

Given determinant A = [ 1 2 4 2 ], then we have to show that | 2 A | = 4 | A |,

So, A = [ 1 2 4 2 ] then, 2 A = 2 [ 1 2 4 2 ] = [ 2 4 8 4 ]

Hence we have | 2 A | = | 2 4 8 4 | = 2 ( 4 ) − 4 ( 8 ) = − 24

So, L.H.S. = |2A| = -24

then calculating R.H.S. 4 | A |

We have,

| A | = | 1 2 4 2 | = 1 ( 2 ) − 2 ( 4 ) = − 6

hence R.H.S becomes 4 | A | = 4 × ( − 6 ) = − 24

Therefore L.H.S. = R.H.S.

Hence proved.

Question 4: If A=[101012004] then show that |3A|=27|A|

Answer:

Given Matrix A = [ 1 0 1 0 1 2 0 0 4 ]

Calculating 3 A = 3 [ 1 0 1 0 1 2 0 0 4 ] = [ 3 0 3 0 3 6 0 0 12 ]

So, | 3 A | = 3 ( 3 ( 12 ) − 6 ( 0 ) ) − 0 ( 0 ( 12 ) − 0 ( 6 ) ) + 3 ( 0 − 0 ) = 3 ( 36 ) = 108

calculating 27 | A | ,

⇒ | A | = | 1 0 1 0 1 2 0 0 4 | = 1 | 1 2 0 4 | − 0 | 0 2 0 4 | + 1 | 0 1 0 0 | = 4 − 0 + 0 = 4

So, 27 | A | = 27 ( 4 ) = 108

Therefore |3A|=27|A| .

Hence proved.

Question 5: Evaluate the determinants.

(i) | 3 − 1 − 2 0 0 − 1 3 − 5 0 |

(ii) | 3 − 4 5 1 1 − 2 2 3 1 |

(iii) | 0 1 2 − 1 0 − 3 − 2 3 0 |

(iv) | 2 − 1 2 0 2 − 1 3 − 5 0 |

Answer:

(i) Given the determinant | 3 − 1 − 2 0 0 − 1 3 − 5 0 | ;

Now, calculating its determinant value,

⇒ | 3 − 1 − 2 0 0 − 1 3 − 5 0 | = 3 | 0 − 1 − 5 0 | − ( − 1 ) | 0 − 1 3 0 | + ( − 2 ) | 0 0 3 − 5 |

= 3 ( 0 − 5 ) + 1 ( 0 + 3 ) − 2 ( 0 − 0 ) = − 15 + 3 − 0 = − 12 .

(ii) Given determinant | 3 − 4 5 1 1 − 2 2 3 1 | ;

Now calculating the determinant value;

⇒ | 3 − 4 5 1 1 − 2 2 3 1 | = 3 | 1 − 2 3 1 | − ( − 4 ) | 1 − 2 2 1 | + 5 | 1 1 2 3 |

= 3 ( 1 + 6 ) + 4 ( 1 + 4 ) + 5 ( 3 − 2 ) = 21 + 20 + 5 = 46 .

(iii) Given determinant | 0 1 2 − 1 0 − 3 − 2 3 0 | ;

Now calculating the determinant value;

⇒ | 0 1 2 − 1 0 − 3 − 2 3 0 | = 0 | 0 − 1 3 0 | − 1 | − 1 − 3 − 2 0 | + 2 | − 1 0 − 2 3 |

= 0 − 1 ( 0 − 6 ) + 2 ( − 3 − 0 ) = 6 − 6 = 0

(iv) Given determinant: | 2 − 1 − 2 0 2 − 1 3 − 5 0 | ,

We now calculate the determinant value:

⇒ | 2 − 1 − 2 0 2 − 1 3 − 5 0 | = 2 | 2 − 1 − 5 0 | − ( − 1 ) | 0 − 1 3 0 | + ( − 2 ) | 0 2 3 − 5 |

= 2 ( 0 − 5 ) + 1 ( 0 + 3 ) − 2 ( 0 − 6 ) = − 10 + 3 + 12 = 5

Question 6: If A=[11−221−354−9] , then find |A|.

Answer:

Given the matrix A = [ 1 1 − 2 2 1 − 3 5 4 − 9 ], then

Finding the determinant value of A.

| A | = 1 | 1 − 3 4 − 9 | − 1 | 2 − 3 5 − 9 | − 2 | 2 1 5 4 |

= 1 ( − 9 + 12 ) − 1 ( − 18 + 15 ) − 2 ( 8 − 5 ) = 3 + 3 − 6 = 0

Question 7: Find values of x, if

(i) | 2 4 5 1 | = | 2 x 4 6 x |

(ii) | 2 3 4 5 | = | x 3 2 x 5 |

Answer:

(i) Given that | 2 4 5 1 | = | 2 x 4 6 x |

First, we solve the determinant value of L.H.S. and equate it to the determinant value of R.H.S,

100 | 2 4 5 1 | = 2 − 20 = − 18 and | 2 x 4 6 x | = 2 x ( x ) − 24 = 2 x 2 − 24

So, we have then,

− 18 = 2 x 2 − 24 or 3 = x 2 or x = ± 3

(ii) Given | 2 3 4 5 | = | x 3 2 x 5 | ;

So, we here equate both sides after calculating each side's determinant values.

L.H.S. determinant value;

100 | 2 3 4 5 | = 10 − 12 = − 2

Similarly R.H.S. determinant value;

| x 3 2 x 5 | = 5 ( x ) − 3 ( 2 x ) = 5 x − 6 x = − x

So, we have then;

− 2 = − x or x = 2 .

Question 8: If |x218x|=|62186| , then x is equal to

(A) 6 (B) ±6 (C) −6 (D) 0

Answer:

Solving the L.H.S. determinant ;

100 | x 2 18 x | = x 2 − 36

and solving R.H.S determinant;

| 6 2 18 6 | = 36 − 36 = 0

So equating both sides;

x 2 − 36 = 0 or x 2 = 36 or x = ± 6

Hence, the answer is (B).

Question 1: Find the area of the triangle with vertices at the point given in each of the following :

(i) ( 1 , 0 ) , ( 6 , 0 ) , ( 4 , 3 )

(ii) ( 2 , 7 ) , ( 1 , 1 ) , ( 10 , 8 )

(iii) ( − 2 , − 3 ) , ( 3 , 2 ) , ( − 1 , − 8 )

Answer:

(i) We can find the area of the triangle with vertices ( 1 , 0 ) , ( 6 , 0 ) , ( 4 , 3 ) by the following determinant relation:

⇒ △ = 1 2 | 1 0 1 6 0 1 4 3 1 |

Expanding using the second column

⇒ 1 2 ( − 3 ) | 1 1 6 1 |

⇒ 15 2 s q u a r e u n i t s .

(ii) We can find the area of the triangle with given coordinates by the following method:

⇒ △ = | 2 7 1 1 1 1 10 8 1 |

⇒ 1 2 | 2 7 1 1 1 1 10 8 1 | = 1 2 [ 2 ( 1 − 8 ) − 7 ( 1 − 10 ) + 1 ( 8 − 10 ) ]

⇒ 1 2 [ 2 ( − 7 ) − 7 ( − 9 ) + 1 ( − 2 ) ] = 1 2 [ − 14 + 63 − 2 ] = 47 2 s q u a r e u n i t s .

(iii) Area of the triangle by the determinant method:

⇒ A r e a △ = 1 2 | − 2 − 3 1 3 2 1 − 1 − 8 1 |

⇒ 1 2 [ − 2 ( 2 + 8 ) + 3 ( 3 + 1 ) + 1 ( − 24 + 2 ) ]

⇒ 1 2 [ − 20 + 12 − 22 ] = 1 2 [ − 30 ] = − 15

Hence the area is equal to | − 15 | = 15 s q u a r e u n i t s.

Question 2: Show that points A(ab+c),B(b,c+a),C(c,a+b) are collinear.

Answer:

If the area formed by the points is equal to zero, then we can say that the points are collinear.

So, we have an area of a triangle given by,

⇒ △ = 1 2 | a b + c 1 b c + a 1 c a + b 1 |

Calculating the area:

⇒ 1 2 [ a | c + a 1 a + b 1 | − ( b + c ) | b 1 c 1 | + 1 | b c + a c a + b | ]

⇒ 1 2 [ a ( c + a − a − b ) − ( b + c ) ( b − c ) + 1 ( b ( a + b ) − c ( c + a ) ) ]

⇒ 1 2 [ a c − a b − b 2 + c 2 + a b + b 2 − c 2 − a c ] = 1 2 [ 0 ] = 0

Hence, the area of the triangle formed by the points is equal to zero.

Therefore, the given points A ( a , b + c ) , B ( b , c + a ) , and C ( c , a + b ) are collinear.

Question 3: Find the values of k if the area of a triangle is 4 sq. units and the vertices are

(i) ( k , 0 ) , ( 4 , 0 ) , ( 0 , 2 )

(ii) ( − 2 , 0 ) , ( 0 , 4 ) , ( 0 , k )

Answer:

(i) We can easily calculate the area by the formula :

⇒ △ = 1 2 | k 0 1 4 0 1 0 2 1 | = 4 s q . u n i t s

⇒ 1 2 [ k | 0 1 2 1 | − 0 | 4 1 0 1 | + 1 | 4 0 0 2 | ] = 4 s q . u n i t s

⇒ 1 2 [ k ( 0 − 2 ) − 0 + 1 ( 8 − 0 ) ] = 1 2 [ − 2 k + 8 ] = 4 s q . u n i t s

⇒ [ − 2 k + 8 ] = 8 s q . u n i t s or − 2 k + 8 = ± 8 s q . u n i t s

or k = 0 or k = 8

Hence, two values are possible for k.

(ii) The area of the triangle is given by the formula:

⇒ △ = 1 2 | − 2 0 1 0 4 1 0 k 1 | = 4 s q . u n i t s.

Now, calculating the area:

⇒ 1 2 | − 2 ( 4 − k ) − 0 ( 0 − 0 ) + 1 ( 0 − 0 ) | = 1 2 | − 8 + 2 k | = 4

or − 8 + 2 k = ± 8

Therefore, we have two possible values of 'k', i.e., k=8 or k=0.

Question 4: Find the equation of the line joining

(i) ( 1 , 2 ) and ( 3 , 6 ) using determinants.

(ii) ( 3 , 1 ) and ( 9 , 3 ) using determinants.

Answer:

(i) As we know, the line joining ( 1 , 2 ) , ( 3 , 6 ) and let say a point on line A ( x , y ) will be collinear.

Therefore area formed by them will be equal to zero.

⇒ △ = 1 2 | 1 2 1 3 6 1 x y 1 | = 0

So, we have:

⇒ 1 ( 6 − y ) − 2 ( 3 − x ) + 1 ( 3 y − 6 x ) = 0

or 6 − y − 6 + 2 x + 3 y − 6 x = 0 ⇒ 2 y − 4 x = 0

Hence, we have the equation of line ⇒y=2x.

(ii) We can find the equation of the line by considering any arbitrary point A ( x , y ) on line.

So, we have three collinear points, and therefore area surrounded by them will be equal to zero.

△ = 1 2 | 3 1 1 9 3 1 x y 1 | = 0

Calculating the determinant:

⇒ 1 2 [ 3 | 3 1 y 1 | − 1 | 9 1 x 1 | + 1 | 9 3 x y | ]

⇒ 1 2 [ 3 ( 3 − y ) − 1 ( 9 − x ) + 1 ( 9 y − 3 x ) ] = 0

⇒ 1 2 [ 9 − 3 y − 9 + x + 9 y − 3 x ] = 1 2 [ 6 y − 2 x ] = 0

Hence, we have the line equation:

3y=x or x−3y=0 .

Question 5: If the area of a triangle is 35 sq units with vertices (2, -6),(5,4), and (k,4). Then k is

(A) 12 (B) −2 (C) −12,−2 (D) 12,−2

Answer:

The area of a triangle is given by:

⇒ △ = 1 2 | 2 − 6 1 5 4 1 k 4 1 | = 35 s q . u n i t s .

or | 2 − 6 1 5 4 1 k 4 1 | = 70 s q . u n i t s .

⇒ 2 | 4 1 4 1 | − ( − 6 ) | 5 1 k 1 | + 1 | 5 4 k 4 | = 70

⇒ 2 ( 4 − 4 ) + 6 ( 5 − k ) + ( 20 − 4 k ) = ± 70

⇒ 50 − 10 k = ± 70

⇒ k = 12 or k = − 2

Hence, the possible values of k are 12 and -2.

Therefore, option (D) is correct.

Question 1: Write Minors and Cofactors of the elements of the following determinants:

(i) | 2 − 4 0 3 |

(ii) | a c b d |

Answer:

(i) GIven determinant: | 2 − 4 0 3 |

Minor of element a i j is M_ij.

Therefore we have

M 11 = minor of element a 11 = 3

M 12 = minor of element a 12 = 0

M 21 = minor of element a 21 = -4

M 22 = minor of element a 22 = 2

and finding cofactors of a i j is A i j = ( − 1 ) i + j M i j .

Therefore, we have:

A 11 = ( − 1 ) 1 + 1 M 11 = ( − 1 ) 2 ( 3 ) = 3

A 12 = ( − 1 ) 1 + 2 M 12 = ( − 1 ) 3 ( 0 ) = 0

A 21 = ( − 1 ) 2 + 1 M 21 = ( − 1 ) 3 ( − 4 ) = 4

A 22 = ( − 1 ) 2 + 2 M 22 = ( − 1 ) 4 ( 2 ) = 2

(ii) Given determinant: | a c b d |

Minor of element a i j is M_ij.

Therefore we have

M 11 = minor of element a 11 = d

M 12 = minor of element a 12 = b

M 21 = minor of element a 21 = c

M 22 = minor of element a 22 = a

and finding cofactors of a ij is A_ij = (− 1)^(i + j) M_ij.

Therefore, we have:

A₁₁ = ( − 1 ) 1 + 1 M 11 = ( − 1 ) 2 ( d ) = d

A 12 = ( − 1 ) 1 + 2 M 12 = ( − 1 ) 3 ( b ) = − b

A 21 = ( − 1 ) 2 + 1 M 21 = ( − 1 ) 3 ( c ) = − c

A 22 = ( − 1 ) 2 + 2 M 22 = ( − 1 ) 4 ( a ) = a

Question 2: Write Minors and Cofactors of the elements of the following determinants:

(i) | 1 0 0 0 1 0 0 0 1 |

(ii) | 1 0 4 3 5 − 1 0 1 2 |

Answer:

(i) Given determinant : | 1 0 0 0 1 0 0 0 1 |

Finding Minors: by the definition,

M 11 = minor of a 11 = | 1 0 0 1 | = 1 M 12 = minor of a 12 = | 0 0 0 1 | = 0

M 13 = minor of a 13 = | 0 1 0 0 | = 0 M 21 = minor of a 21 = | 0 0 0 1 | = 0

M 22 = minor of a 22 = | 1 0 0 1 | = 1 M 23 = minor of a 23 = | 1 0 0 0 | = 0

M 31 = minor of a 31 = | 0 0 1 0 | = 0 M 32 = minor of a 32 = | 1 0 0 0 | = 0

M 33 = minor of a 33 = | 1 0 0 1 | = 1

Finding the cofactors:

A11= cofactor of a11=(−1)1+1M11=1

A12= cofactor of a12=(−1)1+2M12=0

A13= cofactor of a13=(−1)1+3M13=0

A21= cofactor of a21=(−1)2+1M21=0

A22= cofactor of a22=(−1)2+2M22=1

A23= cofactor of a23=(−1)2+3M23=0

A31= cofactor of a31=(−1)3+1M31=0

A32= cofactor of a32=(−1)3+2M32=0

A33= cofactor of a33=(−1)3+3M33=1 .

(ii) Given determinant : | 1 0 4 3 5 − 1 0 1 2 |

Finding Minors: by the definition,

M 11 = minor of a 11 = | 5 − 1 1 2 | = 11 M 12 = minor of a 12 = | 3 − 1 0 2 | = 6

M 13 = minor of a 13 = | 3 5 0 1 | = 3 M 21 = minor of a 21 = | 0 4 1 2 | = − 4

M 22 = minor of a 22 = | 1 4 0 2 | = 2 M 23 = minor of a 23 = | 1 0 0 1 | = 1

M 31 = minor of a 31 = | 0 4 5 − 1 | = − 20

M 32 = minor of a 32 = | 1 4 3 − 1 | = − 1 − 12 = − 13

M 33 = minor of a 33 = | 1 0 3 5 | = 5

Finding the cofactors:

A11= cofactor of a11=(−1)1+1M11=11

A12= cofactor of a12=(−1)1+2M12=−6

A13= cofactor of a13=(−1)1+3M13=3

A21= cofactor of a21=(−1)2+1M21=4

A22= cofactor of a22=(−1)2+2M22=2

A23= cofactor of a23=(−1)2+3M23=−1

A31= cofactor of a31=(−1)3+1M31=−20

A32= cofactor of a32=(−1)3+2M32=13

A33= cofactor of a33=(−1)3+3M33=5 .

Question 3: Using Cofactors of elements of the second row, evaluate. Δ=|538201123|

Answer:

Given determinant : Δ = | 5 3 8 2 0 1 1 2 3 |

First finding Minors of the second rows by the definition,

M 21 = minor of a 21 = | 3 8 2 3 | = 9 − 16 = − 7

M 22 = minor of a 22 = | 5 8 1 3 | = 15 − 8 = 7

M 23 = minor of a 23 = | 5 3 1 2 | = 10 − 3 = 7

Finding the Cofactors of the second row:

A 21 = Cofactor of a 21 = ( − 1 ) 2 + 1 M 21 = 7

A 22 = Cofactor of a 22 = ( − 1 ) 2 + 2 M 22 = 7

A 23 = Cofactor of a 23 = ( − 1 ) 2 + 3 M 23 = − 7

Therefore we can calculate △ by sum of the product of the elements of the second row with their corresponding cofactors.

Therefore we have,

△=a21A21+a22A22+a23A23=2(7)+0(7)+1(−7)=14−7=7

Question 4: Using Cofactors of elements of third column, evaluate Δ=|1xyz1yzx1zxy|

Answer:

Given determinant : Δ = | 1 x y z 1 y z x 1 z x y |

First finding Minors of the third column by the definition,

M 13 = minor of a 13 = | 1 y 1 z | = z − y

M 23 = minor of a 23 = | 1 x 1 z | = z − x

M 33 = minor of a 33 = | 1 x 1 y | = y − x

Finding the Cofactors of the second row:

A 13 = Cofactor of a 13 = ( − 1 ) 1 + 3 M 13 = z − y

A 23 = Cofactor of a 23 = ( − 1 ) 2 + 3 M 23 = x − z

A 33 = Cofactor of a 33 = ( − 1 ) 3 + 3 M 33 = y − x

Therefore we can calculate △ by sum of the product of the elements of the third column with their corresponding cofactors.

Therefore we have,

△=a13A13+a23A23+a33A33

=(z−y)yz+(x−z)zx+(y−x)xy

=yz2−y2z+zx2−xz2+xy2−x2y

=z(x2−y2)+z2(y−x)+xy(y−x)

=(x−y)[zx+zy−z2−xy]

=(x−y)[z(x−z)+y(z−x)]

=(x−y)(z−x)[−z+y]

=(x−y)(y−z)(z−x)

Thus, we have value of △=(x−y)(y−z)(z−x) .

Question 5: If Δ=|a11a12a13a21a22a23a31a32a33| and Aij is Cofactors of aij , then the value of Δ is given by

(A) a11A31+a12A32+a13A33

(B) a11A11+a12A21+a13A31

(C) a21A11+a22A12+a23A13

(D) a11A11+a21A21+a31A31

Answer:

The answer is (D) a 11 A 11 + a 21 A 21 + a 31 A 31 by the definition itself, Δ is equal to the product of the elements of the row/column with their corresponding cofactors.

Question 1: Find the adjoint of each of the matrices.

[ 1 2 3 4 ]

Answer:

Given matrix: [ 1 2 3 4 ] = A

Then we have,

A 11 = 4 , A 12 = − ( 1 ) 3 , A 21 = − ( 1 ) 2 , a n d A 22 = 1

Hence we get:

a d j A = [ A 11 A 12 A 21 A 22 ] T = [ A 11 A 21 A 12 A 22 ] = [ 4 − 2 − 3 1 ]

Question 2: Find the adjoint of each of the matrices

[ 1 − 1 2 2 3 5 − 2 0 1 ]

Answer:

Given the matrix: A = [ 1 − 1 2 2 3 5 − 2 0 1 ]

Then we have,

A 11 = ( − 1 ) 1 + 1 | 3 5 0 1 | = ( 3 − 0 ) = 3

A 12 = ( − 1 ) 1 + 2 | 2 5 − 2 1 | = − ( 2 + 10 ) = − 12

A 13 = ( − 1 ) 1 + 3 | 2 3 − 2 0 | = 0 + 6 = 6

A 21 = ( − 1 ) 2 + 1 | − 1 2 0 1 | = − ( − 1 − 0 ) = 1

A 22 = ( − 1 ) 2 + 2 | 1 2 − 2 1 | = ( 1 + 4 ) = 5

A 23 = ( − 1 ) 2 + 3 | 1 − 1 − 2 0 | = − ( 0 − 2 ) = 2

A 31 = ( − 1 ) 3 + 1 | − 1 2 3 5 | = ( − 5 − 6 ) = − 11

A 32 = ( − 1 ) 3 + 2 | 1 2 2 5 | = − ( 5 − 4 ) = − 1

A 33 = ( − 1 ) 3 + 3 | 1 − 1 2 3 | = ( 3 + 2 ) = 5

Hence we get:

a d j A = [ A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ] = [ 3 1 − 11 − 12 5 − 1 6 2 5 ]

Question 3: Verify A(adjA)=(adjA)A=|A|I .

[ 2 3 − 4 − 6 ]

Answer:

Given the matrix: [ 2 3 − 4 − 6 ]

Let A = [ 2 3 − 4 − 6 ]

Calculating the cofactors;

A 11 = ( − 1 ) 1 + 1 ( − 6 ) = − 6

A 12 = ( − 1 ) 1 + 2 ( − 4 ) = 4

A 21 = ( − 1 ) 2 + 1 ( 3 ) = − 3

A 22 = ( − 1 ) 2 + 2 ( 2 ) = 2

Hence, a d j A = [ − 6 − 3 4 2 ]

Now,

A ( a d j A ) = [ 2 3 − 4 − 6 ] ( [ − 6 − 3 4 2 ] )

[ − 12 + 12 − 6 + 6 24 − 24 12 − 12 ] = [ 0 0 0 0 ]

also,

( a d j A ) A = [ − 6 − 3 4 2 ] [ 2 3 − 4 − 6 ]

= [ − 12 + 12 − 18 + 18 8 − 8 12 − 12 ] = [ 0 0 0 0 ]

Now, calculating |A|;

|A|=−12−(−12)=−12+12=0

So, |A|I=0[1001]=[0000]

Hence we get

A ( a d j A ) = ( a d j A ) A = | A | I

Question 4: Verify A(adjA)=(adjA)A=|A|I .

[ 1 − 1 2 3 0 − 2 1 0 3 ]

Answer:

Given matrix: [ 1 − 1 2 3 0 − 2 1 0 3 ]

Let A = [ 1 − 1 2 3 0 − 2 1 0 3 ]

Calculating the cofactors;

A 11 = ( − 1 ) 1 + 1 | 0 − 2 0 3 | = 0

A 12 = ( − 1 ) 1 + 2 | 3 − 2 1 3 | = − ( 9 + 2 ) = − 11

A 13 = ( − 1 ) 1 + 3 | 3 0 1 0 | = 0

A 21 = ( − 1 ) 2 + 1 | − 1 2 0 3 | = − ( − 3 − 0 ) = 3

A 22 = ( − 1 ) 2 + 2 | 1 2 1 3 | = 3 − 2 = 1

A 23 = ( − 1 ) 2 + 3 | 1 − 1 1 0 | = − ( 0 + 1 ) = − 1

A 31 = ( − 1 ) 3 + 1 | − 1 2 0 − 2 | = 2

A 32 = ( − 1 ) 3 + 2 | 1 2 3 − 2 | = − ( − 2 − 6 ) = 8

A 33 = ( − 1 ) 3 + 3 | 1 − 1 3 0 | = 0 + 3 = 3

Hence, a d j A = [ 0 3 2 − 11 1 8 0 − 1 3 ]

Now,

A ( a d j A ) = [ 1 − 1 2 3 0 − 2 1 0 3 ] [ 0 3 2 − 11 1 8 0 − 1 3 ]

= [ 0 + 11 + 0 3 − 1 − 2 2 − 8 + 6 0 + 0 + 0 9 + 0 + 2 6 + 0 − 6 0 + 0 + 0 3 + 0 − 3 2 + 0 + 9 ] = [ 11 0 0 0 11 0 0 0 11 ]

also,

A ( a d j A ) = [ 0 3 2 − 11 1 8 0 − 1 3 ] [ 1 − 1 2 3 0 − 2 1 0 3 ]

= [ 0 + 9 + 2 0 + 0 + 0 0 − 6 + 6 − 11 + 3 + 8 11 + 0 + 0 − 22 − 2 + 24 0 − 3 + 3 0 + 0 + 0 0 + 2 + 9 ] = [ 11 0 0 0 11 0 0 0 11 ]

Now, calculating |A|;

|A|=1(0−0)+1(9+2)+2(0−0)=11

So, |A|I=11[100010001]=[110001100011]

Hence we get,

A ( a d j A ) = ( a d j A ) A = | A | I .

Question 5: Find the inverse of each of the matrices (if it exists).

[ 2 − 2 4 3 ]

Answer:

Given matrix : [ 2 − 2 4 3 ]

To find the inverse we have to first find adj A then as we know the relation:

A−1=1|A|adjA

So, calculating |A| :

|A| = (6+8) = 14

Now, calculate the cofactors terms and then adj A.

A 11 = ( − 1 ) 1 + 1 ( 3 ) = 3

A 12 = ( − 1 ) 1 + 2 ( 4 ) = − 4

A 21 = ( − 1 ) 2 + 1 ( − 2 ) = 2

A 22 = ( − 1 ) 2 + 2 ( 2 ) = 2

So, we have a d j A = [ 3 2 − 4 2 ]

Therefore inverse of A will be:

A−1=1|A|adjA

=114[32−42]=[31417−2717]

Question 6: Find the inverse of each of the matrices (if it exists).

[ − 1 5 − 3 2 ]

Answer:

Given the matrix : [ − 1 5 − 3 2 ] = A

To find the inverse we have to first find adjA then as we know the relation:

A−1=1|A|adjA

So, calculating |A| :

|A| = (-2+15) = 13

Now, calculate the cofactors terms and then adjA.

A 11 = ( − 1 ) 1 + 1 ( 2 ) = 2

A 12 = ( − 1 ) 1 + 2 ( − 3 ) = 3

A 21 = ( − 1 ) 2 + 1 ( 5 ) = − 5

A 22 = ( − 1 ) 2 + 2 ( − 1 ) = − 1

So, we have a d j A = [ 2 − 5 3 − 1 ]

Therefore inverse of A will be:

A−1=1|A|adjA

=113[2−53−1]=[213−513313−113]

Question 7: Find the inverse of each of the matrices (if it exists).

[ 1 2 3 0 2 4 0 0 5 ]

Answer:

Given the matrix : [ 1 2 3 0 2 4 0 0 5 ] = A

To find the inverse we have to first find adjA then as we know the relation:

A−1=1|A|adjA

So, calculating |A| :

| A | = 1 ( 10 − 0 ) − 2 ( 0 − 0 ) + 3 ( 0 − 0 ) = 10

Now, calculate the cofactors terms and then adjA.

A 11 = ( − 1 ) 1 + 1 ( 10 ) = 10 A 12 = ( − 1 ) 1 + 2 ( 0 ) = 0

A 13 = ( − 1 ) 1 + 3 ( 0 ) = 0 A 21 = ( − 1 ) 2 + 1 ( 10 ) = − 10

A 22 = ( − 1 ) 2 + 2 ( 5 − 0 ) = 5 A 23 = ( − 1 ) 2 + 1 ( 0 − 0 ) = 0

A 31 = ( − 1 ) 3 + 1 ( 8 − 6 ) = 2 A 32 = ( − 1 ) 3 + 2 ( 4 − 0 ) = − 4

A 33 = ( − 1 ) 3 + 3 ( 2 − 0 ) = 2

So, we have a d j A = [ 10 − 10 2 0 5 − 4 0 0 2 ]

Therefore inverse of A will be:

A−1=1|A|adjA

=110[10−10205−4002]

Question 8: Find the inverse of each of the matrices (if it exists).

[ 1 0 0 3 3 0 5 2 − 1 ]

Answer:

Given the matrix : [ 1 0 0 3 3 0 5 2 − 1 ] = A

To find the inverse we have to first find adjA then as we know the relation:

A−1=1|A|adjA

So, calculating |A| :

| A | = 1 ( − 3 − 0 ) − 0 ( − 3 − 0 ) + 0 ( 6 − 15 ) = − 3

Now, calculate the cofactors terms and then adjA.

A 11 = ( − 1 ) 1 + 1 ( − 3 − 0 ) = − 3 A 12 = ( − 1 ) 1 + 2 ( − 3 − 0 ) = 3

A 13 = ( − 1 ) 1 + 3 ( 6 − 15 ) = − 9 A 21 = ( − 1 ) 2 + 1 ( 0 − 0 ) = 0

A 22 = ( − 1 ) 2 + 2 ( − 1 − 0 ) = − 1 A 23 = ( − 1 ) 2 + 1 ( 2 − 0 ) = − 2

A 31 = ( − 1 ) 3 + 1 ( 0 − 0 ) = 0 A 32 = ( − 1 ) 3 + 2 ( 0 − 0 ) = 0

A 33 = ( − 1 ) 3 + 3 ( 3 − 0 ) = 3

So, we have a d j A = [ − 3 0 0 3 − 1 0 − 9 − 2 3 ]

Therefore inverse of A will be:

A−1=1|A|adjA

=−13[−3003−10−9−23]

Question 9: Find the inverse of each of the matrices (if it exists).

[ 2 1 3 4 − 1 0 − 7 2 1 ]

Answer:

Given the matrix : [ 2 1 3 4 − 1 0 − 7 2 1 ] = A

To find the inverse we have to first find adjA then as we know the relation:

A−1=1|A|adjA

So, calculating |A| :

| A | = 2 ( − 1 − 0 ) − 1 ( 4 − 0 ) + 3 ( 8 − 7 ) = − 2 − 4 + 3 = − 3

Now, calculate the cofactors terms and then adjA.

A 11 = ( − 1 ) 1 + 1 ( − 1 − 0 ) = − 1 A 12 = ( − 1 ) 1 + 2 ( 4 − 0 ) = − 4

A 13 = ( − 1 ) 1 + 3 ( 8 − 7 ) = 1 A 21 = ( − 1 ) 2 + 1 ( 1 − 6 ) = 5

A 22 = ( − 1 ) 2 + 2 ( 2 + 21 ) = 23 A 23 = ( − 1 ) 2 + 1 ( 4 + 7 ) = − 11

A 31 = ( − 1 ) 3 + 1 ( 0 + 3 ) = 3 A 32 = ( − 1 ) 3 + 2 ( 0 − 12 ) = 12

A 33 = ( − 1 ) 3 + 3 ( − 2 − 4 ) = − 6

So, we have a d j A = [ − 1 5 3 − 4 23 12 1 − 11 − 6 ]

Therefore inverse of A will be:

A−1=1|A|adjA

A−1=1−3[−153−423121−11−6]

Question 10: Find the inverse of each of the matrices (if it exists).

[ 1 − 1 2 0 2 − 3 3 − 2 4 ]

Answer:

Given the matrix : [ 1 − 1 2 0 2 − 3 3 − 2 4 ] = A

To find the inverse we have to first find adjA then as we know the relation:

A−1=1|A|adjA

So, calculating |A| :

| A | = 1 ( 8 − 6 ) + 1 ( 0 + 9 ) + 2 ( 0 − 6 ) = 2 + 9 − 12 = − 1

Now, calculate the cofactors terms and then adjA.

A 11 = ( − 1 ) 1 + 1 ( 8 − 6 ) = 2 A 12 = ( − 1 ) 1 + 2 ( 0 + 9 ) = − 9

A 13 = ( − 1 ) 1 + 3 ( 0 − 6 ) = − 6 A 21 = ( − 1 ) 2 + 1 ( − 4 + 4 ) = 0

A 22 = ( − 1 ) 2 + 2 ( 4 − 6 ) = − 2 A 23 = ( − 1 ) 2 + 1 ( − 2 + 3 ) = − 1

A 31 = ( − 1 ) 3 + 1 ( 3 − 4 ) = − 1 A 32 = ( − 1 ) 3 + 2 ( − 3 − 0 ) = 3

A 33 = ( − 1 ) 3 + 3 ( 2 − 0 ) = 2

So, we have a d j A = [ 2 0 − 1 − 9 − 2 3 − 6 − 1 2 ]

Therefore inverse of A will be:

A−1=1|A|adjA

A−1=1−1[20−1−9−23−6−12]

A−1=[−20192−361−2]

Question 11: Find the inverse of each of the matrices (if it exists).

[ 1 0 0 0 cos ⁡ α sin ⁡ α 0 sin ⁡ α − cos ⁡ α ]

Answer:

Given the matrix : [ 1 0 0 0 cos ⁡ α sin ⁡ α 0 sin ⁡ α − cos ⁡ α ] = A

To find the inverse we have to first find adjA then as we know the relation:

A−1=1|A|adjA

So, calculating |A| :

| A | = 1 ( − cos 2 ⁡ α − sin 2 ⁡ α ) + 0 ( 0 − 0 ) + 0 ( 0 − 0 )

= − ( cos 2 ⁡ α + sin 2 ⁡ α ) = − 1

Now, calculate the cofactors terms and then adjA.

A 11 = ( − 1 ) 1 + 1 ( − cos 2 ⁡ α − sin 2 ⁡ α ) = − 1 A 12 = ( − 1 ) 1 + 2 ( 0 − 0 ) = 0

A 13 = ( − 1 ) 1 + 3 ( 0 − 0 ) = 0 A 21 = ( − 1 ) 2 + 1 ( 0 − 0 ) = 0

A 22 = ( − 1 ) 2 + 2 ( − cos ⁡ α − 0 ) = − cos ⁡ α A 23 = ( − 1 ) 2 + 1 ( sin ⁡ α − 0 ) = − sin ⁡ α

A 31 = ( − 1 ) 3 + 1 ( 0 − 0 ) = 0 A 32 = ( − 1 ) 3 + 2 ( sin ⁡ α − 0 ) = − sin ⁡ α

A 33 = ( − 1 ) 3 + 3 ( cos ⁡ α − 0 ) = cos ⁡ α

So, we have a d j A = [ − 1 0 0 0 − cos ⁡ α − sin ⁡ α 0 − sin ⁡ α cos ⁡ α ]

Therefore inverse of A will be:

A−1=1|A|adjA

A−1=1−1[−1000−cos⁡α−sin⁡α0−sin⁡αcos⁡α]=[1000cos⁡αsin⁡α0sin⁡α−cos⁡α]

Question 12: Let A=[3725] and B=[6879] . Verify that (AB)−1=B−1A−1.

Answer:

We have A = [ 3 7 2 5 ] and B = [ 6 8 7 9 ] .

then calculating;

A B = [ 3 7 2 5 ] [ 6 8 7 9 ]

= [ 18 + 49 24 + 63 12 + 35 16 + 45 ] = [ 67 87 47 61 ]

Finding the inverse of AB.

Calculating the cofactors fo AB:

AB11=(−1)1+1(61)=61 AB12=(−1)1+2(47)=−47

AB21=(−1)2+1(87)=−87 AB22=(−1)2+2(67)=67

Then we have adj(AB):

adj(AB)=[61−87−4767]

and |AB| = 61(67) - (-87)(-47) = 4087-4089 = -2

Therefore we have the inverse:

( A B ) − 1 = 1 | A B | a d j ( A B ) = − 1 2 [ 61 − 87 − 47 67 ]

= [ − 61 2 87 2 47 2 − 67 2 ] .....................................(1)

Now, calculate the inverses of A and B.

|A| = 15-14 = 1 and |B| = 54- 56 = -2

adjA=[5−7−23] and adjB=[9−8−76]

therefore we have

A−1=1|A|adjA=11[5−7−23] and B−1=1|B|adjB=1−2[9−8−76]=[−92472−3]

Now calculating B − 1 A − 1 .

B − 1 A − 1 = [ − 9 2 4 7 2 − 3 ] [ 5 − 7 − 2 3 ]

= [ − 45 2 − 8 63 2 + 12 35 2 + 6 − 49 2 − 9 ] = [ − 61 2 87 2 47 2 − 67 2 ] ........................(2)

From (1) and (2) we get

(AB)−1=B−1A−1

Hence proved.

Question 13: If A=[31−12] ? , show that A2−5A+7I=O. Hence find A−1

Answer:

Given A = [ 3 1 − 1 2 ] then we have to show the relation A 2 − 5 A + 7 I = 0

So, calculating each term;

⇒ A 2 = [ 3 1 − 1 2 ] [ 3 1 − 1 2 ] = [ 9 − 1 3 + 2 − 3 − 2 − 1 + 4 ] = [ 8 5 − 5 3 ]

therefore A 2 − 5 A + 7 I ;

⇒ [ 8 5 − 5 3 ] − 5 [ 3 1 − 1 2 ] + 7 [ 1 0 0 1 ]

⇒ [ 8 5 − 5 3 ] − [ 15 5 − 5 10 ] + [ 7 0 0 7 ]

⇒ [ 8 − 15 + 7 5 − 5 + 0 − 5 + 5 + 0 3 − 10 + 7 ] = [ 0 0 0 0 ]

Hence A 2 − 5 A + 7 I = 0 .

∴ A . A − 5 A = − 7 I

⇒ A . A ( A − 1 ) − 5 A A − 1 = − 7 I A − 1

[ Post multiplying by A−1 , also |A|≠0 ]

⇒A(AA−1)−5I=−7A−1

⇒AI−5I=−7A−1

⇒−17(AI−5I)=17(5I−A)

∴A−1=17(5[1001]−[31−12])=17[2−113]

Question 14: For the matrix A=[3211] , find the numbers a and b such that A2+aA+bI=0.

Answer:

Given A = [ 3 2 1 1 ] then we have the relation A2+aA+bI=O

So, calculating each term;

⇒ A 2 = [ 3 2 1 1 ] [ 3 2 1 1 ] = [ 9 + 2 6 + 2 3 + 1 2 + 1 ] = [ 11 8 4 3 ]

Therefore A2+aA+bI=O ;

⇒ [ 11 8 4 3 ] + a [ 3 2 1 1 ] + b [ 1 0 0 1 ] = [ 0 0 0 0 ]

⇒ [ 11 + 3 a + b 8 + 2 a 4 + a 3 + a + b ] = [ 0 0 0 0 ]

So, we have equations;

⇒ 11 + 3 a + b = 0 , 8 + 2 a = 0 and 4 + a = 0 , a n d 3 + a + b = 0

We get a = − 4 a n d b = 1 .

Question 15: For the matrix A=[11112−32−13] Show that A3−6A2+5A+11I=O Hence, find A−1.

Answer:

Given matrix: A = [ 1 1 1 1 2 − 3 2 − 1 3 ] ;

To show: A 3 − 6 A 2 + 5 A + 11 I = O

Finding each term:

⇒ A 2 = [ 1 1 1 1 2 − 3 2 − 1 3 ] [ 1 1 1 1 2 − 3 2 − 1 3 ]

⇒ [ 1 + 1 + 2 1 + 2 − 1 1 − 3 + 3 1 + 2 − 6 1 + 4 + 3 1 − 6 − 9 2 − 1 + 6 2 − 2 − 3 2 + 3 + 9 ]

⇒ [ 4 2 1 − 3 8 − 14 7 − 3 14 ]

⇒ A 3 = [ 4 2 1 − 3 8 − 14 7 − 3 14 ] [ 1 1 1 1 2 − 3 2 − 1 3 ]

⇒ [ 4 + 2 + 2 4 + 4 − 1 4 − 6 + 3 − 3 + 8 − 28 − 3 + 16 + 14 − 3 − 24 − 42 7 − 3 + 28 7 − 6 − 14 7 + 9 + 42 ]

⇒ [ 8 7 1 − 23 27 − 69 32 − 13 58 ]

So now we have, A 3 − 6 A 2 + 5 A + 11 I

⇒ [ 8 7 1 − 23 27 − 69 32 − 13 58 ] − 6 [ 4 2 1 − 3 8 − 14 7 − 3 14 ] + 5 [ 1 1 1 1 2 − 3 2 − 1 3 ] + 11 [ 1 0 0 0 1 0 0 0 1 ]

⇒ [ 8 7 1 − 23 27 − 69 32 − 13 58 ] − [ 24 12 6 − 18 48 − 84 42 − 18 84 ] + [ 5 5 5 5 10 − 15 10 − 5 15 ] + [ 11 0 0 0 11 0 0 0 11 ]

⇒ [ 8 − 24 + 5 + 11 7 − 12 + 5 1 − 6 + 5 − 23 + 18 + 5 27 − 48 + 10 + 11 − 69 + 84 − 15 32 − 42 + 10 − 13 + 18 − 5 58 − 84 + 15 + 11 ]

⇒ [ 0 0 0 0 0 0 0 0 0 ] = 0

Now finding the inverse of A;

Post-multiplying by A−1 as, |A|≠0

⇒(AAA)A−1−6(AA)A−1+5AA−1+11IA−1=0

⇒AA(AA−1)−6A(AA−1)+5(AA−1)=−11IA−1

⇒A2−6A+5I=−11A−1

A−1=−111(A2−6A+5I) ...................(1)

Now,

From equation (1) we get;

⇒ A−1=−111([421−38−147−314]−6[11112−32−13]+5[100010001])

⇒ A−1=−111([4−6+52−61−6−3−68−12+5−14+187−12−3+614−18+5]

⇒ A−1=−111([3−4−5−914−531]

Question 16: If A=[2−11−12−11−12] , verify that A3−6A2+9A−4I=O . Hence find A−1.

Answer:

Given matrix: A = [ 2 − 1 1 − 1 2 − 1 1 − 1 2 ] ;

To show: A 3 − 6 A 2 + 9 A − 4 I

Finding each term:

⇒A2=[2−11−12−11−12][2−11−12−11−12]

⇒[4+1+1−2−2−12+1+2−2−2−11+4+1−1−2−22+1+2−1−2−21+1+4]

⇒[6−55−56−55−56]

⇒A3=[6−55−56−55−56][2−11−12−11−12]

⇒[12+5+5−6−10−56+5+10−10−6−55+12+5−5−6−1010+5+6−5−10−65+5+12]

⇒[22−2121−2122−2121−2122]

So now we have, A3−6A2+9A−4I

⇒[22−2121−2122−2121−2122]−6[6−55−56−55−56]+9[2−11−12−11−12]−4[100010001]

⇒[22−2121−2122−2121−2122]−[36−3030−3036−3030−3036]+[18−99−918−99−918]−[400040004]

⇒[22−36+18−4−21+30−921−30+9−21+30−922−36+18−4−21+30−921−30+9−21+30−922−36+18−4]

⇒[000000000]=O

Now finding the inverse of A;

Post-multiplying by A−1 as, |A|≠0

⇒(AAA)A−1−6(AA)A−1+9AA−1−4IA−1=0

⇒AA(AA−1)−6A(AA−1)+9(AA−1)=4IA−1

⇒A2−6A+9I=4A−1

⇒ A−1=14(A2−6A+9I) ...................(1)

Now,

From equation (1) we get;

⇒ A−1=14([6−55−56−55−56]−6[2−11−12−11−12]+9[100010001])

⇒ A−1=14[6−12+9−5+65−6−5+66−12+9−5+65−6−5+66−12+9]

Hence inverse of A is :

⇒ A−1=14[31−1131−113]

Question 17: Let A be a nonsingular square matrix of order 3×3. Then |adjA| is equal to

(A) |A| (B) |A|2 (C) |A|3 (D) 3|A|

Answer:

We know the identity ( a d j A ) A = | A | I

Hence we can determine the value of | ( a d j A ) | .

Taking both sides determinant value we get,

⇒ | ( a d j A ) A | = | | A | I | or | ( a d j A ) | | A | = | | A | | | I |

or taking R.H.S.,

⇒ | | A | | | I | = | | A | 0 0 0 | A | 0 0 0 | A | |

⇒ | A | ( | A | 2 ) = | A | 3

or, we have then | ( a d j A ) | | A | = | A | 3

Therefore | ( a d j A ) | = | A | 2

Hence the correct answer is B.

Question 18: If A is an invertible matrix of order 2, then det (A−1) is equal to

(A) det(A) (B) 1det(A) (C) 1 (D) 0

Answer:

Given that the matrix is invertible hence A − 1 exists and A − 1 = 1 | A | a d j A

Let us assume a matrix of the order of 2;

⇒ A = [ a b c d ] .

Then | A | = a d − b c .

adjA=[d−b−ca] and |adjA|=ad−bc

Now,

⇒ A − 1 = 1 | A | a d j A

Taking determinant both sides;

⇒ | A − 1 | = | 1 | A | a d j A | = [ d | A | − b | A | − c | A | a | A | ]

∴ | A − 1 | = | d | A | − b | A | − c | A | a | A | | = 1 | A | 2 | d − b − c a | = 1 | A | 2 ( a d − b c ) = 1 | A | 2 . | A | = 1 | A |

Therefore we get;

⇒ | A − 1 | = 1 | A |

Hence the correct answer is B.

Question 1: Examine the consistency of the system of equations.

x + 2 y = 2

2 x + 3 y = 3

Answer:

We have given the system of equations:18967

x + 2 y = 2

2 x + 3 y = 3

The given system of equations can be written in the form of the matrix; A X = B

where A = [ 1 2 2 3 ] , X = [ x y ] and B = [ 2 3 ] .

So, we want to check for the consistency of the equations;

⇒ | A | = 1 ( 3 ) − 2 ( 2 ) = − 1 ≠ 0

Here A is non -singular therefore there exists A − 1 .

Hence, the given system of equations is consistent.

Question 2: Examine the consistency of the system of equations

2 x − y = 5

x + y = 4

Answer:

We have given the system of equations:

2 x − y = 5

x + y = 4

The given system of equations can be written in the form of the matrix; A X = B

where A = [ 2 − 1 1 1 ] , X = [ x y ] and B = [ 5 4 ] .

So, we want to check for the consistency of the equations;

⇒ | A | = 2 ( 1 ) − 1 ( − 1 ) = 3 ≠ 0

Here A is non -singular therefore there exists A − 1 .

Hence, the given system of equations is consistent.

Question 3: Examine the consistency of the system of equations.

x + 3 y = 5

2 x + 6 y = 8

Answer:

We have given the system of equations:

x + 3 y = 5

2 x + 6 y = 8

The given system of equations can be written in the form of the matrix; A X = B

where A = [ 1 3 2 6 ] , X = [ x y ] and B = [ 5 8 ] .

So, we want to check for the consistency of the equations;

⇒ | A | = 1 ( 6 ) − 2 ( 3 ) = 0

Here A is a singular matrix therefore now we will check whether the ( a d j A ) B is zero or non-zero.

⇒ a d j A = [ 6 − 3 − 2 1 ]

So, ( a d j A ) B = [ 6 − 3 − 2 1 ] [ 5 8 ] = [ 30 − 24 − 10 + 8 ] = [ 6 − 2 ] ≠ 0

As, ( a d j A ) B ≠ 0 , the solution of the given system of equations does not exist.

Hence, the given system of equations is inconsistent.

Question 4: Examine the consistency of the system of equations.

x + y + z = 1

2 x + 3 y + 2 z = 2

a x + a y + 2 a z = 4

Answer:

We have given the system of equations:

x + y + z = 1

2 x + 3 y + 2 z = 2

a x + a y + 2 a z = 4

The given system of equations can be written in the form of the matrix; A X = B

where A = [ 1 1 1 2 3 2 a a 2 a ] , X = [ x y z ] and B = [ 1 2 4 ] .

So, we want to check for the consistency of the equations;

⇒ | A | = 1 ( 6 a − 2 a ) − 1 ( 4 a − 2 a ) + 1 ( 2 a − 3 a )

⇒ 4 a − 2 a − a = 4 a − 3 a = a ≠ 0

[If zero then it won't satisfy the third equation]

Here A is a non-singular matrix therefore there exists A − 1 .

Hence, the given system of equations is consistent.

Question 5: Examine the consistency of the system of equations.

3 x − y − 2 z = 2

2 y − z = − 1

3 x − 5 y = 3

Answer:

We have given the system of equations:

3 x − y − 2 z = 2

2 y − z = − 1

3 x − 5 y = 3

The given system of equations can be written in the form of the matrix; A X = B

where A = [ 3 − 1 − 2 0 2 − 1 3 − 5 0 ] , X = [ x y z ] and B = [ 2 − 1 3 ] .

So, we want to check for the consistency of the equations;

⇒ | A | = 3 ( 0 − 5 ) − ( − 1 ) ( 0 + 3 ) − 2 ( 0 − 6 )

= − 15 + 3 + 12 = 0

Therefore matrix A is a singular matrix.

So, we will then check ( a d j A ) B ,

( a d j A ) = [ − 5 10 5 − 3 6 3 − 6 12 6 ]

∴ ( a d j A ) B = [ − 5 10 5 − 3 6 3 − 6 12 6 ] [ 2 − 1 3 ] = [ − 10 − 10 + 15 − 6 − 6 + 9 − 12 − 12 + 18 ] = [ − 5 − 3 − 6 ] ≠ 0

As ( a d j A ) B is non-zero thus the solution of the given system of the equation does not exist. Hence, the given system of equations is inconsistent.

Question 6: Examine the consistency of the system of equations.

5 x − y + 4 z = 5

2 x + 3 y + 5 z = 2

5 x − 2 y + 6 z = − 1

Answer:

We have given the system of equations:

5 x − y + 4 z = 5

2 x + 3 y + 5 z = 2

5 x − 2 y + 6 z = − 1

The given system of equations can be written in the form of the matrix; A X = B

where A = [ 5 − 1 4 2 3 5 5 − 2 6 ] , X = [ x y z ] and B = [ 5 2 − 1 ] .

So, we want to check for the consistency of the equations;

⇒ | A | = 5 ( 18 + 10 ) + 1 ( 12 − 25 ) + 4 ( − 4 − 15 )

= 140 − 13 − 76 = 51 ≠ 0

Here A is a non-singular matrix therefore there exists A − 1 .

Hence, the given system of equations is consistent.

Question 7: Solve the system of linear equations, using the matrix method.

5 x + 2 y = 4

7 x + 3 y = 5

Answer:

The given system of equations

5 x + 2 y = 4

7 x + 3 y = 5

can be written in the matrix form of AX =B, where

⇒ A = [ 5 4 7 3 ] , X = [ x y ] and B = [ 4 5 ]

we have,

⇒ | A | = 15 − 14 = 1 ≠ 0 .

So, A is non-singular, Therefore, its inverse A − 1 exists.

as we know A − 1 = 1 | A | ( a d j A )

A − 1 = 1 | A | ( a d j A ) = ( a d j A ) = [ 3 − 2 − 7 5 ]

So, the solutions can be found by X = A − 1 B = [ 3 − 2 − 7 5 ] [ 4 5 ]

⇒ [ x y ] = [ 12 − 10 − 28 + 25 ] = [ 2 − 3 ]

Hence the solutions of the given system of equations;

x = 2 and y = -3 .

Question 8: Solve the system of linear equations, using the matrix method.

2 x − y = − 2

3 x + 4 y = 3

Answer:

The given system of equations

2 x − y = − 2

3 x + 4 y = 3

can be written in the matrix form of AX =B, where

⇒ A = [ 2 − 1 3 4 ] , X = [ x y ] and B = [ − 2 3 ]

we have,

⇒ | A | = 8 + 3 = 11 ≠ 0 .

So, A is non-singular, Therefore, its inverse A − 1 exists.

as we know A − 1 = 1 | A | ( a d j A )

⇒ A − 1 = 1 | A | ( a d j A ) = 1 11 [ 4 1 − 3 2 ]

So, the solutions can be found by X = A − 1 B = 1 11 [ 4 1 − 3 2 ] [ − 2 3 ]

⇒ [ x y ] = 1 11 [ − 8 + 3 6 + 6 ] = 1 11 [ − 5 12 ] = [ − 5 11 − 12 11 ]

Hence the solutions of the given system of equations;

x = − 5 11 a n d y = 12 11.

Question 9: Solve the system of linear equations, using the matrix method.

4 x − 3 y = 3

3 x − 5 y = 7

Answer:

The given system of equations

4 x − 3 y = 3

3 x − 5 y = 7

can be written in the matrix form of AX =B, where

A = [ 4 − 3 3 − 5 ] , X = [ x y ] and B = [ 3 7 ]

we have,

⇒ | A | = − 20 + 9 = − 11 ≠ 0 .

So, A is non-singular, Therefore, its inverse A − 1 exists.

as we know A − 1 = 1 | A | ( a d j A )

⇒ A − 1 = 1 | A | ( a d j A ) = − 1 11 [ − 5 3 − 3 4 ] = 1 11 [ 5 − 3 3 − 4 ]

So, the solutions can be found by X = A − 1 B = 1 11 [ 5 − 3 3 − 4 ] [ 3 7 ]

⇒ [ x y ] = 1 11 [ 15 − 21 9 − 28 ] = 1 11 [ − 6 − 19 ] = [ − 6 11 − 19 11 ]

Hence the solutions of the given system of equations;

x = − 6 11 a n d y = − 19 11 .

Question 10: Solve the system of linear equations, using the matrix method.

5 x + 2 y = 3

3 x + 2 y = 5

Answer:

The given system of equations

5 x + 2 y = 3

3 x + 2 y = 5

can be written in the matrix form of AX =B, where

⇒ A = [ 5 2 3 2 ] , X = [ x y ] and B = [ 3 5 ]

we have,

⇒ | A | = 10 − 6 = 4 ≠ 0 .

So, A is non-singular, Therefore, its inverse A − 1 exists.

as we know A − 1 = 1 | A | ( a d j A )

⇒ A − 1 = 1 | A | ( a d j A ) = 1 4 [ 2 − 2 − 3 5 ]

So, the solutions can be found by X = A − 1 B = 1 4 [ 2 − 2 − 3 5 ] [ 3 5 ]

⇒ [ x y ] = 1 4 [ 6 − 10 − 9 + 25 ] = 1 4 [ − 4 16 ] = [ − 1 4 ]

Hence the solutions of the given system of equations;

x = − 1 a n d y = 4.

Question 11: Solve a system of linear equations, using the matrix method.

2 x + y + z = 1

x − 2 y − z = 3 2

3 y − 5 z = 9

Answer:

The given system of equations

2 x + y + z = 1

x − 2 y − z = 3 2

3 y − 5 z = 9

can be written in the matrix form of AX =B, where

⇒ A = [ 2 1 1 1 − 2 − 1 0 3 − 5 ] , X = [ x y z ] and B = [ 1 3 2 9 ]

we have,

⇒ | A | = 2 ( 10 + 3 ) − 1 ( − 5 − 0 ) + 1 ( 3 − 0 ) = 26 + 5 + 3 = 34 ≠ 0 .

So, A is non-singular, Therefore, its inverse A − 1 exists.

as we know A − 1 = 1 | A | ( a d j A )

Now, we will find the cofactors;

A 11 = ( − 1 ) 1 + 1 ( 10 + 3 ) = 13 A 12 = ( − 1 ) 1 + 2 ( − 5 − 0 ) = 5

A 13 = ( − 1 ) 1 + 3 ( 3 − 0 ) = 3 A 21 = ( − 1 ) 2 + 1 ( − 5 − 3 ) = 8

A 22 = ( − 1 ) 2 + 2 ( − 10 − 0 ) = − 10 A 23 = ( − 1 ) 2 + 3 ( 6 − 0 ) = − 6

A 31 = ( − 1 ) 3 + 1 ( − 1 + 2 ) = 1 A 32 = ( − 1 ) 3 + 2 ( − 2 − 1 ) = 3

A 33 = ( − 1 ) 3 + 3 ( − 4 − 1 ) = − 5

⇒ ( a d j A ) = [ 13 8 1 5 − 10 3 3 − 6 − 5 ]

A − 1 = 1 | A | ( a d j A ) = 1 34 [ 13 8 1 5 − 10 3 3 − 6 − 5 ]

So, the solutions can be found by X = A − 1 B = 1 34 [ 13 8 1 5 − 10 3 3 − 6 − 5 ] [ 1 3 2 9 ]

⇒ [ x y z ] = 1 34 [ 13 + 12 + 9 5 − 15 + 27 3 − 9 − 45 ] = 1 34 [ 34 17 − 51 ] = [ 1 1 2 − 3 2 ]

Hence the solutions of the given system of equations;

x = 1 , y = 1 2 , a n d z = − 3 2 .

Question 12: Solve the system of linear equations, using the matrix method.

x − y + z = 4

2 x + y − 3 z = 0

x + y + z = 2

Answer:

The given system of equations

x − y + z = 4

2 x + y − 3 z = 0

x + y + z = 2

can be written in the matrix form of AX =B, where

⇒ A = [ 1 − 1 1 2 1 − 3 1 1 1 ] , X = [ x y z ] a n d B = [ 4 0 2 ] .

we have,

⇒ | A | = 1 ( 1 + 3 ) + 1 ( 2 + 3 ) + 1 ( 2 − 1 ) = 4 + 5 + 1 = 10 ≠ 0 .

So, A is non-singular, Therefore, its inverse A − 1 exists.

as we know A − 1 = 1 | A | ( a d j A )

Now, we will find the cofactors;

A 11 = ( − 1 ) 1 + 1 ( 1 + 3 ) = 4 A 12 = ( − 1 ) 1 + 2 ( 2 + 3 ) = − 5

A 13 = ( − 1 ) 1 + 3 ( 2 − 1 ) = 1 A 21 = ( − 1 ) 2 + 1 ( − 1 − 1 ) = 2

A 22 = ( − 1 ) 2 + 2 ( 1 − 1 ) = 0 A 23 = ( − 1 ) 2 + 3 ( 1 + 1 ) = − 2

A 31 = ( − 1 ) 3 + 1 ( 3 − 1 ) = 2 A 32 = ( − 1 ) 3 + 2 ( − 3 − 2 ) = 5

A 33 = ( − 1 ) 3 + 3 ( 1 + 2 ) = 3

⇒ ( a d j A ) = [ 4 2 2 − 5 0 5 1 − 2 3 ]

⇒ A − 1 = 1 | A | ( a d j A ) = 1 10 [ 4 2 2 − 5 0 5 1 − 2 3 ]

So, the solutions can be found by X = A − 1 B = 1 10 [ 4 2 2 − 5 0 5 1 − 2 3 ] [ 4 0 2 ]

⇒ [ x y z ] = 1 10 [ 16 + 0 + 4 − 20 + 0 + 10 4 + 0 + 6 ] = 1 10 [ 20 − 10 10 ] = [ 2 − 1 1 ]

Hence the solutions of the given system of equations;

x = 2 , y = − 1 , a n d z = 1.

Question 13: Solve the system of linear equations, using the matrix method.

2 x + 3 y + 3 z = 5

x − 2 y + z = − 4

3 x − y − 2 z = 3

Answer:

The given system of equations

2 x + 3 y + 3 z = 5

x − 2 y + z = − 4

3 x − y − 2 z = 3

can be written in the matrix form of AX =B, where

⇒ A = [ 2 3 3 1 − 2 1 3 − 1 − 2 ] , X = [ x y z ] a n d B = [ 5 − 4 3 ] .

we have,

⇒ | A | = 2 ( 4 + 1 ) − 3 ( − 2 − 3 ) + 3 ( − 1 + 6 ) = 10 + 15 + 15 = 40 .

So, A is non-singular, Therefore, its inverse A − 1 exists.

as we know A − 1 = 1 | A | ( a d j A )

Now, we will find the cofactors;

A 11 = ( − 1 ) 1 + 1 ( 4 + 1 ) = 5 A 12 = ( − 1 ) 1 + 2 ( − 2 − 3 ) = 5

A 13 = ( − 1 ) 1 + 3 ( − 1 + 6 ) = 5 A 21 = ( − 1 ) 2 + 1 ( − 6 + 3 ) = 3

A 22 = ( − 1 ) 2 + 2 ( − 4 − 9 ) = − 13 A 23 = ( − 1 ) 2 + 3 ( − 2 − 9 ) = 11

A 31 = ( − 1 ) 3 + 1 ( 3 + 6 ) = 9 A 32 = ( − 1 ) 3 + 2 ( 2 − 3 ) = 1

A 33 = ( − 1 ) 3 + 3 ( − 4 − 3 ) = − 7

⇒ ( a d j A ) = [ 5 3 9 5 − 13 1 5 11 − 7 ]

⇒ A − 1 = 1 | A | ( a d j A ) = 1 40 [ 5 3 9 5 − 13 1 5 11 − 7 ]

So, the solutions can be found by X = A − 1 B = 1 40 [ 5 3 9 5 − 13 1 5 11 − 7 ] [ 5 − 4 3 ]

⇒ [ x y z ] = 1 40 [ 25 − 12 + 27 25 + 52 + 3 25 − 44 − 21 ] = 1 40 [ 40 80 − 40 ] = [ 1 2 − 1 ]

Hence the solutions of the given system of equations;

x = 1 , y = 2 , a n d z = − 1.

Question 14: Solve the system of linear equations, using the matrix method.

x − y + 2 z = 7

3 x + 4 y − 5 z = − 5

2 x − y + 3 z = 12

Answer:

The given system of equations

x − y + 2 z = 7

3 x + 4 y − 5 z = − 5

2 x − y + 3 z = 12

can be written in the matrix form of AX =B, where

⇒ A = [ 1 − 1 2 3 4 − 5 2 − 1 3 ] , X = [ x y z ] a n d B = [ 7 − 5 12 ] .

we have,

⇒ | A | = 1 ( 12 − 5 ) + 1 ( 9 + 10 ) + 2 ( − 3 − 8 ) = 7 + 19 − 22 = 4 ≠ 0 .

So, A is non-singular, Therefore, its inverse A − 1 exists.

as we know A − 1 = 1 | A | ( a d j A )

Now, we will find the cofactors;

A 11 = ( − 1 ) 12 − 5 = 7 A 12 = ( − 1 ) 1 + 2 ( 9 + 10 ) = − 19

A 13 = ( − 1 ) 1 + 3 ( − 3 − 8 ) = − 11 A 21 = ( − 1 ) 2 + 1 ( − 3 + 2 ) = 1

A 22 = ( − 1 ) 2 + 2 ( 3 − 4 ) = − 1 A 23 = ( − 1 ) 2 + 3 ( − 1 + 2 ) = − 1

A 31 = ( − 1 ) 3 + 1 ( 5 − 8 ) = − 3 A 32 = ( − 1 ) 3 + 2 ( − 5 − 6 ) = 11

A 33 = ( − 1 ) 3 + 3 ( 4 + 3 ) = 7

⇒ ( a d j A ) = [ 7 1 − 3 − 19 − 1 11 − 11 − 1 7 ]

⇒ A − 1 = 1 | A | ( a d j A ) = 1 4 [ 7 1 − 3 − 19 − 1 11 − 11 − 1 7 ]

So, the solutions can be found by X = A − 1 B = 1 4 [ 7 1 − 3 − 19 − 1 11 − 11 − 1 7 ] [ 7 − 5 12 ]

⇒ [ x y z ] = 1 4 [ 49 − 5 − 36 − 133 + 5 + 132 − 77 + 5 + 84 ] = 1 4 [ 8 4 12 ] = [ 2 1 3 ]

Hence the solutions of the given system of equations;

x = 2 , y = 1 , a n d z = 3.

Question 15: If A=[2−3532−411−2] , find A−1 . Using A−1 solve the system of equations

2 x − 3 y + 5 z = 11

3 x + 2 y − 4 z = − 5

x + y − 2 z = − 3

Answer:

The given system of equations

2 x − 3 y + 5 z = 11

3 x + 2 y − 4 z = − 5

x + y − 2 z = − 3

can be written in the matrix form of AX =B, where

⇒ A = [ 2 − 3 5 3 2 − 4 1 1 − 2 ] , X = [ x y z ] a n d B = [ 11 − 5 − 3 ] .

we have,

⇒ | A | = 2 ( − 4 + 4 ) + 3 ( − 6 + 4 ) + 5 ( 3 − 2 ) = 0 − 6 + 5 = − 1 ≠ 0 .

So, A is non-singular, Therefore, its inverse A − 1 exists.

as we know A − 1 = 1 | A | ( a d j A )

Now, we will find the cofactors;

A 11 = ( − 1 ) − 4 + 4 = 0 A 12 = ( − 1 ) 1 + 2 ( − 6 + 4 ) = 2

A 13 = ( − 1 ) 1 + 3 ( 3 − 2 ) = 1 A 21 = ( − 1 ) 2 + 1 ( 6 − 5 ) = − 1

A 22 = ( − 1 ) 2 + 2 ( − 4 − 5 ) = − 9 A 23 = ( − 1 ) 2 + 3 ( 2 + 3 ) = − 5

A 31 = ( − 1 ) 3 + 1 ( 12 − 10 ) = 2 A 32 = ( − 1 ) 3 + 2 ( − 8 − 15 ) = 23

A 33 = ( − 1 ) 3 + 3 ( 4 + 9 ) = 13

⇒ ( a d j A ) = [ 0 − 1 2 2 − 9 23 1 − 5 13 ]

⇒ A − 1 = 1 | A | ( a d j A ) = − 1 [ 0 − 1 2 2 − 9 23 1 − 5 13 ] = [ 0 1 − 2 − 2 9 − 23 − 1 5 − 13 ]

So, the solutions can be found by X = A − 1 B = [ 0 1 − 2 − 2 9 − 23 − 1 5 − 13 ] [ 11 − 5 − 3 ]

⇒ [ x y z ] = [ 0 − 5 + 6 − 22 − 45 + 69 − 11 − 25 + 39 ] = [ 1 2 3 ]

Hence the solutions of the given system of equations;

x = 1 , y = 2 , a n d z = 3.

Question 16: The cost of 4 kg onion, 3 kg wheat, and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat, and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find the cost of each item per kg by matrix method.

Answer:

So, let us assume the cost of onion, wheat, and rice to be x, ,y and z respectively.

Then we have the equations for the given situation :

4 x + 3 y + 2 z = 60

2 x + 4 y + 6 z = 90

6 x + 2 y + 3 y = 70

We can find the cost of each item per Kg by the matrix method as follows;

Taking the coefficients of x, y, and z as a matrix A .

We have;

⇒ A = [ 4 3 2 2 4 6 6 2 3 ] , X = [ x y z ] a n d B = [ 60 90 70 ] .

⇒ | A | = 4 ( 12 − 12 ) − 3 ( 6 − 36 ) + 2 ( 4 − 24 ) = 0 + 90 − 40 = 50 ≠ 0

Now, we will find the cofactors of A;

A 11 = ( − 1 ) 1 + 1 ( 12 − 12 ) = 0 A 12 = ( − 1 ) 1 + 2 ( 6 − 36 ) = 30

A 13 = ( − 1 ) 1 + 3 ( 4 − 24 ) = − 20 A 21 = ( − 1 ) 2 + 1 ( 9 − 4 ) = − 5

A 22 = ( − 1 ) 2 + 2 ( 12 − 12 ) = 0 A 23 = ( − 1 ) 2 + 3 ( 8 − 18 ) = 10

A 31 = ( − 1 ) 3 + 1 ( 18 − 8 ) = 10 A 32 = ( − 1 ) 3 + 2 ( 24 − 4 ) = − 20

A 33 = ( − 1 ) 3 + 3 ( 16 − 6 ) = 10

Now we have adjA;

⇒ a d j A = [ 0 − 5 10 30 0 − 20 − 20 10 10 ]

⇒ A − 1 = 1 | A | ( a d j A ) = 1 50 [ 0 − 5 10 30 0 − 20 − 20 10 10 ] s

So, the solutions can be found by X = A − 1 B = 1 50 [ 0 − 5 10 30 0 − 20 − 20 10 10 ] [ 60 90 70 ]

⇒ [ x y z ] = [ 0 − 450 + 700 1800 + 0 − 1400 − 1200 + 900 + 700 ] = 1 50 [ 250 400 400 ] = [ 5 8 8 ]

Hence the solutions of the given system of equations;

x = 5 , y = 8 , a n d z = 8.

Therefore, we have the cost of onions is Rs.5 per Kg, the cost of wheat is Rs.8 per Kg, and the cost of rice is Rs.8 per kg.

Question 1: Prove that the determinant |xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x| is independent of θ.

Answer:

Calculating the determinant value of | x sin ⁡ θ cos ⁡ θ − sin ⁡ θ − x 1 cos ⁡ θ 1 x | ;

⇒ x [ − x 1 1 x ] − sin ⁡ Θ [ − sin ⁡ Θ 1 cos ⁡ Θ x ] + cos ⁡ Θ [ − sin ⁡ Θ − x cos ⁡ Θ 1 ]

⇒ x ( − x 2 − 1 ) − sin ⁡ Θ ( − x sin ⁡ Θ − cos ⁡ Θ ) + cos ⁡ Θ ( − sin ⁡ Θ + x cos ⁡ Θ )

⇒ − x 3 − x + x sin 2 ⁡ Θ + sin ⁡ Θ cos ⁡ Θ − cos ⁡ Θ sin ⁡ Θ + x cos 2 ⁡ Θ

⇒ − x 3 − x + x ( sin 2 ⁡ Θ + cos 2 ⁡ Θ )

⇒ − x 3 − x + x = − x 3

Clearly, the determinant is independent of θ .

Question 2: Evaluate |cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α−sin⁡βcos⁡β0sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α|.

Answer:

Given determinant | cos ⁡ α cos ⁡ β cos ⁡ α sin ⁡ β − sin ⁡ α − sin ⁡ β cos ⁡ β 0 sin ⁡ α cos ⁡ β sin ⁡ α sin ⁡ β cos ⁡ α | ;

⇒ cos ⁡ α cos ⁡ β | cos ⁡ β 0 sin ⁡ α sin ⁡ β cos ⁡ α | − cos ⁡ α sin ⁡ β | − sin ⁡ β 0 sin ⁡ α cos ⁡ β cos ⁡ α | − sin ⁡ α | − sin ⁡ β cos ⁡ β sin ⁡ α cos ⁡ β sin ⁡ α sin ⁡ β | = cos ⁡ α cos ⁡ β ( cos ⁡ β cos ⁡ α − 0 ) − cos ⁡ α sin ⁡ β ( − cos ⁡ α sin ⁡ β − 0 ) − sin ⁡ α ( − sin ⁡ α sin 2 ⁡ β − sin ⁡ α cos 2 ⁡ β )

⇒ cos 2 ⁡ α cos 2 ⁡ β + cos 2 ⁡ α sin 2 ⁡ β + sin 2 ⁡ α sin 2 ⁡ β + sin 2 ⁡ α cos 2 ⁡ β

⇒ cos 2 ⁡ α ( cos 2 ⁡ β + sin 2 ⁡ β ) + sin 2 ⁡ α ( sin 2 ⁡ β + cos 2 ⁡ β )

⇒ cos 2 ⁡ α ( 1 ) + sin 2 ⁡ α ( 1 ) = 1 .

Question 3: If A−1=[3−11−156−55−22] and B=[12−2−1300−21] , find (AB)−1.

Answer:

We know from the identity that;

( A B ) − 1 = B − 1 A − 1 .

Then we can find easily,

Given A − 1 = [ 3 − 1 1 − 15 6 − 5 5 − 2 2 ] and B = [ 1 2 − 2 − 1 3 0 0 − 2 1 ]

Then we have to find the B − 1 matrix.

So, Given matrix B = [ 1 2 − 2 − 1 3 0 0 − 2 1 ]

| B | = 1 ( 3 − 0 ) − 2 ( − 1 − 0 ) − 2 ( 2 − 0 ) = 3 + 2 − 4 = 1 ≠ 0

Hence its inverse B − 1 exists;

Now, as we know that

B − 1 = 1 | B | a d j B

So, calculating cofactors of B,

B 11 = ( − 1 ) 1 + 1 ( 3 − 0 ) = 3 B 12 = ( − 1 ) 1 + 2 ( − 1 − 0 ) = 1

B 13 = ( − 1 ) 1 + 3 ( 2 − 0 ) = 2 B 21 = ( − 1 ) 2 + 1 ( 2 − 4 ) = 2

B 22 = ( − 1 ) 2 + 2 ( 1 − 0 ) = 1 B 23 = ( − 1 ) 2 + 3 ( − 2 − 0 ) = 2

B 31 = ( − 1 ) 3 + 1 ( 0 + 6 ) = 6 B 32 = ( − 1 ) 3 + 2 ( 0 − 2 ) = 2

B 33 = ( − 1 ) 3 + 3 ( 3 + 2 ) = 5

a d j B = [ 3 2 6 1 1 2 2 2 5 ]

B − 1 = 1 | B | a d j B = 1 1 [ 3 2 6 1 1 2 2 2 5 ]

Now, We have both A − 1 as well as B − 1 ;

Putting in the relation we know; ( A B ) − 1 = B − 1 A − 1

( A B ) − 1 = 1 1 [ 3 2 6 1 1 2 2 2 5 ] [ 3 − 1 1 − 15 6 − 5 5 − 2 2 ]

= [ 9 − 30 + 30 − 3 + 12 − 12 3 − 10 + 12 3 − 15 + 10 − 1 + 6 − 4 1 − 5 + 4 6 − 30 + 25 − 2 + 12 − 10 2 − 10 + 10 ]

= [ 9 − 3 5 − 2 1 0 1 0 2 ]

Question 4: Let A=[121231115] . Verify that,

(i) 100 [ a d j A ] − 1 = a d j ( A − 1 )

(ii) ( A − 1 ) − 1 = A

Answer:

(i) Given that A = [ 1 2 1 2 3 1 1 1 5 ] ;

So, let us assume that A − 1 = B matrix and a d j A = C then;

| A | = 1 ( 15 − 1 ) − 2 ( 10 − 1 ) + 1 ( 2 − 3 ) = 14 − 18 − 1 = − 5 ≠ 0

Hence its inverse exists;

A − 1 = 1 | A | a d j A or B = 1 | A | C ;

so, we now calculate the value of a d j A

Cofactors of A;

A 11 = ( − 1 ) 1 + 1 ( 15 − 1 ) = 14 A 12 = ( − 1 ) 1 + 2 ( 10 − 1 ) = − 9

A 13 = ( − 1 ) 1 + 3 ( 2 − 3 ) = − 1 A 21 = ( − 1 ) 2 + 1 ( 10 − 1 ) = − 9

A 22 = ( − 1 ) 2 + 2 ( 5 − 1 ) = 4 A 23 = ( − 1 ) 2 + 3 ( 1 − 2 ) = 1

A 31 = ( − 1 ) 3 + 1 ( 2 − 3 ) = − 1 A 32 = ( − 1 ) 3 + 2 ( 1 − 2 ) = 1

A 33 = ( − 1 ) 3 + 3 ( 3 − 4 ) = − 1

⇒ a d j A = C = [ 14 − 9 − 1 − 9 4 1 − 1 1 − 1 ]

A − 1 = B = 1 | A | a d j A = 1 − 5 [ 14 − 9 − 1 − 9 4 1 − 1 1 − 1 ]

Finding the inverse of C;

| C | = 14 ( − 4 − 1 ) + 9 ( 9 + 1 ) − 1 ( − 9 + 4 ) = − 70 + 90 + 5 = 25 ≠ 0

Hence its inverse exists;

C − 1 = 1 | C | a d j C

Now, finding the a d j C ;

C 11 = ( − 1 ) 1 + 1 ( − 4 − 1 ) = − 5 C 12 = ( − 1 ) 1 + 2 ( 9 + 1 ) = − 10

C 13 = ( − 1 ) 1 + 3 ( − 9 + 4 ) = − 5 C 21 = ( − 1 ) 2 + 1 ( 9 + 1 ) = − 10

C 22 = ( − 1 ) 2 + 2 ( − 14 − 1 ) = − 15 C 23 = ( − 1 ) 2 + 3 ( 14 − 9 ) = − 5

C 31 = ( − 1 ) 3 + 1 ( − 9 + 4 ) = − 5 C 32 = ( − 1 ) 3 + 2 ( 14 − 9 ) = − 5

C 33 = ( − 1 ) 3 + 3 ( 56 − 81 ) = − 25

a d j C = [ − 5 − 10 − 5 − 10 − 15 − 5 − 5 − 5 − 25 ]

C − 1 = 1 | C | a d j C = 1 25 [ − 5 − 10 − 5 − 10 − 15 − 5 − 5 − 5 − 25 ] = [ − 1 5 − 2 5 − 1 5 − 2 5 − 3 5 − 1 5 − 1 5 − 1 5 − 1 ]

or L . H . S . = C − 1 = [ a d j A ] − 1 = [ − 1 5 − 2 5 − 1 5 − 2 5 − 3 5 − 1 5 − 1 5 − 1 5 − 1 ]

Now, finding the R.H.S.

a d j ( A − 1 ) = a d j B

A − 1 = B = [ − 14 5 9 5 1 5 9 5 − 4 5 − 1 5 1 5 − 1 5 1 5 ]

Cofactors of B;

B 11 = ( − 1 ) 1 + 1 ( − 4 25 − 1 25 ) = − 1 5

B 12 = ( − 1 ) 1 + 2 ( 9 25 + 1 25 ) = − 2 5

B 13 = ( − 1 ) 1 + 3 ( − 9 25 + 4 25 ) = − 1 5

B 21 = ( − 1 ) 2 + 1 ( 9 25 + 1 25 ) = − 2 5

B 22 = ( − 1 ) 2 + 2 ( − 14 25 − 1 25 ) = − 3 5

B 23 = ( − 1 ) 2 + 3 ( 14 25 − 9 25 ) = − 1 5

B 31 = ( − 1 ) 3 + 1 ( − 9 25 + 4 25 ) = − 1 5

B 32 = ( − 1 ) 3 + 2 ( 14 25 − 9 25 ) = − 1 5

B 33 = ( − 1 ) 3 + 3 ( 56 25 − 81 25 ) = − 1

R . H . S . = a d j B = a d j ( A − 1 ) = [ − 1 5 − 2 5 − 1 5 − 2 5 − 3 5 − 1 5 − 1 5 − 1 5 − 1 ]

Hence L.H.S. = R.H.S. proved.

(ii) Given that A = [ 1 2 1 2 3 1 1 1 5 ] ;

So, let us assume that A − 1 = B

| A | = 1 ( 15 − 1 ) − 2 ( 10 − 1 ) + 1 ( 2 − 3 ) = 14 − 18 − 1 = − 5 ≠ 0

Hence its inverse exists;

A − 1 = 1 | A | a d j A or B = 1 | A | C ;

so, we now calculate the value of a d j A

Cofactors of A;

A 11 = ( − 1 ) 1 + 1 ( 15 − 1 ) = 14 A 12 = ( − 1 ) 1 + 2 ( 10 − 1 ) = − 9

A 13 = ( − 1 ) 1 + 3 ( 2 − 3 ) = − 1 A 21 = ( − 1 ) 2 + 1 ( 10 − 1 ) = − 9

A 22 = ( − 1 ) 2 + 2 ( 5 − 1 ) = 4 A 23 = ( − 1 ) 2 + 3 ( 1 − 2 ) = 1

A 31 = ( − 1 ) 3 + 1 ( 2 − 3 ) = − 1 A 32 = ( − 1 ) 3 + 2 ( 1 − 2 ) = 1

A 33 = ( − 1 ) 3 + 3 ( 3 − 4 ) = − 1

⇒ a d j A = C = [ 14 − 9 − 1 − 9 4 1 − 1 1 − 1 ]

A − 1 = B = 1 | A | a d j A = 1 − 5 [ 14 − 9 − 1 − 9 4 1 − 1 1 − 1 ] = [ − 14 5 9 5 1 5 9 5 − 4 5 − 1 5 1 5 − 1 5 1 5 ]

Finding the inverse of B ;

| B | = − 14 5 ( − 4 25 − 1 25 ) − 9 5 ( 9 25 + 1 25 ) + 1 5 ( − 9 25 + 4 25 )

= 70 125 − 90 125 − 5 125 = − 25 125 = − 1 5 ≠ 0

Hence its inverse exists;

B − 1 = 1 | B | a d j B

Now, finding the a d j B ;

A − 1 = B = 1 | A | a d j A = 1 − 5 [ 14 − 9 − 1 − 9 4 1 − 1 1 − 1 ] = [ − 14 5 9 5 1 5 9 5 − 4 5 − 1 5 1 5 − 1 5 1 5 ]

B 11 = ( − 1 ) 1 + 1 ( − 4 25 − 1 25 ) = − 1 5 B 12 = ( − 1 ) 1 + 2 ( 9 25 + 1 25 ) = − 2 5

B 13 = ( − 1 ) 1 + 3 ( − 9 25 + 4 25 ) = − 1 5 B 21 = ( − 1 ) 2 + 1 ( 9 25 + 1 25 ) = − 2 5

B 22 = ( − 1 ) 2 + 2 ( − 14 25 − 1 25 ) = − 3 5 B 23 = ( − 1 ) 2 + 3 ( 14 25 − 9 25 ) = − 1 5

B 31 = ( − 1 ) 3 + 1 ( − 9 25 + 4 25 ) = − 1 5

B 32 = ( − 1 ) 3 + 2 ( 14 25 − 9 25 ) = − 1 5

B 33 = ( − 1 ) 3 + 3 ( 56 25 − 81 25 ) = − 25 25 = − 1

a d j B = [ − 1 5 − 2 5 − 1 5 − 2 5 − 3 5 − 1 5 − 1 5 − 1 5 − 1 ]

B − 1 = 1 | B | a d j B = − 5 1 [ − 1 5 − 2 5 − 1 5 − 2 5 − 3 5 − 1 5 − 1 5 − 1 5 − 1 ] = [ 1 2 1 2 3 1 1 1 5 ]

L . H . S . = B − 1 = ( A − 1 ) − 1 = [ 1 2 1 2 3 1 1 1 5 ]

R . H . S . = A = [ 1 2 1 2 3 1 1 1 5 ]

Hence proved L.H.S. = R.H.S.

Question 5: Evaluate |xyx+yyx+yxx+yxy|

Answer:

We have determinant △ = | x y x + y y x + y x x + y x y |

Applying row transformations; R 1 → R 1 + R 2 + R 3 , we have then;

△ = | 2 ( x + y ) 2 ( x + y ) 2 ( x + y ) y x + y x x + y x y |

Take out the common factor 2(x+y) from the row first.

= 2 ( x + y ) | 1 1 1 y x + y x x + y x y |

Now, applying the column transformation; C 1 → C 1 − C 2 and C 2 → C 2 − C 1 we have ;

= 2 ( x + y ) | 0 0 1 − x y x y x − y y |

Expanding the remaining determinant;

= 2 ( x + y ) ( − x ( x − y ) − y 2 ) = 2 ( x + y ) [ − x 2 + x y − y 2 ]

= − 2 ( x + y ) [ x 2 − x y + y 2 ] = − 2 ( x 3 + y 3 ) .

Question 6: Evaluate |1xy1x+yy1xx+y|

Answer:

We have determinant △ = | 1 x y 1 x + y y 1 x x + y |

Applying row transformations; R 1 → R 1 − R 2 and R 2 → R 2 − R 3 then we have then;

△ = | 0 − y 0 0 y − x 1 x x + y |

Taking out the common factor-y from the row first.

△ = − y | 0 1 0 0 y − x 1 x x + y |

Expanding the remaining determinant;

− y [ 1 ( − x − o ) ] = x y

Question 7: Solve the system of equations

2 x + 3 y + 10 z = 4

4 x − 6 y + 5 z = 1

6 x + 9 y − 20 z = 2

Answer:

We have a system of equations;

2 x + 3 y + 10 z = 4

4 x − 6 y + 5 z = 1

6 x + 9 y − 20 z = 2

So, we will convert the given system of equations into a simple form to solve the problem by the matrix method;

Let us take, 1 x = a , 1 y = b a n d 1 z = c

Then we have the equations;

2 a + 3 b + 10 c = 4

4 a − 6 b + 5 c = 1

6 a + 9 b − 20 c = 2

We can write it in the matrix form as A X = B , where

A = [ 2 3 10 4 − 6 5 6 9 − 20 ] , X = [ a b c ] a n d B = [ 4 1 2 ] .

Now, Finding the determinant value of A;

| A | = 2 ( 120 − 45 ) − 3 ( − 80 − 30 ) + 10 ( 36 + 36 )

= 150 + 330 + 720

= 1200 ≠ 0

Hence we can say that A is non-singular ∴ its inverse exists;

Finding cofactors of A;

A 11 = 75 , A 12 = 110 , A 13 = 72

A 21 = 150 , A 22 = − 100 , A 23 = 0

A 31 = 75 , A 31 = 30 , A 33 = − 24

∴ as we know A − 1 = 1 | A | a d j A

= 1 1200 [ 75 150 75 110 − 100 30 72 0 − 24 ]

Now we will find the solutions by relation X = A − 1 B .

⇒ [ a b c ] = 1 1200 [ 75 150 75 110 − 100 30 72 0 − 24 ] [ 4 1 2 ]

= 1 1200 [ 300 + 150 + 150 440 − 100 + 60 288 + 0 − 48 ]

= 1 1200 [ 600 400 240 ] = [ 1 2 1 3 1 5 ]

Therefore we have the solutions a = 1 2 , b = 1 3 , a n d c = 1 5 .

Or in terms of x, y, and z;

x=2, y=3, and z=5

Question 8: Choose the correct answer.

If x, y, z are nonzero real numbers, then the inverse of matrix A=[x000y000z] is

( A ) [ x − 1 0 0 0 y − 1 0 0 0 z − 1 ] ( B ) x y z [ x − 1 0 0 0 y − 1 0 0 0 z − 1 ]

( C ) 1 x y z [ x 0 0 0 y 0 0 0 z ] ( D ) 1 x y z [ 1 0 0 0 1 0 0 0 1 ]

Answer:

Given Matrix A = [ x 0 0 0 y 0 0 0 z ] ,

| A | = x ( y z − 0 ) = x y z

As we know,

A − 1 = 1 | A | a d j A

So, we will find the a d j A ,

Determining its cofactor first,

A 11 = y z A 12 = 0 A 13 = 0

A 21 = 0 A 22 = x z A 23 = 0

A 31 = 0 A 32 = 0 A 33 = x y

Hence A − 1 = 1 | A | a d j A = 1 x y z [ y z 0 0 0 x z 0 0 0 x y ]

A − 1 = [ 1 x 0 0 0 1 y 0 0 0 1 z ]

Therefore the correct answer is (A)

Question 9: Choose the correct answer.

Let A=|1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1|, where 0≤θ≤2π . Then

(A) Det(A)=0 ; (B) Det(A)∈(2,∞)

(C) Det(A)∈(2,4) (D) Det(A)∈[2,4]

Answer:

Given determinant A = | 1 sin ⁡ θ 1 − sin ⁡ θ 1 sin ⁡ θ − 1 − sin ⁡ θ 1 |

| A | = 1 ( 1 + sin 2 ⁡ Θ ) − sin ⁡ Θ ( − sin ⁡ Θ + sin ⁡ Θ ) + 1 ( sin 2 ⁡ Θ + 1 )

= 1 + sin 2 ⁡ Θ + sin 2 ⁡ Θ + 1

= 2 + 2 sin 2 ⁡ Θ = 2 ( 1 + sin 2 ⁡ Θ )

Now, given the range of Θ from 0 ≤ Θ ≤ 2 π

⇒ 0 ≤ sin ⁡ Θ ≤ 1

⇒ 0 ≤ sin 2 ⁡ Θ ≤ 1

⇒ 1 ≤ 1 + sin 2 ⁡ Θ ≤ 2

⇒ 2 ≤ 2 ( 1 + sin 2 ⁡ Θ ) ≤ 4

Therefore the | A | ϵ [ 2 , 4 ] .

Hence, the correct answer is D.