Q. 205.0( 3 Votes )

# Using properties

Answer :

Let

Given: To find: the value of x using the properties of determinants Now we will apply the operation, R1 R1+R2+R3, we get  Taking (3a-x) 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=(3a-x)[1{(0)(-2x)-(-2x)(2x)}]

Δ=(3a-x)[1(0+4x2)]

Δ=4x2 (3a-x)

But Given Δ=0

So,

4x2 (3a-x)=0

4x2=0 or (3a-x)=0

x=0 or x=3a

Hence this is the required values of x

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