Exercise 7.3-Additional Questions - Chapter 7 - Matrices and Determinants - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Problems
Question 1.
Prove that $\left|\begin{array}{lll}\mathbf{1} & \boldsymbol{a} & a^3 \\ \mathbf{1} & b & b^3 \\ \mathbf{1} & c & c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$.
Solution:
Let $\Delta=\left|\begin{array}{lll}1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{array}\right|$. Put $a=b, \Delta=\left|\begin{array}{ccc}1 & b & b^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{array}\right|=0\left[\because \mathrm{R}_1\right.$ is identical to $\left.\mathrm{R}_2\right]$
$\therefore(\mathrm{a}-\mathrm{b})$ is a factor of $\Delta$.
Similarly we observe that $\Delta$ is symmetric in a, b, c by putting $\mathrm{b}=\mathrm{c}, \mathrm{c}=\mathrm{a}$, we get $\Delta=0$. Hence $(\mathrm{b}-\mathrm{c})$ and $(\mathrm{c}-\mathrm{a})$ are also factors of $\Delta$.
$\therefore$ The product $(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})$ is a factor of $\Delta$. The degree of this product is 3 . The product of leading diagonal elements is $1 . \mathrm{bc}^3$. The degree of this product is 4 .
$\therefore$ By cyclic and symmetric properties, the remaining symmetric factor of first degree must be $\mathrm{k}(\mathrm{a}+\mathrm{b}+$

c), where $\mathrm{k}$ is any non-zero constant.
Thus $\left|\begin{array}{lll}1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{array}\right|=(a-b)(b-c)(c-a) k(a+b+c)$
To find the value of $k$, give suitable values for $a, b, c$ so that both sides do not becom Take $a=0, b=1, c=2$.

Question 2.
Using factor method show 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:
Let $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$ Put $a=b$,
then $\Delta=\left|\begin{array}{ccc}1 & b & b^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|=0 \quad\left(\because \mathrm{R}_1=\mathrm{R}_2\right)$
$\Rightarrow(\mathrm{a}-\mathrm{b})$ is a factor of $\Delta$.
similarly $(b-c)$ and $(c-a)$ are factors of $\Delta$.
The product of leading diagonal elements is $b c^2$. The degree of the product is $1+2=3$.
$\therefore$ there will be three factors for $\Delta$. We got 3 factors for $\Delta$ as $(a-b),(b-c)$ and $(c-a)$. Its degree $=3 . \therefore$ $\mathrm{m}=3-3=0$

$\therefore$ there can be a constant $\mathrm{k}$ as a factor of $\Delta$.
(i.e.)
$
\Delta=\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|=k(a-b)(b-c)(c-a)
$
To find $k$ : put $a=0, b=1, c=2$
we get, $\left|\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4\end{array}\right|=k(0-1)(1-2)(2-0)$
(i.e.) $2=2 k \Rightarrow k=1$
$\therefore \Delta=1(a-b)(b-c)(c-a)$
(i.e.) $\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|=(a-b)(b-c)(c-a)$

Question 3.
Factorise $\left|\begin{array}{ccc}a & b & c \\ a^2 & b^2 & c^2 \\ b c & c a & a b\end{array}\right|$
Solution:
Let $\Delta=\left|\begin{array}{ccc}a & b & c \\ a^2 & b^2 & c^2 \\ b c & c a & a b\end{array}\right|$
Put $a=b$, then
$
\Delta=\left|\begin{array}{ccc}
b & b & c \\
b^2 & b^2 & c^2 \\
b c & b c & b^2
\end{array}\right|=0\left(\because C_1=\mathrm{C}_2\right)
$
$\Rightarrow(\mathrm{a}-\mathrm{b})$ is a factor of $A$.
Similarly $(b-c)$ and $(c-a)$ are factors of $\Delta$.
The degree of $\Delta=5$ and degree of product of factors $=3$.

$
\begin{aligned}
& \therefore m=5-3=2 \\
& \therefore \text { The other factor is } k\left(a^2+b^2+c^2\right)+l(a b+b c+c a) \\
& \therefore\left|\begin{array}{ccc}
a & b & c \\
a^2 & b^2 & c^2 \\
b c & c a & a b
\end{array}\right|=\left[k\left(a^2+b^2+c^2\right)+l(a b+b c+c a)\right](a-b)(b-c)(c-a)
\end{aligned}
$
To determine $k$ and $l$ give suitable values for $a, b$ and $c$ so that both sides do not become zer put $a=0, b=1, c=2$
$
\begin{aligned}
\therefore\left|\begin{array}{lll}
0 & 1 & 2 \\
0 & 1 & 4 \\
2 & 0 & 0
\end{array}\right|=[k(0+1+4)+l(0 & +2+0)](0-1)(1-2)(2-0) \\
\therefore \quad 2(4-2) & =[k(5)+l(2)](-1)(-1)(2) \\
2(2) & =(5 k+2 l)(2) \\
4 & =2(5 k+2 l) \\
2 & =5 k+2 l
\end{aligned}
$
Again put $a=0, b=-1, c=1$
$
\begin{aligned}
\therefore \quad\left|\begin{array}{ccc}
0 & -1 & 1 \\
0 & 1 & 1 \\
-1 & 0 & 0
\end{array}\right| & =[k(0+1+1)+l(0-1+0)](0+1)(-1-1)(1-0) \\
-1(-1-1) & =(2 k-l)(1)(-2)(1) \\
2 & =-2(2 k-l) \\
(\div 2) \quad \therefore \quad 1 & =-(2 k-l) \\
1 & =-2 k+l
\end{aligned}
$