Question:
If \(\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}12 \\ 15 \\ 13\end{array}\right]\), then the value of \(x^2+y^2+z^2=\)
If \(\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}12 \\ 15 \\ 13\end{array}\right]\), then the value of \(x^2+y^2+z^2=\)
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Write the three equations and subtract them to get z, y and then x.
Updated On: Nov 13, 2026
- 6
- 12
- 3
- 14
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Multiply out the matrix to get \(x+3y+3z=12\), \(x+4y+4z=15\), \(x+3y+4z=13\).
Step 2: Eliminate
Third minus first: \(z=1\).
Second minus third: \(y=2\).
Step 3: Find x
\(x=12-3(2)-3(1)=3\).
Step 4: Compute
\(x^2+y^2+z^2=9+4+1=14\).
Final Answer:
The value is 14. \[ \boxed{\text{(D) }14} \]
Multiply out the matrix to get \(x+3y+3z=12\), \(x+4y+4z=15\), \(x+3y+4z=13\).
Step 2: Eliminate
Third minus first: \(z=1\).
Second minus third: \(y=2\).
Step 3: Find x
\(x=12-3(2)-3(1)=3\).
Step 4: Compute
\(x^2+y^2+z^2=9+4+1=14\).
Final Answer:
The value is 14. \[ \boxed{\text{(D) }14} \]
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