\[D=\left|\begin{array}{ccc}p & q & p \alpha-q \\q & r & q_{\alpha}-r

1.	\[D=\left\|\begin{array}{ccc}p & q & p \alpha-q \\q & r & q_{\alpha}-r \\2 & 1 & 0\end{array}\right\|\]If \( q^{2} \neq P_{r} \), got what value of \( \alpha, D \) becomes zero
Answer» \(D=\begin{vmatrix}p&q&p\alpha-q\\q&r&q\alpha-r\\2&1&0\end{vmatrix}=0\) Applying c₃ → c₃ - (αc₁- c₂) \(D=\begin{vmatrix}p&q&0\\q&r&0\\2&1&-(2\alpha-1)\end{vmatrix}=0\) ⇒ -(2α - 1) \(\begin{vmatrix}p&q\\q&r\end{vmatrix}=0\) (By expanding determinant along column c₃) ⇒ -(2α - 1) (pr - q²) = 0 ⇒ 2α - 1 = 0 (∵ pr ≠ q² (given)) ⇒ α = 1/2.

\[D=\left|\begin{array}{ccc}p & q & p \alpha-q \\q & r & q_{\alpha}-r \\2 & 1 & 0\end{array}\right|\]If \( q^{2} \neq P_{r} \), got what value of \( \alpha, D \) becomes zero

Answer»

\(D=\begin{vmatrix}p&q&p\alpha-q\\q&r&q\alpha-r\\2&1&0\end{vmatrix}=0\)

Applying c₃ → c₃ - (αc₁- c₂)

\(D=\begin{vmatrix}p&q&0\\q&r&0\\2&1&-(2\alpha-1)\end{vmatrix}=0\)

⇒ -(2α - 1) \(\begin{vmatrix}p&q\\q&r\end{vmatrix}=0\) (By expanding determinant along column c₃)

⇒ -(2α - 1) (pr - q²) = 0

⇒ 2α - 1 = 0 (∵ pr ≠ q² (given))

⇒ α = 1/2.

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\[D=\left|\begin{array}{ccc}p & q & p \alpha-q \\q & r & q_{\alpha}-r \\2 & 1 & 0\end{array}\right|\]If \( q^{2} \neq P_{r} \), got what value of \( \alpha, D \) becomes zero

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