Let D = diag [d1, d2, d3], where none of d1, d2, d3 is 0; prove that D-1 = diag [d1-1, d2-1, d3-1].
Given: D = diag [d1, d2, d3]
It is also given that d1 ≠ 0, d2 ≠ 0, d3 ≠ 0
A diagonal matrix D = diag(d1, d2, …dn) is invertible iff all diagonal entries are non – zero, i.e. di ≠ 0 for 1 ≤ i ≤ n
If D is invertible then D-1 = diag(d1-1, …dn-1)
By the Inverting Diagonal Matrices Theorem, which states that
Here, it is given that d1 ≠ 0, d2 ≠ 0, d3 ≠ 0
∴ D is invertible
⇒ D-1 = diag [d1-1, d2-1, d3-1]
Hence Proved.