Q. 52 D5.0( 1 Vote )

Solve the followi

Answer :


Let


We need to find the roots of Δ = 0.


Recall that the value of a determinant remains same if we apply the operation Ri Ri + kRj or Ci Ci + kCj.


Applying R2 R2 – R1, we get




Applying R3 R3 – R1, we get





Taking (a – x) and (b – x) common from R2 and R3, we get



Expanding the determinant along C1, we have


Δ = (a – x)(b – x)(1)[(1)(b + x) – (1)(a + x)]


Δ = (a – x)(b – x)[b + x – a – x]


Δ = (a – x)(b – x)(b – a)


The given equation is Δ = 0.


(a – x)(b – x)(b – a) = 0


However, a ≠ b according to the given condition.


(a – x)(b – x) = 0


Case – I:


a – x = 0 x = a


Case – II:


b – x = 0 x = b


Thus, a and b are the roots of the given determinant equation.


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