if \(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}

1.	if \(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\), find a, b, c and d.[(2a+b, 3a-b) (c+2d) (2c-d)] = [(2, 3) (4, -1)]
Answer» \(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\) ∴ By equality of matrices, we get 2a + b = 2 ….(i) 3a – b = 3 ….(ii) c + 2d = 4 ….(iii) 2c – d = -1 ….(iv) Adding (i) and (ii), we get 5a = 5 ∴ a = 1 Substituting a = 1 in (i), we get 2(1) + b = 2 ∴ b = 0 By (iii) + (iv) x = 2, we get 5c = 2 ∴ c = 2/5 Substituting c = 2/5 in (iii), we get 2/5 + 2d =4 ∴ 2d = 4 - 2/5 ∴ 2d = 18/5 ∴ d = 9/5

if \(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\), find a, b, c and d.[(2a+b, 3a-b) (c+2d) (2c-d)] = [(2, 3) (4, -1)]

Answer»

\(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\)

∴ By equality of matrices, we get 2a + b = 2 ….(i)

3a – b = 3 ….(ii)

c + 2d = 4 ….(iii)

2c – d = -1 ….(iv)

Adding (i) and (ii), we get 5a = 5

∴ a = 1 Substituting a = 1 in (i), we get 2(1) + b = 2

∴ b = 0

By (iii) + (iv) x = 2, we get 5c = 2

∴ c = 2/5

Substituting c = 2/5 in (iii), we get 2/5 + 2d =4

∴ 2d = 4 - 2/5

∴ 2d = 18/5

∴ d = 9/5

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if \(\begin{vmatrix}2a+b&amp;3a-b\\c+2d&amp;2c-d\end{vmatrix}=\begin{vmatrix}2&amp;3\\4&amp;-1\end{vmatrix}\), find a, b, c and d.[(2a+b, 3a-b) (c+2d) (2c-d)] = [(2, 3) (4, -1)]

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if \(\begin{vmatrix}2a+b&3a-b\\c+2d&2c-d\end{vmatrix}=\begin{vmatrix}2&3\\4&-1\end{vmatrix}\), find a, b, c and d.[(2a+b, 3a-b) (c+2d) (2c-d)] = [(2, 3) (4, -1)]