Using properties of determinants, prove that
$|\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| = 9 (3 xyz + xy + yz + zx)$

Solution

$L . H . S = |\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| C_{1} \to C_{1} - C_{2}; C_{3} \to C_{3} - C_{2} = |\begin{matrix} 0 & 1 & 3 x \\ 3 y & 1 & 0 \\ - 3 z & 1 + 3 z & - 3 z \end{matrix}| = (3 x 3) |\begin{matrix} 0 & 1 & x \\ y & 1 & 0 \\ - z & 1 + 3 z & - z \end{matrix}| = 9 [- 1 (- yz - 0) + x (y + 3 zy + z) = 9 (yz + xy + 3 xyz + xz) = 9 (yz + 3 xyz + xz) = 9 (3 xyz + xy + yz + zx) = R . H . S$

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Determinants

Using properties of determinants, prove that
$|\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| = 9 (3 xyz + xy + yz + zx)$

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Determinants

Using properties of determinants, prove that111+3x1+ 3y1111+3z1 = 9 (3xyz + xy + yz + zx)

Some More Questions From Determinants Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Using properties of determinants, prove that
$|\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| = 9 (3 xyz + xy + yz + zx)$