The given matrix is
\(u=\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}\)
So 2A = 2\(\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}\)
\(= \begin{bmatrix}2 & 4\\8 & 4\end{bmatrix}\)
so L.H.S. = |2A| \(= \begin{bmatrix}2 & 4\\8 & 4\end{bmatrix}\)
= 2 x 4 - 4 x 8
= 8 - 32
= -24
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y = Ax : xy' = y (x ≠ 0)
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is (A) 2 (B) 1 (C) 0 (D) not defined
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Plzz give all difficult question.