Exercise 7.3 - Chapter 7 Matrices and Determinants 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$
\text { Ex } 7.3
$

Solve the following problems by using factor theorem

Question 1.
Show that $<\left|\begin{array}{lll}\boldsymbol{x} & \boldsymbol{a} & \boldsymbol{a} \\ \boldsymbol{a} & \boldsymbol{x} & \boldsymbol{a} \\ \boldsymbol{a} & \boldsymbol{a} & \boldsymbol{x}\end{array}\right|=(\mathrm{x}-\mathrm{a})^2(\mathrm{x}+2 \mathrm{a})$
Solution:
$
\Delta=\left|\begin{array}{lll}
x & a & a \\
a & x & a \\
a & a & x
\end{array}\right|
$
put $x=a$ we get
$
\left|\begin{array}{lll}
a & a & a \\
a & a & a \\
a & a & a
\end{array}\right|=0
$

Since all the three rows are identical we get $(x-a)^2$ as a factor Now, put $x=-2 a$.
We get
$
\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
-2 a & a & a \\
a & -2 a & a \\
a & a & -2 a
\end{array}\right| \\
& =\left|\begin{array}{ccc}
0 & a & a \\
0 & -2 a & a \\
0 & a & -2 a
\end{array}\right| \mathrm{C}_1=\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3=0
\end{aligned}
$
$\Rightarrow(\mathrm{x}+2 \mathrm{~d})$ is a factor of $\mathrm{A}$.
Now degree of $\Delta$ is $3\left(x \times x \times x=x^3\right)$ and we have 3 factors for $A$ $\therefore$ There can be a constant as a factor for $\mathrm{A}$.
$
\text { (i.e.) } \Delta=\mathrm{k}(\mathrm{x}-\mathrm{a})^2(\mathrm{x}+2 \mathrm{~d})
$
equating coefficient of $x^3$ on either sides we get $k=1$
$
\therefore \Delta=(x-a)^2(x+2 a)
$

Question 2.
$
\left|\begin{array}{lll}
b+c & a-c & a-b \\
b-c & c+a & b-a \\
c-b & c-a & a+b
\end{array}\right|=8 a b c
$
Solution:
Let $\Delta=\left|\begin{array}{lll}b+c & a-c & a-b \\ b-c & c+a & b-a \\ c-b & c-a & a+b\end{array}\right|$
Put $a=0$ in $\Delta$
then $\Delta=\left|\begin{array}{ccc}b+c & -c & -b \\ b-c & c & b \\ c-b & c & b\end{array}\right|=0$
$\left(\because C_2\right.$ and $C_3$ are proportional)
$\therefore a-0$ (i.e.,) $a$ is a factor.
Similarly $b$ and $\mathrm{c}$ are factors of $\Delta$.
The product of the leading diagonal elements is $(b+c)(c+a)(a+b)$
The degree is 3 . And we got 3 factors for $\Delta \therefore \mathrm{m}=3-3=0$
$\therefore$ there can be a constant $\mathrm{k}$ as a factor for $\Delta$.
$
\text { (i.e.) } \Delta=\left|\begin{array}{lll}
b+c & a-c & a-b \\
b-c & c+a & b-a \\
c-b & c-a & a+b
\end{array}\right|=k a b c
$
To find $k$ :
put $a=1, b=1, c=1$
$
\therefore \quad \Delta=\left|\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right|=k(1)(1)(1)
$
$
\begin{aligned}
\Rightarrow \quad 2(4-0) 0-0 & =k \\
8 & =k
\end{aligned}
$
So, $\Delta=8 a b c$ (i.e.,) $\left|\begin{array}{lll}b+c & a-c & a-b \\ b-c & c+a & b-a \\ c-b & c-a & a+b\end{array}\right|=8 a b c$

Question 3.
Solve $\left|\begin{array}{ccc}x+a & b & c \\ a & x+b & c \\ a & b & x+c\end{array}\right|=0$
Solution:
Put $x=-(a+b+c)$
$
\text { Then } \begin{aligned}
\Delta & =\left|\begin{array}{ccc}
-(a+b+c)+a & b & c \\
a & -(a+b+c)+b & c \\
a & b & -(a+b+c)+c
\end{array}\right| \\
& =\left|\begin{array}{ccc}
-b-c & b & c \\
a & -a-c & c \\
a & b & -a-b
\end{array}\right| \mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \\
& =\left|\begin{array}{ccc}
0 & b & c \\
0 & -a-c & c \\
0 & b & -a-b
\end{array}\right|=0 \text { (expanding along } \mathrm{C}_1 \text { ) } \\
& \Rightarrow x=-(a+b+c) \text { is a root. }
\end{aligned}
$
Put $x=0$, then
$
\Delta=\left|\begin{array}{lll}
a & b & c \\
a & b & c \\
a & b & c
\end{array}\right|=0 \quad\left(\because \mathrm{R}_1=\mathrm{R}_2=\mathrm{R}_3\right)
$
$\Rightarrow \mathrm{x}=0,0$ are roots.
Now the degree of the leading diagonal elements is 3 .
$\therefore$ the equation is of degree 3 , so the roots are $0,0,-(a+b+c)$

Question 4 .
$
\left|\begin{array}{lll}
\boldsymbol{b}+\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{a}^2 \\
\boldsymbol{c}+\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{b}^2 \\
\boldsymbol{a}+\boldsymbol{b} & \boldsymbol{c} & \boldsymbol{c}^2
\end{array}\right|=(a+b+c)(a-b)(b-c)(c-a)
$
Solution:
Let Put $a=b$, then $\Delta=\left|\begin{array}{ccc}b+c & b & b^2 \\ b+c & b & b^2 \\ 2 b & c & c^2\end{array}\right|=0\left(\because \mathrm{R}_1=\mathrm{R}_2\right)$

$\Rightarrow(a-b)$ is a factor of $\Delta$.
Similarly $(b-c)$ and $(\mathrm{c}-\mathrm{a})$ are factors of $\Delta$.
The degree of the product of elements along leading diagonal is $1+1+2=4$ and we got 3 factors for $\Delta$. $\mathrm{m}=4-3=1$
$\therefore$ There can be one more factor symmetric with $\mathrm{a}, \mathrm{b}, \mathrm{c}$ which is of the form $\mathrm{k}(\mathrm{a}+\mathrm{b}+\mathrm{c})$.
$
\text { Now } \Delta=\left|\begin{array}{lll}
b+c & a & a^2 \\
c+a & b & b^2 \\
a+b & c & c^2
\end{array}\right|=k(a+b+c)(a-b)(b-c)(c-a)
$
To find $k:$ Put $a=0, b=1, c=2$
We get $\Delta=\left|\begin{array}{lll}3 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 2 & 4\end{array}\right|=k(0+1+2)(0-1)(1-2)(2-0)$
(i.e.,) $3(4-2)=k(3)(-1)(-1)(2)($ i.e.,) $6=6 k \Rightarrow k=1$
$
\therefore \Delta=\left|\begin{array}{lll}
b+c & a & a^2 \\
c+a & b & b^2 \\
a+b & c & c^2
\end{array}\right|=(a+b+c)(a-b)(b-c)(c-a)
$
Question 5.
$
\left|\begin{array}{lll}
4-x & 4+x & 4+x \\
4+x & 4-x & 4+x \\
4+x & 4+x & 4-x
\end{array}\right|=0
$
Solution:

Let $\Delta=\left|\begin{array}{ccc}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|=\left|\begin{array}{ccc}12+x & 12+x & 12+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|\left(\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3\right)$
Put $x=-12$. We get
Put $x=0$ we get
$
\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
0 & 0 & 0 \\
-8 & 16 & -8 \\
-8 & -8 & 16
\end{array}\right|=0 \\
\Rightarrow x & =-12 \text { is a root. } \\
\Delta & =\left|\begin{array}{ccc}
4 & 4 & 4 \\
4 & 4 & 4 \\
4 & 4 & 4
\end{array}\right|=0
\end{aligned}
$
all the 3 rows are identical
So,
$
\begin{aligned}
\Rightarrow \quad(x-0)^2 & =x^2 \text { is a factor } \\
\Delta & =0 \Rightarrow x^2(x+12)=0 \\
\Rightarrow \quad x & =0,0,-12 .
\end{aligned}
$
Question 6.
$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
\boldsymbol{x} & y & z \\
x^2 & y^2 & z^2
\end{array}\right|_{=(x-y)(y-z)(z-x)}
$
Solution:
Let $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2\end{array}\right|$
Put $\quad x=y$
We get $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ y & y & z \\ y^2 & y^2 & z^2\end{array}\right|=0\left(\because \mathrm{C}_1=\mathrm{C}_2\right)$
$\Rightarrow(\mathrm{x}-\mathrm{y})$ is a factor of $\Delta$.
Similarly $(y-z)$ and $(z-x)$ are factors of $\Delta$.
Now degree of $\Delta=0+1+2=3$ and we have 3 factors of $\Delta$.

and so there can be a constant $\mathrm{k}$ as a factor of $\Delta$.
So, $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2\end{array}\right|=k(x-y)(y-z)(z-x)$
put $x=0, y=1$ and $z=2$ we get
$
\Delta=\left|\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 1 & 4
\end{array}\right|=k(0-1)(1-2)(2-0)
$
$\Rightarrow 2 k=2 \Rightarrow k=1$
$
\Delta=\left|\begin{array}{ccc}
1 & 1 & 1 \\
x & y & z \\
x^2 & y^2 & z^2
\end{array}\right|=(x-y)(y-z)(z-x)
$