Question

what will be the value of x, pls explain the process​

Answer 1

Given,

$\longrightarrow\left|\begin{array}{ccc}x+2&x+6&x-1\\x+6&x-1&x+2\\x-1&x+2&x+6\end{array}\right|=0$

Performing $R_1\to R_1-R_3$ and $R_2\to R_2-R_3,$

$\longrightarrow\left|\begin{array}{ccc}3&4&-7\\7&-3& -4\\x-1&x+2&x+6\end{array}\right|=0$

Performing $C_1\to C_1+C_2+C_3,$

$\longrightarrow\left|\begin{array}{ccc}0&4&-7\\0&-3&-4\\3x+7&x+2&x+6 \end{array}\right|=0$

Expanding along $C_1,$

$\longrightarrow(3x+7)(-16-21)=0$

Anyways,

$\longrightarrow3x+7=0$

$\longrightarrow\underline{\underline{x=-\dfrac{7}{3}}}$

Answer 2

Answer:-

$\begin{gathered}\left|\begin{array}{ccc}x+2&x+6&x-1\\x+6&x-1&x+2\\x-1&x+2&x+6\end{array}\right|=0\end{gathered}$

Operate$R_1\to R_1+R_2+R_3$

$\begin{gathered}\ = > \left|\begin{array}{ccc}x+7&x+7&x+7\\x+6&x-1& x+2\\x-1&x+2&x+6\end{array}\right|=0\end{gathered}$

$\begin{gathered}\ = > (x+7)\left|\begin{array}{ccc}1&1&1\\x+6&x-1& x+2\\x-1&x+2&x+6\end{array}\right|=0\end{gathered}$

Operate $\bf C_2→ C_2-C_1\\\bf and\\\bf C_3→C_3-C_1$

$\begin{gathered}=>(x+7)\left|\begin{array}{ccc}1&0&0\\x+6&-7&-4\\x-1&3&7 \end{array}\right|=0\end{gathered}$

$\bf = > (x+7)(−49+12)=0$

$\bf = > x= -7$

@BengaliBeauty

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