If \(p, q, r\) are 3 real numbers satisfying the matrix equation, \([p\ q\ r]\begin{bmatrix}3&4&1\\3&2&3\\2&0&2\end{bmatrix} = [3\ \ 0\ \ 1]\) then \(2p + q - r\) equals :
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- \(-3\)
- \(-1\)
- \(4\)
- \(2\)
The Correct Option is A
Solution and Explanation
\(3p + 3q + 2r = 3\) (first column)
\(4p + 2q = 0\) (second column)
\(p + 3q + 2r = 1\) (third column)
From the second equation \(q = -2p\).
First: \(3p - 6p + 2r = 3\), so \(2r = 3 + 3p\). Third: \(p - 6p + 2r = 1\), so \(2r = 1 + 5p\).
Equating: \(3 + 3p = 1 + 5p\), so \(p = 1\), \(q = -2\), \(r = 3\).
\(2p + q - r = 2 - 2 - 3 = -3\).
Answer: (1) \(-3\)
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