Exercise 7.2-Additional Questions - Chapter 7 Matrices and Determinants 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Prove that $\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|=(a-b)(b-c)(c-a)$.
Solution:
$
\left|\begin{array}{ccc}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|=\left|\begin{array}{ccc}
0 & a-b & a^2-b^2 \\
0 & b-c & b^2-c^2 \\
1 & c & c^2
\end{array}\right| \begin{aligned}
& \mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3
\end{aligned}
$
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^2\end{array}\right|$ Take $(a-b)$ and $(b-c)$
$
=(a-b)(b-c)[(1)(b+c)-(1)(a+b)]=(a-b)(b-c)(c-a)
$
Question 2.
$
\text { Prove that }\left|\begin{array}{ccc}
a+b+c & -c & -b \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|=2(a+b)(b+c)(c+a)
$

Solution:
$
\begin{aligned}
& =(a+b)(b+c)\left|\begin{array}{ccc}
0 & 2 & 0 \\
-1 & 1 & 1 \\
-b & -a & a+b+c
\end{array}\right| \mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2 \\
& =(a+b)(b+c) \times(-2)\left|\begin{array}{cc}
-1 & 1 \\
-b & a+b+c
\end{array}\right| \\
& =(a+b)(b+c) \times(-2)[-(a+b+c)+b] \\
& =(a+b)(b+c) \times(-2)[-a-c] \\
& \Delta=2(a+b)(b+c)(c+a) \\
&
\end{aligned}
$
Question 3.
$
\text { that }\left|\begin{array}{lll}
1 / a^2 & b c & b+c \\
1 / b^2 & c a & c+a \\
1 / c^2 & a b & a+b
\end{array}\right|=0
$
Solution:

$
\begin{aligned}
& \left|\begin{array}{lll}
1 / a^2 & b c & b+c \\
1 / b^2 & c a & c+a \\
1 / c^2 & a b & a+b
\end{array}\right|=\frac{1}{a b c}\left|\begin{array}{ccc}
1 / a & a b c & a(b+c) \\
1 / b & a b c & b(c+a) \\
1 / c & a b c & c(a+b)
\end{array}\right| \text { Multiply } \mathrm{R}_1, R_2, R_3 \\
& =\frac{a b c}{a b c}\left|\begin{array}{lll}
1 / a & 1 & a(b+c) \\
1 / b & 1 & b(c+a) \\
1 / c & 1 & c(a+b)
\end{array}\right| \text { Take } a b c \text { from } \mathrm{C}_2 \\
& =\frac{1}{a b c}\left|\begin{array}{lll}
b c & 1 & a(b+c) \\
c a & 1 & b(c+a) \\
a b & 1 & c(a+b)
\end{array}\right| \text { Multiply } \mathrm{C}_1 \text { by } a b c \\
& =\frac{1}{a b c}\left|\begin{array}{lll}
b c & 1 & a b+b c+c a \\
c a & 1 & a b+b c+c a \\
a b & 1 & a b+b c+c a
\end{array}\right| \mathrm{C}_3 \rightarrow \mathrm{C}_3+\mathrm{C}_1 \\
& =\frac{(a b+b c+c a)}{a b c}\left|\begin{array}{lll}
b c & 1 & 1 \\
c a & 1 & 1 \\
a b & 1 & 1
\end{array}\right| \text { Take }(a b+b c+c a) \text { from } \mathrm{C}_3 \\
& =\frac{(a b+b c+c a)}{a b c}(0) \\
& =0 \\
&
\end{aligned}
$
Question 4.
$
\text { Prove that }\left|\begin{array}{ccc}
a^2+\lambda & a b & a c \\
a b & b^2+\lambda & b c \\
a c & b c & c^2+\lambda
\end{array}\right|=\lambda^2\left(a^2+b^2+c^2+\lambda\right)
$
Solution:

Let $\Delta=\left|\begin{array}{ccc}a^2+\lambda & a b & a c \\ a b & b^2+\lambda & b c \\ a c & b c & c^2+\lambda\end{array}\right|$
Multiply $\mathrm{R}_1, \mathrm{R}_2$ and $\mathrm{R}_3$, by $a, b$ and $c$ respectively
$
\Delta=\frac{1}{a b c}\left|\begin{array}{ccc}
a\left(a^2+\lambda\right) & a^2 b & a^2 c \\
a b^2 & b\left(b^2+\lambda\right) & b^2 c \\
a c^2 & b c^2 & c\left(c^2+\lambda\right)
\end{array}\right|
$
Take $a, b$ and $c$ from $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ respectively
$
\begin{aligned}
& \Delta=\frac{a b c}{a b c}\left|\begin{array}{ccc}
a^2+\lambda & a^2 & a^2 \\
b^2 & b^2+\lambda & b^2 \\
c^2 & c^2 & c^2+\lambda
\end{array}\right| \\
& =\left|\begin{array}{ccc}
a^2+b^2+c^2+\lambda & a^2+b^2+c^2+\lambda & a^2+b^2+c^2+\lambda \\
b^2 & b^2+\lambda & b^2 \\
c^2 & c^2 & c^2+\lambda
\end{array}\right| \mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R} \\
& =\left(a^2+b^2+c^2+\lambda\right)\left|\begin{array}{ccc}
1 & 1 & 1 \\
b^2 & b^2+\lambda & b^2 \\
c^2 & c^2 & c^2+\lambda
\end{array}\right| \\
& =\left(a^2+b^2+c^2+\lambda\right)\left|\begin{array}{lll}
1 & 0 & 0 \\
b^2 & \lambda & 0 \\
c^2 & 0 & \lambda
\end{array}\right| \begin{array}{l}
\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1 \\
\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1
\end{array} \\
& =\left(a^2+b^2+c^2+\lambda\right)\left|\begin{array}{ll}
\lambda & 0 \\
0 & \lambda
\end{array}\right| \\
& \therefore\left|\begin{array}{ccc}
a^2+\lambda & a b & a c \\
a b & b^2+\lambda & b c \\
a c & b c & c^2+\lambda
\end{array}\right|=\lambda^2\left(a^2+b^2+c^2+\lambda\right) \\
&
\end{aligned}
$

Question 5.
Prove that $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|=x y$.
Solution:
$
\begin{aligned}
\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+y
\end{array}\right| & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
0 & x & 0 \\
0 & 0 & y
\end{array}\right| \begin{array}{l}
\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1 \\
\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1
\end{array} \\
& =x y[\because \text { upper diagonal matrix }]
\end{aligned}
$
Question 6.
Prove that $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^3$
Solution:

$
\begin{aligned}
& =(a+b+c)\left|\begin{array}{ccc}
0 & 0 & 1 \\
a+b+c & -(a+b+c) & 2 b \\
0 & a+b+c & c-a-b
\end{array}\right| \\
&
\end{aligned}
$
Expanding along $\mathrm{R}_1$
$
\begin{aligned}
& =(a+b+c)\left\{0()-0()+1\left|\begin{array}{cc}
a+b+c & -(a+b+c) \\
0 & a+b+c
\end{array}\right|\right\} \\
& =(a+b+c)\left[1(a+b+c)^2-0\right] \\
& =(a+b+c)^3=\text { RHS }
\end{aligned}
$
Question 7.
Prove that $\left|\begin{array}{lll}1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$.

Solution:
$
\begin{aligned}
& \mathrm{LHS}=\left|\begin{array}{ccc}
1 & a & a^3 \\
1 & b & b^3 \\
1 & c & c^3
\end{array}\right| \begin{array}{l}
\mathrm{C}_1 \rightarrow \mathrm{C}_1-\mathrm{C}_2 \\
\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_3
\end{array} \\
& =\left|\begin{array}{ccc}
0 & a-b & a^3-b^3 \\
0 & b-c & b^3-c^3 \\
1 & c & c^3
\end{array}\right| \text { Expanding along } \mathrm{C}_1 \\
& =(0)-(0)+1\left|\begin{array}{ll}
a-b & a^3-b^3 \\
b-c & b^3-c^3
\end{array}\right| \\
& =(a-b)\left(b^3-c^3\right)-(b-c)\left(a^3-b^3\right) \\
& =(a-b)(b-c)\left(b^2+b c+c^2\right)-(b-c)(a-b)\left(a^2+a b+b^2\right) \\
& =(a-b)(b-c)\left[\left(b^2+b c+c^2\right)-\left(a^2+a b+b^2\right)\right] \\
& =(a-b)(b-c)\left[b^2+b c+c^2-a^2-a b-b^2\right] \\
& =(a-b)(b-c)\left[(b c-a b)+\left(c^2-a^2\right)\right] \\
& =(a-b)(b-c)[(b(c-a)+(c+a)(c-a)] \\
& =(a-b)(b-c)[b+c+a](c-a) \\
& =(a-b)(b-c)(c-a)(a+b+c)=\text { RHS } \\
&
\end{aligned}
$
Question 8 .
Prove that $\left|\begin{array}{lll}y+z & x & y \\ z+x & z & x \\ x+y & y & z\end{array}\right|=(x+y+z)(x-z)^2$.
Solution:

$
\begin{aligned}
& \mathrm{LHS}=\left|\begin{array}{lll}
y+z & x & y \\
z+x & z & x \\
x+y & y & z
\end{array}\right| \quad \mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3 \\
& =\left|\begin{array}{ccc}
2(x+y+z) & (x+y+z) & (x+y+z) \\
z+x & z & x \\
x+y & y & z
\end{array}\right| \\
& =(x+y+z)\left|\begin{array}{ccc}
2 & 1 & 1 \\
z+x & z & x \\
x+y & y & z
\end{array}\right|\left[\text { Take }(x+y+z) \text { from } \mathrm{R}_1\right] \\
& =(x+y+z)\left|\begin{array}{ccc}
0 & 1 & 1 \\
0 & z & x \\
x-z & y & z
\end{array}\right| \mathrm{C}_1 \rightarrow \mathrm{C}_1-\mathrm{C}_2-\mathrm{C}_3 \\
&
\end{aligned}
$
Expand along $\mathrm{C}_1$
$
\begin{aligned}
& =(x+y+z)\left\{(x-z)\left|\begin{array}{ll}
1 & 1 \\
z & x
\end{array}\right|\right\} \\
& =(x+y+z)\{(x-z)(x-z)\} \\
& =(x+y+z)(x-z)^2 \\
& =\text { RHS }
\end{aligned}
$
Question 9.
Prove that $\left|\begin{array}{ccc}-a^2 & a b & a c \\ a b & -b^2 & b c \\ a c & b c & -c^2\end{array}\right|=4 a^2 b^2 c^2$.
Solution:

$
\text { LHS }=\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right| .
$
Taking $a$ from $\mathrm{R}_1, b$ from $\mathrm{R}_2$ and $c$ from $\mathrm{R}_3$ as common factors we get,
$
=(a b c)\left|\begin{array}{ccc}
-a & b & c \\
a & -b & c \\
a & b & -c
\end{array}\right|
$
Again taking $a$ from $\mathrm{C}_1, b$ from $\mathrm{C}_2$ and $c$ from $\mathrm{C}_3$ we get,
$
\begin{aligned}
& =(a b c)(a b c)\left|\begin{array}{ccc}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right| \\
& =(a b c)^2\left|\begin{array}{ccc}
0 & 0 & 2 \\
2 & 0 & 0 \\
1 & 1 & -1
\end{array}\right| \begin{array}{l}
\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2 \\
\mathrm{R}_2 \rightarrow \mathrm{R}_2+\mathrm{R}_3
\end{array} \\
& =(a b c)^2\left[0+0+2\left|\begin{array}{ll}
2 & 0 \\
1 & 1
\end{array}\right|\right] \\
& =(a b c)^2(4)=4 a^2 b^2 c^2=\mathrm{RHS}
\end{aligned}
$