Using properties

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