Question

If $AX = B$, where $A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 12 \\ 15 \\ 13 \end{bmatrix}$, then $x^2 + y^2 + z^2 =$

• A. $14$
• B. $19$
• C. $21$
• D. $6$

Hint:

Solve matrix equations by converting them into simultaneous linear equations.

Answer 1

The Correct Option is A

Step 1: Writing the system of equations.
From $AX = B$, we get: \[ x + 3y + 3z = 12 \] \[ x + 4y + 4z = 15 \] \[ x + 3y + 4z = 13 \]
Step 2: Solving the equations.
Subtracting the first equation from the second: \[ y + z = 3 \] Subtracting the first equation from the third: \[ z = 1 \] Substituting $z = 1$ into $y + z = 3$, we get $y = 2$.
Step 3: Finding $x$.
Substitute $y = 2$ and $z = 1$ into the first equation: \[ x + 6 + 3 = 12 \Rightarrow x = 3 \]
Step 4: Computing $x^2 + y^2 + z^2$.
\[ x^2 + y^2 + z^2 = 3^2 + 2^2 + 1^2 = 9 + 4 + 1 = 14 \]
Step 5: Conclusion.
The required value is $14$.