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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