Q. 125.0( 2 Votes )

# Using properties

Answer :

Given: To find: the value of x using the properties of determinants

Let Now we will apply the operation, R1 R1+R2+R3, we get  Taking (3x+a) common we get Now we will apply the operation, C1 C1-C3, we get  Now we will apply the operation, C2 C2-C3, we get  Now we will expand along R1, we get

Δ=(3x+a)[1{(0)(-a)-(-a)(a)}]

Δ=(3x+a)[1(0+a2)]

Δ=a2 (3x+a)

Δ=a2 (3x+a)

But Given Δ=0

So,

a2 (3x+a)=0

(3x+a)=0 Hence this is the required values of x

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