Matrix Determinant Calculation

Given the matrix $ M = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 3 & 0 \\ -1 & a & b \end{bmatrix} $. The task is to determine the conditions on variables $a$ and $b$ for which the determinant of $M$, denoted as $\det(M)$, is non-zero ($\det(M) \neq 0$).

Calculate Matrix Determinant

The determinant of a 3x3 matrix $ \begin{bmatrix} p & q & r \\ s & t & u \\ v & w & x \end{bmatrix} $ is calculated using the formula:

$ \det = p(tx - uw) - q(sx - uv) + r(sw - tv) $

Applying this formula to matrix $M$:

$ \det(M) = 2 \begin{vmatrix} 3 & 0 \\ a & b \end{vmatrix} - 1 \begin{vmatrix} 1 & 0 \\ -1 & b \end{vmatrix} + 1 \begin{vmatrix} 1 & 3 \\ -1 & a \end{vmatrix} $

First, calculate the determinants of the 2x2 submatrices (minors):

• $ \begin{vmatrix} 3 & 0 \\ a & b \end{vmatrix} = (3 \times b) - (0 \times a) = 3b $
• $ \begin{vmatrix} 1 & 0 \\ -1 & b \end{vmatrix} = (1 \times b) - (0 \times -1) = b $
• $ \begin{vmatrix} 1 & 3 \\ -1 & a \end{vmatrix} = (1 \times a) - (3 \times -1) = a + 3 $

Substitute these values back into the determinant expansion:

$ \det(M) = 2(3b) - 1(b) + 1(a + 3) $

Simplify the expression:

$ \det(M) = 6b - b + a + 3 $

$ \det(M) = a + 5b + 3 $

Determinant Condition

The condition given is that the determinant must be non-zero:

$ \det(M) \neq 0 $

This translates to the condition on $a$ and $b$:

$ a + 5b + 3 \neq 0 $

Option Verification

We now evaluate each option to see if it satisfies the condition $ a + 5b + 3 \neq 0 $.

• Option 1: $a = -\frac{1}{2}$, $b = -\frac{1}{2}$
  Substitute values: $ -\frac{1}{2} + 5(-\frac{1}{2}) + 3 = -\frac{1}{2} - \frac{5}{2} + 3 = -\frac{6}{2} + 3 = -3 + 3 = 0 $
  Result: $ \det(M) = 0 $. This option is FALSE because the determinant is zero.
• Option 2: $a = \frac{1}{2}$, $b = \frac{1}{2}$
  Substitute values: $ \frac{1}{2} + 5(\frac{1}{2}) + 3 = \frac{1}{2} + \frac{5}{2} + 3 = \frac{6}{2} + 3 = 3 + 3 = 6 $
  Result: $ \det(M) = 6 $. Since $ 6 \neq 0 $, this option is TRUE.
• Option 3: $a = -3$, $b = 0$
  Substitute values: $ -3 + 5(0) + 3 = -3 + 0 + 3 = 0 $
  Result: $ \det(M) = 0 $. This option is FALSE because the determinant is zero.
• Option 4: $a = \frac{1}{2}$, $b = -3$
  Substitute values: $ \frac{1}{2} + 5(-3) + 3 = \frac{1}{2} - 15 + 3 = \frac{1}{2} - 12 = \frac{1 - 24}{2} = -\frac{23}{2} $
  Result: $ \det(M) = -\frac{23}{2} $. Since $ -\frac{23}{2} \neq 0 $, this option is TRUE.

Based on the analysis, the options where the determinant is non-zero are Option 2 and Option 4.