CUET · MATHS · PYQ PAPER 2023
\(\left|\begin{array}{lll}1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9\end{array}\right|=x\left|\begin{array}{ll}2 & 3 \\8 & 9\end{array}\right|+y\left|\begin{array}{ll}1 & 3 \\7 & 9\end{array}\right|+z\left|\begin{array}{ll}1 & 2 \\7 & 8\end{array}\right|\)
Then \(x+y+z\) is:
- A 15
- B 5
- C -5
- D 0
Answer & Solution
Correct Answer
(C) -5
Step-by-step Solution
Detailed explanation
\( \left|\begin{array}{lll}1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9\end{array}\right| = 1(5 \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - 5 \cdot 7) = 1(45-48) - 2(36-42) + 3(32-35) = -3 - 2(-6) + 3(-3) = -3 + 12 - 9 = 0 \) The right side is an expansion using minors…
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