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Solve the following equations:\(\begin{vmatrix}\text{x} & -6 & -1 \\[0.3em]2 & -3\text{x} & \text {x} - 3 \\[0.3em]-3 &2\text{x} & \text{x}+ 2 \end{vmatrix}\) = 0|(x, -6, -1)(2, -3x, x-3)(-3, 2x, x + 2)| = 0 |
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Answer» Expanding with R1 0= x(-3x2- 6x - 2x2 + 6x) + 6(2x+4+3x-9) -1(4x - 9x) 0=x (-5x2) + 6(5x - 5) -1(-5x) 0= -5x3 + 30x - 30 + 5x 0= -5x3 + 35x - 30 x3- 7x + 6 = 0 x3 - x - 6x + 6 = 0 x(x2-1) - 6(x - 1) = 0 x(x-1)(x+1) - 6(x - 1) = 0 (x - 1)(x2 + x - 6) = 0 (x-1)(x2 + 3x - 2x - 6) = 0 (x-1)(x(x + 3) - 2(x + 3) =0 (x-1)(x+3)(x-2) = 0 Either x - 1 = 0 or x + 3 = 0 or x - 2 = 0 ∴ x = 1 or x = -3 or x = 2 |
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