Class 12 ncert solution math exercise 4.4

 EXERCISE 4.4 (Determinants)

Write Minors and Cofactors of the elements of following determinants:(Class 12 ncert solution math exercise 4.4)

Question 1: (i) \left|\begin{array}{cc}2 & -4 \\ 0 & 3\end{array}\right|
(ii) \left|\begin{array}{ll}a & c \\ b & d\end{array}\right|

Solution: (i) The given determinant is \left|\begin{array}{cc}2 & -4 \\ 0 & 3\end{array}\right|

Minor of element a_{i j} is M_{i j}

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

Cofactor of a_{i j} is A_{i j}=(-1)^{i+j} M_{i j}

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) The given determinant is \left|\begin{array}{ll}a & c \\ b & d\end{array}\right|

Minor of element a_{i j} is M_{i j}.

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

Cofactor of a_{i j} is A_{i j}=(-1)^{i+j} M_{i j}

A_{11}=(-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: (i) \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|

(ii) \left|\begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{array}\right|

Solution: (i) The given determinant is \left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|

Minor of element a_{i j} is M_{i j}.

M_{11}=\text { minor of element } a_{11}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1

M_{12}=\text { minor of element } a_{12}=\left|\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right|=0

M_{13}=\text { minor of element } a_{13}=\left|\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right|=0

M_{21}=\text { minor of element } a_{21}=\left|\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right|=0

M_{22}=\text { minor of element } a_{22}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1

M_{23}=\text { minor of element } a_{23}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right|=0

 

M_{31}=\text { minor of element } a_{31}=\left|\begin{array}{ll}0 & 0 \\1 & 0\end{array}\right|=0

M_{32}=\text { minor of element } a_{32}=\left|\begin{array}{ll}1 & 0 \\0 & 0\end{array}\right|=0

M_{33}=\text { minor of element } a_{33}=\left|\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right|=1

Cofactor of a_{i j} is A_{i j}=(-1)^{i+j} M_{i j}

A_{11}=(-1)^{1+1} M_{11}=(-1)^{2}(1)=1
A_{12}=(-1)^{1+2} M_{12}=(-1)^{3}(0)=0
A_{13}=(-1)^{1+3} M_{13}=(-1)^{4}(0)=0
A_{21}=(-1)^{2+1} M_{21}=(-1)^{3}(0)=0
A_{22}=(-1)^{2+2} M_{22}=(-1)^{4}(1)=1
A_{23}=(-1)^{2+3} M_{23}=(-1)^{5}(0)=0
A_{31}=(-1)^{3+1} M_{31}=(-1)^{4}(0)=0
A_{32}=(-1)^{3+2} M_{32}=(-1)^{5}(0)=0
A_{33}=(-1)^{3+3} M_{33}=(-1)^{6}(1)=1

(ii) The given determinant is \left|\begin{array}{ccc}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|

Minor of element a_{i j} is M_{i j}.

M_{11}=\text { minor of element } a_{11}=\left|\begin{array}{cc} 5 & -1 \\ 1 & 2 \end{array}\right|=11

M_{12}=\text { minor of element } a_{12}=\left|\begin{array}{cc} 3 & -1 \\ 0 & 2 \end{array}\right|=6

M_{13}=\text { minor of element } a_{13}=\left|\begin{array}{ll} 3 & 5 \\ 0 & 1 \end{array}\right|=3

M_{21}=\text { minor of element } a_{21}=\left|\begin{array}{ll} 0 & 4 \\ 1 & 2 \end{array}\right|=-4

M_{22}=\text { minor of element } a_{22}=\left|\begin{array}{ll} 1 & 4 \\ 0 & 2 \end{array}\right|=2

 

M_{23}=\text { minor of element } a_{23}=\left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right|=1

M_{31}=\text { minor of element } a_{31}=\left|\begin{array}{cc} 0 & 4 \\ 5 & -1 \end{array}\right|=-20

M_{32}=\text { minor of element } a_{32}=\left|\begin{array}{cc} 1 & 4 \\ 3 & -1 \end{array}\right|=-13

M_{33}=\text { minor of element } a_{33}=\left|\begin{array}{ll} 1 & 0 \\ 3 & 5 \end{array}\right|=5

Cofactor of a_{i j} is A_{i j}=(-1)^{i+j} M_{i j}

A_{11}=(-1)^{1+1} M_{11}=(-1)^{2}(11)=11
A_{12}=(-1)^{1+2} M_{12}=(-1)^{3}(6)=-6
A_{13}=(-1)^{1+3} M_{13}=(-1)^{4}(3)=3
A_{21}=(-1)^{2+1} M_{21}=(-1)^{3}(-4)=4
A_{22}=(-1)^{2+2} M_{22}=(-1)^{4}(2)=2
A_{23}=(-1)^{2+3} M_{23}=(-1)^{5}(1)=-1
A_{31}=(-1)^{3+1} M_{31}=(-1)^{4}(-20)=-20
A_{32}=(-1)^{3+2} M_{32}=(-1)^{5}(-13)=13
A_{33}=(-1)^{3+3} M_{33}=(-1)^{6}(5)=5

Question 3: Using Cofactors of elements of second row, evaluate

\Delta=\left|\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right|

Solution: The given determinant is \left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|

M_{21}= minor of element a_{21}=\left|\begin{array}{ll}3 & 8 \\ 2 & 3\end{array}\right|=-7

A_{21}=(-1)^{2+1} M_{21}=(-1)^{3}(-7)=7

M_{22}=\text { minor of element } a_{22}=\left|\begin{array}{ll} 5 & 8 \\ 1 & 3 \end{array}\right|=15-8=7

A_{22}=(-1)^{2+2} M_{22}=(-1)^{4}(7)=7
M_{23}=\text { minor of element } a_{23}=\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|=7

A_{23}=(-1)^{2+3} M_{21}=(-1)^{5}(7)=-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 =a_{21} A_{21}+a_{22} A_{22}+a_{23} A_{23}
=2(7)+0(7)+1(-7)
=14-7
=7

Question 4: Using Cofactors of elements of second row, evaluate

\Delta=\left|\begin{array}{ccc} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{array}\right|

Solution: The given determinant is \left|\begin{array}{lll}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|

Therefore,

M_{13}=\left|\begin{array}{ll} 1 & y \\ 1 & z \end{array}\right|=z-y

M_{23}=\left|\begin{array}{ll} 1 & x \\ 1 & z \end{array}\right|=z-x

M_{33}=\left|\begin{array}{ll} 1 & x \\ 1 & y \end{array}\right|=y-x

 

A_{13}=(-1)^{1+3} M_{13}=(-1)^{4}(z-y)=z-y
A_{23}=(-1)^{2+3} M_{23}=(-1)^{5}(z-x)=-(z-x)=x-z
A_{33}=(-1)^{3+3} M_{33}=(-1)^{6}(y-x)=y-x

We know that \Delta is equal to the sum of the product of the elements of the third column with their corresponding cofactors.

Therefore,

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

=y z(z-y)+z x(x-z)+x y(y-x)
=y z^{2}-y^{2} z+x^{2} z-x z^{2}+x y^{2}-x^{2} y
=\left(x^{2} z-y^{2} z\right)+\left(y z^{2}-x z^{2}\right)+\left(x y^{2}-x^{2} y\right)
=z\left(x^{2}-y^{2}\right)+z^{2}(y-x)+x y(y-x)
=z(x-y)(x+y)+z^{2}(y-x)+x y(y-x)
=(x-y)\left[z x+z y-z^{2}-x y\right]
=(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)

Question 5:  If \Delta=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| and A_{i j} is Cofactors of a_{ij}, then value of  Δ  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 \Delta is equal to the sum of the product of the elements of a column or row with their corresponding cofactors.

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

Thus, the correct option is D.

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