First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here |A| = [(2) {1×3 – 1 × (0)} – (0) {3 × 5 – 0 × 0} + (-1) {5 × 1 – 1 × 0}]

= [2{3} - 0 {15} - 1 {5}]

= [6-0-5]

= 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [A : I]

→

Apply row operation- R₁→ R₁

→

Apply row operation- R₂→ R₂ -5R₁

→

Apply row operation- R₃→ R₃ – R₂

→

Apply row operation- R₁→ R₁ + R₃

→

Apply row operation- R₂→ R₂ –5R₃

→

Apply row operation- R₃ → 2R₃

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→