1.

What will be the value of x if \(\begin{vmatrix}2-x & 2 & 3 \\2 & 5-x & 6 \\3 & 4 & 10-x \end {vmatrix}\) = 0?(a) 8 ±√37(b) -8 ± √37(c) 8 ± √35(d) -8 ± √35I had been asked this question by my school principal while I was bunking the class.Enquiry is from Application of Determinants in chapter Determinants of Mathematics – Class 12

Answer»

The correct option is (a) 8 ±√37

For explanation: Here, we have, \(\begin{vmatrix}2-x & 2 & 3 \\2 & 5-x & 6 \\3 & 4 & 10-x \end {vmatrix}\) = 0

Now, replacing C3 = C3 – 3C1, we get,

\(\begin{vmatrix}2-x & 2 & 3-6 + 3x \\2 & 5-x & 6-6 \\3 & 4 & 10-x -9 \end {vmatrix}\) = 0

\(\begin{vmatrix}2-x & 2 & 3(x-1) \\2 & 5-x & 0 \\3 & 4 & -(x-1) \end {vmatrix}\) = 0

Or, (x – 1)\(\begin{vmatrix} 2-x & 2 & 3 \\2 & 5-x & 0 \\3 & 4 & -1 \end {vmatrix}\) = 0

Now, replacing R1 = R1 + 3R3, we get,

Or, (x – 1)\(\begin{vmatrix}11-x & 14 & 0 \\2 & 5-x & 0 \\3 & 4 & -1 \end {vmatrix}\) = 0

Or, (x – 1)[-1 {(11 – x)(5 – x) – 28}] = 0

Or, -(x – 1)(55 – 11x – 5x + x^2 – 28)

Or, (x – 1)(x^2 – 16x + 27) = 0

Thus, either x – 1 = 0 i.e. x = 1 or x^2 – 16x + 27 = 0

THEREFORE, solving x^2 – 16x + 27 = 0 further, we get,

x = 8 ± √37


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