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: Build the augmented matrix.
Stack $A$ next to $b$:
\[ \left[\begin{array}{ccc|c}1 & 0 & 2 & 1\\ -1 & 1 & 0 & 3\\ 0 & 1 & 2 & 0\end{array}\right] \]

Step 2: Row reduce.
Add row 1 to row 2, to clear the $-1$ in the first column:
\[ R_2\to R_2+R_1:\quad (0,\ 1,\ 2\ |\ 4) \]
The matrix now reads
\[ \left[\begin{array}{ccc|c}1 & 0 & 2 & 1\\ 0 & 1 & 2 & 4\\ 0 & 1 & 2 & 0\end{array}\right] \]

Step 3: Eliminate the last row using row 2.
Subtract row 2 from row 3:
\[ R_3\to R_3-R_2:\quad (0,\ 0,\ 0\ |\ -4) \]
This row reads $0=-4$, which cannot hold for any values of $x_1,x_2,x_3$.

Step 4: Compare ranks.
The row-reduced coefficient part has only $2$ nonzero rows, so $\text{rank}(A)=2$. But the augmented matrix has a nonzero entry in that all-zero coefficient row, so $\text{rank}([A|b])=3$. Since the two ranks differ, the consistency condition fails.

Step 5: Conclude.
A linear system is consistent only when $\text{rank}(A)=\text{rank}([A|b])$. Here $2\neq3$, so the system has no solution at all.
\[ \boxed{0 \text{ solutions}} \]