Question

pls answer fast
determinants lesson​

Answer 1

$\huge{\underline{\sf{Given\: question:-}}}$

$\left|\begin{array}{ccc}a + x & a - x & a - x \\a - x&a + x&a - x\\a-x &a-x &a+x \end{array}\right| = 0$

$\huge{\underline{\sf{Solution:-}}}$

$=\left|\begin{array}{ccc}a + x & a - x & a - x \\a - x&a + x&a - x\\a-x &a-x &a+x \end{array}\right|$

$=\left|\begin{array}{ccc}3a + x & 3a - x & 3a - x \\a - x&a + x&a - x\\a-x &a-x &a+x \end{array}\right|$

$\to {C'}_{1} = {C}_{1} + {C}_{2} + {C}_{3}$

$= (3a - x)\left|\begin{array}{ccc}1 & 1 & 1 \\a - x&a + x&a - x\\a-x &a-x &a+x \end{array}\right|$

$=(3a - x)\left|\begin{array}{ccc}1 & 1 & 1 \\a - x&a + x&a - x\\0 &0 &2x \end{array}\right|$

$\to { R'}_{3} = { R}_{3} - { R}_{1}$

$=(3a - x)\bigg[2x \left| \begin{array}{c c} 1 & a - x \\ 1 & a + x \end{array}\right| \bigg]$

$=\sf (3a - x) \bigg[2x \left| \begin{array}{c c} 1 & a - x \\ 1 & a + x \end{array} \right| \bigg]$

$= \sf (3a - x) \bigg[2x \times 2x\bigg]$

$\implies \: (3a - x) \times {4x}^{2} = 0$

$\therefore \: {x}^{2} = 0 \: \: or \: 3a - x = 0$

→ x = 0, 0 or x = 3a

$x=\begin{cases}0 \\ 0 \\ 3a\end{cases}$

Hence, proved!