Question

If \(\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 7 \\ 3 & 1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix}\) then \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\)

• A. \(\begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}\)
• B. \(\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\)
• C. \(\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\)
• D. \(\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\)

Hint:

Write the matrix equation as three linear equations and test which option satisfies all of them.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The matrix equation means \(x+y+z=1\), \(2x-y+7z=7\) and \(3x+y+2z=2\).

Step 2: Check the determinant
The coefficient determinant is \(1(-2-7)-1(4-21)+1(2+3)=13\ne 0\), so the solution is unique.

Step 3: Test the options
For \((0,0,1)\): first equation gives \(1\), second gives \(7\), third gives \(2\). All three are satisfied.

Final Answer:
The solution is \(x=0,\ y=0,\ z=1\). \[ \boxed{\text{(B) }\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}} \]