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Find the integral value of x, if \(\begin{vmatrix}x^2 & x & 1 \\[0.3em]0& 2 & 1 \\[0.3em]3 &1 &4\end{vmatrix}\) = 28 |
Answer» |A| = \(\begin{vmatrix}x^2 & x & 1 \\[0.3em]0& 2 & 1 \\[0.3em]3 &1 &4\end{vmatrix}\) Expanding along the first row, |A| = x2\(\begin{vmatrix}2 & 1 \\[0.3em]1&4 \\[0.3em]\end{vmatrix}\) - x\(\begin{vmatrix}0 & 1 \\[0.3em]3&4 \\[0.3em]\end{vmatrix}\) + 1\(\begin{vmatrix}0 & 2 \\[0.3em]3&1 \\[0.3em]\end{vmatrix}\) = x2(2×4 – 1×1) – x(0×4 – 1×3) + 1(0×1 – 2×3) = x2(8 – 1) – x(0 – 3) + 1(0 – 6) = 7x2 + 3x –6 Also, |A| = 28 ⇒ 7x2 + 3x – 6 =28 ⇒ 7x2 + 3x – 34 = 0 ⇒ 7x2 + 17x – 14x – 34 = 0 ⇒ x(7x+ 17) – 2(7x +17) = 0 ⇒ (x – 2)(7x +17) = 0 x = 2, \(-\frac{17}{7}\) Integer value of x is 2. |
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