Determinants 4.3 NCERT Solutions

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

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

Solution

For a determinant of order 2 \times 2 , the minor of an element is obtained by deleting the row and column containing that element.

Minor of 2

Deleting 1st row and 1st column, we get

M_{11} = 3

Cofactor:

A_{11} = (-1)^{1+1} M_{11} = (+1)(3)=3

Minor of -4

Deleting 1st row and 2nd column, we get

M_{12} = 0

Cofactor:

A_{12} = (-1)^{1+2} M_{12} = (-1)(0)=0

Minor of 0

Deleting 2nd row and 1st column, we get

M_{21} = -4

Cofactor:

A_{21} = (-1)^{2+1} M_{21} = (-1)(-4)=4

Minor of 3

Deleting 2nd row and 2nd column, we get

M_{22} = 2

Cofactor:

A_{22} = (-1)^{2+2} M_{22} = (+1)(2)=2

Hence, the minors are:

M_{11}=3,\quad M_{12}=0,\quad M_{21}=-4,\quad M_{22}=2

And the cofactors are:

A_{11}=3,\quad A_{12}=0,\quad A_{21}=4,\quad A_{22}=2

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

Solution

Minor of a

Deleting 1st row and 1st column, we get

M_{11}=d

Cofactor:

A_{11}=(-1)^{1+1}M_{11}=(+1)(d)=d

Minor of c

Deleting 1st row and 2nd column, we get

M_{12}=b

Cofactor:

A_{12}=(-1)^{1+2}M_{12}=(-1)(b)=-b

Minor of b

Deleting 2nd row and 1st column, we get

M_{21}=c

Cofactor:

A_{21}=(-1)^{2+1}M_{21}=(-1)(c)=-c

Minor of d

Deleting 2nd row and 2nd column, we get

M_{22}=a

Cofactor:

A_{22}=(-1)^{2+2}M_{22}=(+1)(a)=a

Hence, the minors are:

M_{11}=d,\quad M_{12}=b,\quad M_{21}=c,\quad M_{22}=a

And the cofactors are:

A_{11}=d,\quad A_{12}=-b,\quad A_{21}=-c,\quad A_{22}=a

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

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

Solution

The minor of an element is obtained by deleting the row and column containing that element.

Finding Minors

M_{11}=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1

M_{12}=\begin{vmatrix}0&0\\0&1\end{vmatrix}=0

M_{13}=\begin{vmatrix}0&1\\0&0\end{vmatrix}=0

M_{21}=\begin{vmatrix}0&0\\0&1\end{vmatrix}=0

M_{22}=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1

M_{23}=\begin{vmatrix}1&0\\0&0\end{vmatrix}=0

M_{31}=\begin{vmatrix}0&0\\1&0\end{vmatrix}=0

M_{32}=\begin{vmatrix}1&0\\0&0\end{vmatrix}=0

M_{33}=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1

Finding Cofactors using A_{ij}=(-1)^{i+j}M_{ij}

A_{11}=1,\quad A_{12}=0,\quad A_{13}=0

A_{21}=0,\quad A_{22}=1,\quad A_{23}=0

A_{31}=0,\quad A_{32}=0,\quad A_{33}=1

Hence, the minors are:

M_{11}=1,\ M_{12}=0,\ M_{13}=0,\ M_{21}=0,\ M_{22}=1,\ M_{23}=0,\ M_{31}=0,\ M_{32}=0,\ M_{33}=1

And the cofactors are:

A_{11}=1,\ A_{12}=0,\ A_{13}=0,\ A_{21}=0,\ A_{22}=1,\ A_{23}=0,\ A_{31}=0,\ A_{32}=0,\ A_{33}=1

💡Quick Note:

Minors and cofactors differ only by sign. The numerical value remains the same, but cofactors include the sign factor (-1)^{i+j} . Because of this, elements at positions having odd value of i+j get a negative sign, while even positions keep the same sign.

The sign pattern is:

\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}

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

Solution

Finding Minors

M_{11}=\begin{vmatrix}5&-1\\1&2\end{vmatrix}=10+1=11

M_{12}=\begin{vmatrix}3&-1\\0&2\end{vmatrix}=6

M_{13}=\begin{vmatrix}3&5\\0&1\end{vmatrix}=3

M_{21}=\begin{vmatrix}0&4\\1&2\end{vmatrix}=0-4=-4

M_{22}=\begin{vmatrix}1&4\\0&2\end{vmatrix}=2

M_{23}=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1

M_{31}=\begin{vmatrix}0&4\\5&-1\end{vmatrix}=0-20=-20

M_{32}=\begin{vmatrix}1&4\\3&-1\end{vmatrix}=-1-12=-13

M_{33}=\begin{vmatrix}1&0\\3&5\end{vmatrix}=5

Finding Cofactors using A_{ij}=(-1)^{i+j}M_{ij}

A_{11}=11,\quad A_{12}=-6,\quad A_{13}=3

A_{21}=4,\quad A_{22}=2,\quad A_{23}=-1

A_{31}=-20,\quad A_{32}=13,\quad A_{33}=5

Hence, the minors are:

M_{11}=11,\ M_{12}=6,\ M_{13}=3,\ M_{21}=-4,\ M_{22}=2,\ M_{23}=1,\ M_{31}=-20,\ M_{32}=-13,\ M_{33}=5

And the cofactors are:

A_{11}=11,\ A_{12}=-6,\ A_{13}=3,\ A_{21}=4,\ A_{22}=2,\ A_{23}=-1,\ A_{31}=-20,\ A_{32}=13,\ A_{33}=5

3. Using Cofactors of elements of second row, evaluate

\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}

Solution

We expand the determinant using cofactors of the second row.

\Delta = a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}

Here,

a_{21}=2,\quad a_{22}=0,\quad a_{23}=1

Finding Cofactor A_{21}

M_{21}=\begin{vmatrix}3&8\\2&3\end{vmatrix}=9-16=-7

A_{21}=(-1)^{2+1}(-7)=7

Finding Cofactor A_{22}

M_{22}=\begin{vmatrix}5&8\\1&3\end{vmatrix}=15-8=7

A_{22}=(-1)^{2+2}(7)=7

Finding Cofactor A_{23}

M_{23}=\begin{vmatrix}5&3\\1&2\end{vmatrix}=10-3=7

A_{23}=(-1)^{2+3}(7)=-7

Substituting these values:

\Delta = 2(7)+0(7)+1(-7)

=14+0-7

=7

Hence, the value of determinant is 7

4. Using Cofactors of elements of third column, evaluate

\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}

Solution

We expand the determinant using cofactors of the third column.

\Delta = a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}

Here,

a_{13}=yz,\quad a_{23}=zx,\quad a_{33}=xy

Finding Cofactor A_{13}

M_{13}=\begin{vmatrix}1&y\\1&z\end{vmatrix}=z-y

A_{13}=(-1)^{1+3}(z-y)=z-y

Finding Cofactor A_{23}

M_{23}=\begin{vmatrix}1&x\\1&z\end{vmatrix}=z-x

A_{23}=(-1)^{2+3}(z-x)=x-z

Finding Cofactor A_{33}

M_{33}=\begin{vmatrix}1&x\\1&y\end{vmatrix}=y-x

A_{33}=(-1)^{3+3}(y-x)=y-x

Substituting these values:

\Delta = yz(z-y)+zx(x-z)+xy(y-x)

Expanding the terms:

\Delta = yz^2-y^2z+x^2z-z^2x+xy^2-x^2y

Now grouping the terms of x^2, x and without x, step-by-step:

\Delta = (x^2z-x^2y)+(xy^2-z^2x)+(yz^2-y^2z)

Now taking x^2, x and yz common from three groups resp., we get

\Delta = x^2(z-y)+x(y^2-z^2)+yz(z-y)

Writing y^2-z^2 = (y-z)(y+z) only in next step, we get

\Delta = x^2(z-y)+x(y-z)(y+z)+yz(z-y)

Now taking (z-y) common from all the terms, we get

\Delta = (z-y)(x^2-x(y+z)+yz)

or \Delta = (z-y)(x^2-xy-xz+yz)

Now factorising the second expression:

\Delta = (z-y)\{x(x-y)-z(x-y)\}

Therefore,

\Delta = (z-y)(x-y)(x-z)

Rearranging the terms, we get

\Delta = (x-y)(y-z)(z-x)

Hence, the value of determinant is

\Delta = (x-y)(y-z)(z-x)

5. If \Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} and A_{ij} are cofactors of a_{ij} , then value of \Delta is given by

• (A) a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}
• (B) a_{11}A_{11}+a_{12}A_{21}+a_{13}A_{31}
• (C) a_{21}A_{11}+a_{22}A_{12}+a_{23}A_{13}
• (D) a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}

Solution

We know that the value of a determinant is obtained by multiplying the elements of a row (or column) with their corresponding cofactors.

Expanding along the first column, we get

\Delta = a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}

This matches option (D).

Hence, the correct answer is (D).