Exercise 4.3 (Revised) - Chapter 4 - Determinants - Ncert Solutions class 12 - Maths

You can Download the Exercise 4.3 (Revised) - Chapter 4 - Determinants - Ncert Solutions class 12 - Maths with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Chapter 4 - Determinants NCERT Solutions Class 12 Maths | Step-by-Step Solutions & Explanations

Ex 4.3 Question 1:
Write Minors and Cofactors of the elements of following determinants:
(i) $\left|\begin{array}{rr}2 & -4 \\ 0 & 3\end{array}\right|_{\text {(ii) }}\left|\begin{array}{ll}a & c \\ b & d\end{array}\right|$

Answer
(i) The given determinant is $\left|\begin{array}{rr}2 & -4 \\ 0 & 3\end{array}\right|$.

Minor of element $a_{i j}$ is $M_{i j}$.
$\therefore \mathrm{M}_{11}=$ 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 $\mathrm{A}_{i j}=(-1)^{i+j} \mathrm{M}_{i j}$.
$
\therefore \mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=(-1)^2(3)=3
$

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

Minor of element $a_{i j}$ is $M_{i j}$.
$
\therefore \mathrm{M}_{11}=\text { 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 $\mathrm{A}_{i j}=(-1)^{i+j} \mathrm{M}_{i j}$.
$
\therefore \mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=(-1)^2(d)=d
$

$\begin{aligned}
& \mathrm{A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=(-1)^3(b)=-b \\
& \mathrm{~A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=(-1)^3(c)=-c \\
& \mathrm{~A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=(-1)^4(a)=a
\end{aligned}$

Ex 4.3 Question 2:
(i) $\left|\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|_{\text {(ii) }}\left|\begin{array}{rrr}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$

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

By the definition of minors and cofactors, we have:

$\begin{aligned}
& M_{11}=\text { minor of } a_{11}=\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|=1 \\
& M_12=\text { minor of } a_12=\left|\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right|=0
\end{aligned}$

$\begin{aligned}
& M_{13}=\text { minor of } a_{13}=\left|\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right|=0 \\
& M_{21}=\text { minor of } a_{21}=\left|\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right|=0 \\
& M_{22}=\text { minor of } a_{22}=\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|=1
\end{aligned}$

$\begin{aligned}
& M_{23}=\text { minor of } a_{23}=\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|=0 \\
& M_{31}=\text { minor of } a_{31}=\left|\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right|=0 \\
& M_{32}=\text { minor of } a_{32}=\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|=0 \\
& M_{33}=\text { minor of } a_{33}=\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|=1
\end{aligned}$

$\begin{aligned}
& \mathrm{A}_{11}=\text { cofactor of } a_{11}=(-1)^{1+1} \mathrm{M}_{11}=1 \\
& \mathrm{~A}_{12}=\text { cofactor of } a_{12}=(-1)^{1+2} \mathrm{M}_{12}=0 \\
& \mathrm{~A}_{13}=\text { cofactor of } a_{13}=(-1)^{1+3} \mathrm{M}_{13}=0 \\
& \mathrm{~A}_{21}=\text { cofactor of } a_{21}=(-1)^{2+1} \mathrm{M}_{21}=0 \\
& \mathrm{~A}_{22}=\text { cofactor of } a_{22}=(-1)^{2+2} \mathrm{M}_{22}=1 \\
& \mathrm{~A}_{23}=\text { cofactor of } a_{23}=(-1)^{2+3} \mathrm{M}_{23}=0 \\
& \mathrm{~A}_{31}=\text { cofactor of } a_{31}=(-1)^{3+1} \mathrm{M}_{31}=0 \\
& \mathrm{~A}_{32}=\text { cofactor of } a_{32}=(-1)^{3+2} \mathrm{M}_{32}=0 \\
& \mathrm{~A}_{33}=\text { cofactor of } a_{33}=(-1)^{3+3} \mathrm{M}_{33}=1
\end{aligned}$

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

By definition of minors and cofactors, we have:
$
\begin{aligned}
& \mathrm{M}_{11}=\text { minor of } a_{11}=\left|\begin{array}{cc}
5 & -1 \\
1 & 2
\end{array}\right|=10+1=11 \\
& \mathrm{M}_{12}=\text { minor of } a_{12}=\left|\begin{array}{cc}
3 & -1 \\
0 & 2
\end{array}\right|=6-0=6 \\
& \mathrm{M}_{13}=\text { minor of } a_{13}=\left|\begin{array}{ll}
3 & 5 \\
0 & 1
\end{array}\right|=3-0=3 \\
& \mathrm{M}_{21}=\text { minor of } a_{21}=\left|\begin{array}{ll}
0 & 4 \\
1 & 2
\end{array}\right|=0-4=-4
\end{aligned}
$

$\begin{aligned}
& M_{22}=\text { minor of } a_{22}=\left|\begin{array}{ll}
1 & 4 \\
0 & 2
\end{array}\right|=2-0=2 \\
& M_{23}=\text { minor of } a_{23}=\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|=1-0=1 \\
& M_{31}=\text { minor of } a_{31}=\left|\begin{array}{cc}
0 & 4 \\
5 & -1
\end{array}\right|=0-20=-20 \\
& M_{32}=\text { minor of } a_{32}=\left|\begin{array}{cc}
1 & 4 \\
3 & -1
\end{array}\right|=-1-12=-13 \\
& M_{33}=\text { minor of } a_{33}=\left|\begin{array}{ll}
1 & 0 \\
3 & 5
\end{array}\right|=5-0=5
\end{aligned}$

$\begin{aligned}
& \mathrm{A}_{11}=\text { cofactor of } a_{11}=(-1)^{1+1} \mathrm{M}_{11}=11 \\
& \mathrm{~A}_{12}=\text { cofactor of } a_{12}=(-1)^{1+2} \mathrm{M}_{12}=-6 \\
& \mathrm{~A}_{13}=\text { cofactor of } a_{13}=(-1)^{1+3} \mathrm{M}_{13}=3 \\
& \mathrm{~A}_{21}=\text { cofactor of } a_{21}=(-1)^{2+1} \mathrm{M}_{21}=4 \\
& \mathrm{~A}_{22}=\text { cofactor of } a_{22}=(-1)^{2+2} \mathrm{M}_{22}=2 \\
& \mathrm{~A}_{23}=\text { cofactor of } a_{23}=(-1)^{2+3} \mathrm{M}_{23}=-1 \\
& \mathrm{~A}_{31}=\text { cofactor of } a_{31}=(-1)^{3+1} \mathrm{M}_{31}=-20 \\
& \mathrm{~A}_{32}=\text { cofactor of } a_{32}=(-1)^{3+2} \mathrm{M}_{32}=13 \\
& \mathrm{~A}_{33}=\text { cofactor of } a_{33}=(-1)^{3+3} \mathrm{M}_{33}=5
\end{aligned}$

Ex 4.3 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| .
$

Answer

The given determinant is $\left|\begin{array}{lll}2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$.
We have:
$
M_{21}=\left|\begin{array}{ll}
3 & 8 \\
2 & 3
\end{array}\right|=9-16=-7
$
$\therefore \mathrm{A}_{21}=$ cofactor of $a_{21}=(-1)^{2+1} \mathrm{M}_{21}=7$

$
\begin{aligned}
& M_{23}=\left|\begin{array}{ll}
5 & 3 \\
1 & 2
\end{array}\right|=10-3=7 \\
& \therefore A_{23}=\text { cofactor of } a_{23}=(-1)^{2+3} M_{23}=-7
\end{aligned}
$

We know that $\Delta$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
$
A \Delta=a_{21} A_{21}+a_{22} A_{22}+a_{23} A_{23}=2(7)+0(7)+1(-7)=14-7=7
$

Ex 4.3 Question 4:

Using Cofactors of elements of third column, evaluate
$
\Delta=\left|\begin{array}{lll}
1 & x & y z \\
1 & y & z x \\
1 & z & x y
\end{array}\right|
$

Answer

The given determinant is
$
\left|\begin{array}{lll}
1 & x & y z \\
1 & y & z x \\
1 & z & x 9
\end{array}\right| .
$

We have:
$
\begin{aligned}
& 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 \\
&
\end{aligned}
$
$
a_{13}=\text { cofactor of } a_{13}=(-1)^{1+3} M_{13}=(z-y)
$

$
\begin{aligned}
& A_{23}=\text { cofactor of } a_{23}=(-1)^{2+3} M_{23}=-(z-x)=(x-z) \\
& A_{23}=\text { cofactor of } a_{33}=(-1)^{3+3} M_{33}=(y-x)
\end{aligned}
$

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

$\begin{aligned}
\Delta & =a_{13} A_{13}+a_{23} A_{23}+a_{33} A_{33} \\
& =y z(z-y)+z x(x-z)+x y(y-x) \\
& =y z^2-y^2 z+z x^2-z^2 x+x y^2-x^2 y \\
& =\left(y z^2-y^2 z\right)+\left(x y^2-z^2 x\right)+\left(z x^2-x^2 y\right) \\
& =y z(z-y)+x\left(y^2-z^2\right)+x^2(z-y) \\
& =-y z(y-z)+x\left(y^2-z^2\right)-x^2(y-z) \\
& =-y z(y-z)+x(y+z)(y-z)-x^2(y-z) \\
& =(y-z)\left(-y z+x(y+z)-x^2\right) \\
& =(y-z)\left(-y z+x y+x z-x^2\right) \\
& =(y-z)(z(x-y)+x(y-x)) \\
& =(y-z)(z(x-y)-x(x-y)) \\
& =(y-z)((z-x)(x-y)) \\
& =(x-y)(y-z)(z-x)
\end{aligned}$

Ex 4.3 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 Cofactor of $a_{i j}$ 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) $a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}$

If $\Delta=\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|$
$\Delta=$ Sum of products of elements of row (or column) with their corresponding cofactors

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

So, option $\mathbf{D}$ is correct