Question

If \(\begin{vmatrix} 1&2&1\\0&x&-3\\2&-1&x \end{vmatrix}=0\), then the value of x are

• A. 5,-3
• B. 5,3
• C. -5,3
• D. 2,3
• E. -2,-3

Answer 1

The Correct Option is A

We are given the determinant of the matrix:

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

To solve for \(x\), we will expand the determinant along the first row:

\(\text{det} = 1 \cdot \begin{vmatrix} x & -3 \\ -1 & x \end{vmatrix} - 2 \cdot \begin{vmatrix} 0 & -3 \\ 2 & x \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & x \\ 2 & -1 \end{vmatrix}\)

Now, calculate the 2x2 determinants:

\(\begin{vmatrix} x & -3 \\ -1 & x \end{vmatrix} = (x \cdot x) - (-3 \cdot -1) = x^2 - 3\)

\(\begin{vmatrix} 0 & -3 \\ 2 & x \end{vmatrix} = (0 \cdot x) - (-3 \cdot 2) = 6\)

\(\begin{vmatrix} 0 & x \\ 2 & -1 \end{vmatrix} = (0 \cdot -1) - (x \cdot 2) = -2x\)

Now substitute these values back into the determinant expansion:

\(\text{det} = 1 \cdot (x^2 - 3) - 2 \cdot 6 + 1 \cdot (-2x) = x^2 - 3 - 12 - 2x = x^2 - 2x - 15\)

Set this equal to 0 to solve for \(x\):

\(x^2 - 2x - 15 = 0\)

Factor the quadratic equation:

\((x - 5)(x + 3) = 0\)

Thus, the solutions are:

\(x = 5 \) or \( x = -3\)

The values of \(x\) are 5 and -3.