Question

The system of equations \( \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \) has

• A. infinite number of solutions when \(a+b+c \neq 0\)
• B. no solution when \(a+b+c = 0\)
• C. unique solution for any values of \(a,b,c\)
• D. unique solution for no value of \(a,b,c\)

Hint:

If \(\det(A) \neq 0\), the system \(A\mathbf{x}=\mathbf{d}\) has a unique solution. 
If \(\det(A) = 0\), the system has either no solution or infinitely many solutions. It never has a unique solution.

Answer 1

The Correct Option is D

Let \(A = \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{bmatrix}\). The determinant of \(A\) is:

\(\det(A) = -2((-2)(-2) - (1)(1)) - 1((1)(-2) - (1)(1)) + 1((1)(1) - (1)(-2))\)

\(= -2(4-1) - 1(-2-1) + 1(1+2)\) 

\(= -2(3) - 1(-3) + 1(3) = -6 + 3 + 3 = 0\).

Since \(\det(A) = 0\), the system cannot have a unique solution for any \(a,b,c\). Thus, the statement "unique solution for no value of \(a,b,c\)" means that there are no values of \(a,b,c\) for which a unique solution exists. This is true because \(\det(A)=0\).

For this system, if we add the three equations:

\((-2x + y + z) + (x - 2y + z) + (x + y - 2z) = a + b + c\)

\(0x + 0y + 0z = a + b + c \Rightarrow 0 = a + b + c\).

For a solution to exist (either no solution or infinitely many), we must have \(a + b + c = 0\).

If \(a + b + c = 0\), there are infinitely many solutions. If \(a + b + c \neq 0\), there is no solution. In either case (solution exists or not), a unique solution never occurs.

Final Answer: \[ \boxed{\text{unique solution for no value of } a,b,c} \]