1.

If \(\begin{bmatrix}a + b&2\\[0.3em]7&ab\\[0.3em]-3&4\end{bmatrix} = \begin{bmatrix}6&2\\[0.3em]7&8\\[0.3em]-3&4\end{bmatrix}\) then find ‘a’ and ‘b’.

Answer»

 \(\begin{bmatrix}a + b&2\\[0.3em]7&ab\\[0.3em]-3&4\end{bmatrix} = \begin{bmatrix}6&2\\[0.3em]7&8\\[0.3em]-3&4\end{bmatrix}\) 

On comparing,

a + b = 6 ……(i)

ab = 8

From equation (i) and (ii),

a(6 – a)= 8

⇒ 6a – a2 – 8 = 0

⇒ a2 – 2a – 4a + 8 = 0

⇒ a2 – 2a – 4a + 8 = 0

(a – 2)(a – 4) = 0

So, a = 2, 4

From ab = 8 we get b = 4, 2



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