Question

Let \(A^{T}=\begin{bmatrix} 1 & 1 & 2 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{bmatrix}\), \(B^{T}=[\begin{matrix} 6 & 5 & 1 \end{matrix}]\) and \(X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}\). If \(AX=B\), then the values of \(x,y,z\) are..........

• A. \(x=2,y=1,z=3\)
• B. \(x=3,y=2,z=1\)
• C. \(x=1,y=2,z=3\)
• D. \(x=2,y=3,z=1\)

Hint:

Transpose the given matrix to get A and solve the three linear equations.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The system \(AX=B\) is solved by writing the three equations and solving them (or using \(X=A^{-1}B\)).

Step 2: Form A and B
Transposing, \(A=\begin{bmatrix}1&1&1\\1&-1&2\\2&1&-1\end{bmatrix}\) and \(B=\begin{bmatrix}6\\5\\1\end{bmatrix}\).

Step 3: Write the equations
\(x+y+z=6\), \(x-y+2z=5\), \(2x+y-z=1\).

Step 4: Solve
Adding the first and third equations: \(3x+2y=7\).
Adding the second and third: \(3x+z=6\), so \(z=6-3x\).
Putting in the first: \(x+y+6-3x=6\Rightarrow y=2x\).
Then \(3x+4x=7\Rightarrow x=1\), \(y=2\), \(z=3\).

Final Answer:
The solution is x = 1, y = 2, z = 3. \[ \boxed{\text{(C) }x=1,y=2,z=3} \]