Ncert Solutions For Class 12 Maths Ex 4.4

Ncert Solutions For Class 12 Maths Chapter 4 Ex 4.4

Q1: Write Minors and Cofactors of the elements of following determinants:

(i) \(\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}\)

(ii) \(\begin{vmatrix} a & c \\ b & d \end{vmatrix}\)

Sol:

(i) The given determinant is \(\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}\)

Minor of element aij is Mij .

Therefore, M11 = minor of element a11 = 3

M12 = minor of element a12 = 0

M21 = minor of element a21 = -4

M22 = minor of element a22 = 2

Cofactor of aij is Aij = (-1) i+j Mij .

Therefore, A11 = (-1)1+1 M11 = (-1)2 (3) = 3

A12 = ( -1 )1 + 2 M12 = ( -1 )3 ( 0 ) = 0

A21 = ( – 1 )2 + 1 M21 = ( -1 )3 ( -4 ) = 4

A22 = ( -1 )2 + 2 M22 ( -1 )4 ( 2 ) = 2

 

(ii) The given determinant is \(\begin{vmatrix} a & c \\ b & d \end{vmatrix}\)

Minor of element aij is Mij .

Therefore, M11 = minor of element a11 is Mij .

M12 = minor of element a12 = b

M21 = minor of element a21 = c

M22 = minor of element a22 = a

Cofactor of aij is Aij = (-1)I + j Mij .

Therefore, A11 = (-1)1 + 1 M11 = (-1)2 (d) = d

A12 = ( -1 )1 + 2 M12 = ( -1 )3 ( b ) = -b

A21 = ( -1 )2 + 1 M21 = ( -1 )3 ( c ) = -c

A22 = ( -1 ) 2 + 2 M22 = ( -1 )­4 ( a ) = a

Q2: (i) \(\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\)

(ii) \(\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}\)

Sol:

(i) The given determinant is \(\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\)

By the definition of minors and cofactors, we have :

M11 = minor of a11 = \(\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\) = 1

M12 = minor of a12 = \(\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix}\) = 0

M13 = minor of a13 = \(\begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix}\) = 0

M21 = minor of a21 = \(\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix}\) = 0

M22 = minor of a22 = \(\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\) = 1

M23 = minor of a23 = \(\begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix}\) = 0

M31 = minor of a31 = \(\begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix}\) = 0

M32 = minor of a32 = \(\begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix}\) = 0

M33 = minor of a33 = \(\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\) = 1

A11 = cofactor of a11 = ( -1 )1 + 1 M11 = 1

A12 = cofactor of a12 = ( -1 )1 + 2 M12 = 0

A13 = cofactor of a13 = ( -1 )1 + 3 M13 = 0

A21 = cofactor of a21 = ( -1 )2 + 1 M21 = 0

A22 = cofactor of a22 = ( -1 )2 + 2 M22 = 1

A23 = cofactor of a23 = ( -1 )2 + 3 M23 = 0

A31 = cofactor of a31 = ( -1 )3 + 1 M31 = 0

A32 = cofactor of a32 = ( -1 )3 + 2 M32 = 0

A33 = cofactor of a33 = ( -1 )3 + 3 M33 = 1

(ii) The given determinant is \(\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}\)

By definition of minors and cofactors , we have:

M11 = minor of a11 = \(\begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix}\) = 10 + 1 = 11

M12 = minor of a12 = \(\begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix}\) = 6 – 0 = 6

M13 = minor of a13 = \(\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}\) = 3 – 0 = 3

M21 = minor of a21 = \(\begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix}\) = 0 – 4 = -4

M22 = minor of a22 = \(\begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix}\) = 2 – 0 = 2

M23 = minor of a23 = \(\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}\) = 1 – 0 = 1

M31 = minor of a31 = \(\begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix}\) = 0 – 20 = -20

M32 = minor of a32 = \(\begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix}\) = -1 – 12 = -13

M33 = minor of a33 = \(\begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix}\) = 5 – 0 = 5

A11 = cofactor of a11 = ( -1 )1 + 1 M11 = 11

A12 = cofactor of a12 = ( -1 )1 + 2 M12 = -6

A13 = cofactor of a13 = ( -1 )1 + 3 M13 = 3

A21 = cofactor of a21 = ( -1 )2 + 1 M21 = 4

A22 = cofactor of a22 = ( -1 )2 + 2 M22 = 2

A23 = cofactor of a23 = ( -1 )2 + 3 M23 = -1

A31 = cofactor of a31 = ( -1 )3 + 1 M31 = -20

A32 = cofactor of a32 = ( -1 )3 + 2 M32 = 13

A33 = cofactor of a33 = ( -1 )3 + 3 M33 = 5

Q.3: Using Cofactors of elements of second row, evaluate \(\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}\) .

Sol:

The given determinant is \(\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}\)

We have:

M21 = \(\begin{vmatrix} 3 & 8\\ 2 & 3 \end{vmatrix}\) = 9 – 16 = -7

Therefore, A21 = cofactor of a21 = ( -1 )2 + 1 M21 = 7

M22 = \(\begin{vmatrix} 5 & 8\\ 1 & 3 \end{vmatrix}\) = 15 – 8 = 7

Therefore, A22 = cofactor of a22 = ( -1 )2 + 2 M22 = 7

M23 = \(\begin{vmatrix} 5 & 3\\ 1 & 2 \end{vmatrix}\) = 10 – 3 = 7

Therefore, A23 = cofactor of a23 = ( -1 )2 + 3 M23 = -7

We know that \(\Delta\) is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

Therefore, \(\Delta\) = a21A21 + a22A22 + a23A23 =

2 ( 7 ) + 0 ( 7 ) + 1 ( -7 ) = 14 – 7 = 7

Q4: Using Cofactors of elements of third column, evaluate \(\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}\) .

Sol:

The given determinant is \(\begin{vmatrix} 1 & x & yz\\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}\) .

We have:

M13 = \(\begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix}\) = z – y

M23 = \(\begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix}\) = z – x

M33 = \(\begin{vmatrix} 1 & x \\ 1 & \end{vmatrix}\) = y – x

Therefore, A13 = cofactor of a13 = ( -1 )1 + 3 M13 = ( z – y )

A23 = cofactor of a23 = ( -1 )2 + 3 M23 = – ( z – x ) = ( x – z )

A33 = cofactor of a33 = ( -1 )3 + 3 M33 = ( y – x )

We know that \(\Delta\) is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

Therefore, \(\Delta\) = a13A13 + a23A23 + a33A33

= yz ( z – y ) + zx ( x – z ) + xy ( y – x )

= yz2 – y2 z + x2 z + x2 z – xz2 + xy2 – x2 y

= ( x2 z – y2 z ) + ( yz2 – xz2 ) + ( xy2 – x2 y )

= z ( x2 – y2 ) + z2 ( y – x ) + xy ( y – x )

= z ( x – y ) ( x + y ) + 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 )

Hence, \(\Delta\) = ( x – y ) ( y – z ) ( z – x )

 

 

Q5: For the matrices A and B, verify that (AB)’ = B’A’ where

(i) A = \(\begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}\) ,

B = \(\begin{bmatrix} -1 & 2 & 1 \end{bmatrix}\)

(ii) A = = \(\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\) ,

B = \(\begin{bmatrix} 1 & 5 & 7 \end{bmatrix}\)

Sol:

(i) AB = \(\begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}\)

Therefore, ( AB )’ = \(\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}\)

Now, A’ = \(\begin{bmatrix} -1 & -4 & 3 \end{bmatrix}\) ,

B’ = \(\begin{bmatrix} -1\\ 2 \\ 1 \end{bmatrix}\)

Therefore, B’A’= \(\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}\)

Hence, we have verified that ( AB )’ = B’A’ .

(ii) AB = \(\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}\)

Therefore, ( AB )’ = \(\begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}\)

Now, A’ = \(\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}\) ,

B’ = \(\begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}\)

Therefore, B’A’= \(\begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}\)

Hence, we have verified that (AB)’ = B’A’

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