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Find the values of a, b, c and d from the equation
\(\left[\begin{array}{cc} a-b & 3 a+c \\ 2 a-b & 3 c+d \end{array}\right]\) = \(\left[\begin{array}{cc} -1 & 5 \\ 0 & 13 \end{array}\right]\)
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Comparing the corresponding elements of

\(\left[\begin{array}{cc} a-b & 3 a+c \\ 2 a-b & 3 c+d \end{array}\right]\) = \(\left[\begin{array}{cc} -1 & 5 \\ 0 & 13 \end{array}\right]\), we get

a – b = – 1, 2a – b = 0,

2a + c – 5, 3c + d = 13

Subtracting a – b = – 1 from 2a – b = 0, we get

a = 1. Therefore, b = 2.

Putting a = 1 in

2a + c = 5, we get

2 + c = 5

∴ c = 3

From

3c + d = 13

9 + d = 13

∴ = 4.

∴ a = 1, b = 2, c = 3, d = 4.
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