Question

Under which of the following condition(s) does(do) the system of equations $\begin{bmatrix} 1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & a-4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 4 \\ a \end{bmatrix}$ possess(possess) a unique solution?

• A. ∀aεR
• B. a=8
• C. for all integral values of a
• D. a≠8

Answer 1

The Correct Option is D

Given system of equations: $$ \begin{bmatrix} 1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & a-4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 4 \\ a \end{bmatrix} $$ We need to find the condition under which this system possesses a unique solution. Step 1: Check the determinant of the coefficient matrix. For the system to have a unique solution, the determinant of the coefficient matrix must be non-zero. The coefficient matrix is: $$ A = \begin{bmatrix} 1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & a-4 \end{bmatrix} $$ Step 2: Calculate the determinant of matrix A. The determinant of a 3x3 matrix is given by: $$ \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & 2 \\ 2 & a-4 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 2 \\ 1 & a-4 \end{vmatrix} + 4 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} $$ Step 3: Compute the minors. The minors are the determinants of the 2x2 matrices formed by deleting the row and column of the element. 1. $$ \begin{vmatrix} 1 & 2 \\ 2 & a-4 \end{vmatrix} = (1)(a-4) - (2)(2) = a - 4 - 4 = a - 8 $$ 2. $$ \begin{vmatrix} 2 & 2 \\ 1 & a-4 \end{vmatrix} = (2)(a-4) - (2)(1) = 2a - 8 - 2 = 2a - 10 $$ 3. $$ \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = (2)(2) - (1)(1) = 4 - 1 = 3 $$ Step 4: Substitute these minors back into the determinant formula. $$ \text{det}(A) = 1 \cdot (a - 8) - 2 \cdot (2a - 10) + 4 \cdot 3 $$ Simplifying: $$ \text{det}(A) = a - 8 - 2(2a - 10) + 12 $$ $$ \text{det}(A) = a - 8 - 4a + 20 + 12 $$ $$ \text{det}(A) = -3a + 24 $$ Step 5: Find when the determinant is non-zero. For a unique solution, we need: $$ \text{det}(A) \neq 0 $$ So, $$ -3a + 24 \neq 0 $$ Solving for \(a\): $$ -3a \neq -24 $$ $$ a \neq 8 $$ Conclusion: The system has a unique solution for all values of \(a\) except \(a = 8\). Hence, the correct condition is: $$ a \neq 8 $$ Therefore, the answer is \(a \neq 8\).