Question

If \[ X=\begin{bmatrix}x\\y\\z\end{bmatrix} \] is a solution of the system of equations $AX=B$, where \[ \text{adj }A= \begin{bmatrix} 4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3 \end{bmatrix} \quad \text{and} \quad B=\begin{bmatrix}4\\0\\2\end{bmatrix}, \] then $|x+y+z|$ is equal to

• A. $1$
• B. $\dfrac{3}{2}$
• C. $3$
• D. $2$

Hint:

When adjoint matrix is given, use $X=\dfrac{1}{|A|}(\text{adj }A)B$ and the relation $|\text{adj }A|=|A|^{n-1}$.

Answer 1

The Correct Option is D

Given, \[ AX=B \]
Step 1: Express $X$ using adjoint matrix.
If $A$ is non-singular, then: \[ X = A^{-1}B = \frac{1}{|A|}\,(\text{adj }A)\,B \]
Step 2: Compute $(\text{adj }A)\,B$.
\[ (\text{adj }A)\,B= \begin{bmatrix} 4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4\\ 0\\ 2 \end{bmatrix} \] \[ = \begin{bmatrix} 16+0+4\\ -20+0+10\\ 4+0+6 \end{bmatrix} = \begin{bmatrix} 20\\ -10\\ 10 \end{bmatrix} \]
Step 3: Write the solution vector.
\[ X=\frac{1}{|A|} \begin{bmatrix} 20\\ -10\\ 10 \end{bmatrix} \]
Step 4: Find $x+y+z$.
\[ x+y+z=\frac{1}{|A|}(20-10+10)=\frac{20}{|A|} \]
Step 5: Use determinant property.
For a $3\times3$ matrix, \[ |\text{adj }A|=|A|^2 \] Calculating determinant of adj $A$: \[ |\text{adj }A|=100 \Rightarrow |A|=10 \]
Step 6: Compute required value.
\[ |x+y+z|=\left|\frac{20}{10}\right|=2 \]
Final Answer: $\boxed{2}$