If $\left[\begin{array}{lll}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right]=0$, then find the value of $x$.
Given, $$\left|\begin{array}{lll} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|=0$$
$\Rightarrow \quad\left|\begin{array}{ccc}12+x & 12+x & 12+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|=0 \quad\left[\because R_1 \rightarrow R_1+R_2+R_3\right]$
$\Rightarrow \quad(12+x)\left|\begin{array}{ccc}1 & 1 & 1 \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|=0 \quad$ [taking $(12+x)$ common from $R_1$ ]
$$ \begin{aligned} \Rightarrow \quad & (12+x)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & 8 & 4+x \\ 2 x & 8 & 4-x \end{array}\right| =0 \quad\left[\because C_1 \rightarrow C_1-C_3 \text { and } C_2 \rightarrow C_2+C_3\right] \\ \Rightarrow \quad & (12+x)[1 \cdot(-16 x)] =0 \\ \Rightarrow \quad & (12+x)(-16 x) =0 \\ \therefore \quad & x =-12,0 \end{aligned}$$
If $a_1, a_2, a_3, \ldots, a_r$ are in GP, then prove that the determinant $\left|\begin{array}{ccc}a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21}\end{array}\right|$ is independent of $r$.
We know that, $$a_{r+1}=A R^{(r+1)-1}=A ^r $$
where $r=r$ th term of a GP, A = First term of a GP and $R=$ Common ratio of GP
$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} &\left|\begin{array}{ccc} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{array}\right| \\ &=\left|\begin{array}{ccc} A R^r & A R^{r+4} & A R^{r+8} \\ A R^{r+6} & A R^{r+10} & A R^{r+14} \\ A R^{r+10} & A R^{r+16} & A R^{r+20} \end{array}\right| \\ &=A R^r \cdot A R^{r+6} \cdot A R^{r+10}\left|\begin{array}{ccc} 1 & A R^4 & A R^8 \\ 1 & A R^4 & A R^8 \\ 1 & A R^6 & A R^{10} \end{array}\right| \end{aligned} \end{aligned}$$
$\begin{array}{r}\text { [taking } A R^r, A R^{r+6} \text { and } A R^{r+10} \text { common from } R_1, R_2 \text { and } R_3, \text { respectively] } \\ =0 \text { [since, } R_1 \text { and } R_2 \text { are identicals] }\end{array}$
Show that the points $(a+5, a-4),(a-2, a+3)$ and $(a, a)$ do not lie on a straight line for any value of $a$.
$$\begin{aligned} &\text { Given, the points are }(a+5, a-4),(a-2, a+3) \text { and }(a, a) \text {. }\\ &\therefore \quad \Delta=\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ a-2 & a+3 & 1 \\ a & a & 1 \end{array}\right| \end{aligned}$$
$$\begin{aligned} &\begin{aligned} & =\frac{1}{2}\left|\begin{array}{rrr} 5 & -4 & 0 \\ -2 & 3 & 0 \\ a & a & 1 \end{array}\right| \quad\left[\because R_1 \rightarrow R_1-R_3 \text { and } R_2 \rightarrow R_2-R_3\right] \\ & =\frac{1}{2}[1(15-8)] \\ \Rightarrow\quad & =\frac{7}{2} \neq 0 \end{aligned}\\ &\text { Hence, given points form a triangle i.e., points do not lie in a straight line. } \end{aligned}$$
Show that $\triangle A B C$ is an isosceles triangle, if the determinant $$ \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos ^2 A+\cos A & \cos ^2 B+\cos B & \cos ^2 C+\cos C \end{array}\right|=0 .$$
We have, $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos ^2 A+\cos A & \cos ^2 B+\cos B & \cos ^2 C+\cos C\end{array}\right|=0$
$\Delta=\left|\begin{array}{ccc}0 & 0 & 1 \\ \cos A-\cos C & \cos B-\cos C & 1+\cos C \\ \cos ^2 A+\cos A-\cos ^2 C-\cos C & \cos ^2 B+\cos B-\cos ^2 C-\cos C \cos ^2 C+\cos C\end{array}\right|=0\quad \left[\begin{array}{ll} \therefore & \left.C_1 \rightarrow C_1-C_3 \text { and } C_2 \rightarrow C_2-C_3\right] \end{array}\right.$
$\Rightarrow(\cos A-\cos C) \cdot(\cos B-\cos C)$
$\left|\begin{array}{ccc}0 & 0 & 1 \\ 1 & 1 & 1+\cos C \\ \cos A+\cos C+1 & \cos B+\cos C+1 & \cos ^2 C+\cos C\end{array}\right|=0$
[taking $(\cos A-\cos C)$ common from $C_1$ and $(\cos B-\cos C)$ common from $C_2$ ]
$$\begin{array}{lrl} \Rightarrow & (\cos A-\cos C) \cdot(\cos B-\cos C)[(\cos B+\cos C+1)-(\cos A+\cos C+1)] & =0 \\ \Rightarrow & (\cos A-\cos C) \cdot(\cos B-\cos C)(\cos B+\cos C+1-\cos A-\cos C-1) & =0 \\ \Rightarrow & (\cos A-\cos C) \cdot(\cos B-\cos C)(\cos B-\cos A) & =0 \\ \text { i.e., } & \cos A=\cos C \text { or } \cos B=\cos C \text { or } \cos B=\cos A \\ \Rightarrow & A=C \text { or } B=C \text { or } B=A \end{array}$$
Hence, $A B C$ is an isosceles triangle.
Find $A^{-1}$, if $A=\left|\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right|$ and show that $A^{-1}=\frac{A^2-3 I}{2}$.
We have, $\quad A=\left|\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right|$
$$\begin{array}{ll} \therefore & A_{11}=-1, A_{12}=1, A_{13}=1, A_{21}=1, A_{22}=-1, A_{23}=1, A_{31}=1, A_{32}=1 \text { and } A_{33}=-1 \\ \therefore & \text { adj } A=\left|\begin{array}{rrr} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right|^{\top}=\left|\begin{array}{rrr} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \end{array}$$
$$\begin{aligned} &\begin{aligned} \text { and }\quad & |A|=-1(-1)+1 \cdot 1=2 \\ \therefore\quad & A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{1}{2}\left[\begin{array}{rrr} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]\quad\text{.... (i)} \end{aligned} \end{aligned}$$
and $$ A^2=\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right] \cdot\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]=\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]\quad\text{..... (ii)}$$
$\begin{aligned} \therefore \quad \frac{A^2-3 I}{2} & \left.=\frac{1}{2}\left\{\left|\begin{array}{lll}2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{array}\right|-\left|\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right|\right\}\right\}=\frac{1}{2}\left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right| \\ & =A^{-1}\end{aligned}\quad$ [using Eq. (i)]
Hence proved.