Question

Consider a matrix
\[ A=\begin{bmatrix}1 & 0 & 2\\ -1 & 1 & 0\\ 0 & 1 & 2\end{bmatrix} \]
Let \(b=\begin{bmatrix}1\\3\\0\end{bmatrix}\) and \(x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\).
The number of solutions to the linear system of equations \(Ax=b\) is _____.

• A. \(0\)
• B. \(1\)
• C. \(6\)
• D. infinitely many

Hint:

First check \(\det(A)\): if it is zero, the system cannot have a unique solution. Then substitute one equation into another to see whether the remaining equations are consistent or contradict each other.

Answer 1

The Correct Option is A

Step 1: Write out the system of equations.
The matrix equation \(Ax=b\) expands to the three scalar equations
\[ x_1+0x_2+2x_3=1 \] \[ -x_1+x_2+0x_3=3 \] \[ 0x_1+x_2+2x_3=0 \]

Step 2: Check whether the coefficient matrix is singular.
Compute the determinant of \(A\) by expanding along the first row:
\[ \det(A)=1(1\times2-0\times1)-0(-1\times2-0\times0)+2(-1\times1-1\times0) \] \[ =1(2)-0(-2)+2(-1)=2-0-2=0 \] Since \(\det(A)=0\), the matrix \(A\) is singular, so the system either has no solution or infinitely many; it cannot have exactly one unique solution. This already rules out option (B).

Step 3: Eliminate \(x_1\) using the first two equations.
From equation 1:
\[ x_1=1-2x_3 \] Substitute into equation 2:
\[ -(1-2x_3)+x_2=3 \] \[ -1+2x_3+x_2=3 \] \[ x_2=4-2x_3 \]

Step 4: Substitute into the third equation and check consistency.
Equation 3 says
\[ x_2+2x_3=0 \] Substituting \(x_2=4-2x_3\):
\[ (4-2x_3)+2x_3=0 \] \[ 4=0 \] This is a false statement, independent of \(x_3\). No choice of \(x_1,x_2,x_3\) can make all three equations hold at once.

Step 5: Interpret the contradiction.
A singular coefficient matrix together with an inconsistent right-hand side (the rank of \(A\) is less than the rank of the augmented matrix \([A\,|\,b]\)) means the system is inconsistent, so it has no solution at all, not infinitely many.

Step 6: Rule out the remaining options.

(B) \(1\): Already ruled out, since \(\det(A)=0\) means a unique solution is impossible.

(C) \(6\): A linear system never has exactly \(6\) solutions; the solution count for a linear system is always \(0\), \(1\), or infinite. Incorrect.

(D) infinitely many: This would need the system to be consistent, but Step 4 showed the direct contradiction \(4=0\). Incorrect.

Final Answer:
The system \(Ax=b\) has
\[ \boxed{0 \text{ solutions}} \]