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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