Question

Prove that the determinant of 3×3 matrix which is x+5 x x
x x+5 x. =25(3x+5)
x x x+5

Answer 1

QUESTION:

Prove that the determinant of 3 × 3 matrix which is

$\left|\begin{array}{ccc} x+5 &x &x\\ \\ x& x+5 &x\\ \\ x &x &x+5\end{array}\right|=25(3x+5)$

GIVEN:

$\left|\begin{array}{ccc} x+5 &x &x\\ \\ x& x+5 &x\\ \\ x &x &x+5\end{array}\right|$

TO PROVE:

$\left|\begin{array}{ccc} x+5 &x &x\\ \\ x& x+5 &x\\ \\ x &x &x+5\end{array}\right|=25(3x+5)$

PROOF:

$\left|\begin{array}{ccc} x+5 &x &x\\ \\ x& x+5 &x\\ \\ x &x &x+5\end{array}\right| \\ \\ \\ \\ \sf{R_1 \implies R_1 + R_2 + R_3} \\ \\ \\ \\ \left|\begin{array}{ccc} 3x+5 &3x + 5 &3x + 5\\ \\ x& x+5 &x\\ \\ x &x &x+5\end{array}\right| \\ \\ \\ \\ \sf{Take \ 3x + 5 \ as \ common.} \\ \\ \\ \\ 3x + 5\left|\begin{array}{ccc} 1 &1 &1\\ \\ x& x+5 &x\\ \\ x &x &x+5\end{array}\right| \\ \\\\ \\ \sf{C_1 \implies C_1 - C_3} \\ \\ \\ \\ 3x + 5\left|\begin{array}{ccc} 1 - 1 &1 &1\\ \\ x - x& x+5 &x\\ \\ x - x - 5 &x &x+5\end{array}\right| \\ \\ \\ \\3x + 5\left|\begin{array}{ccc} 0 &1 &1\\ \\ 0& x+5 &x\\ \\ - 5 &x &x+5\end{array}\right|$

$\sf{C_2 \implies C_2 - C_3} \\ \\ \\ \\ 3x + 5\left|\begin{array}{ccc} 0 &1 - 1 &1\\ \\ 0& x+5 - x &x\\ \\ - 5 &x - x - 5&x+5\end{array}\right| \\ \\ \\ \\ 3x + 5\left|\begin{array}{ccc} 0 &0 &1\\ \\ 0&5 &x\\ \\ - 5 &- 5&x+5\end{array}\right| \\ \\ \\ \\ \sf\Delta = (3x + 5) \bigg(0 + 0 + 1\Big(0 ( - 5) - ( - 5)(5)\Big) \bigg) \\ \\ \\ \\ \sf\Delta= (3x + 5) \Big(1(0+ 25)\Big) \\ \\ \\ \\ \sf \Delta = (3x + 5)25 \\ \\ \\ \\ \sf \Delta =25 (3x + 5) \\ \\ \\ \\ \left|\begin{array}{ccc} x+5 &x &x\\ \\ x& x+5 &x\\ \\ x &x &x+5\end{array}\right|=25(3x+5) \\ \\ \\ \\ \sf \boldsymbol{HENCE \ PROVED.}$