If A is invertible matrix and B is any matrix, then

Question

If A is invertible matrix and B is any matrix, then (A) Rank (AB) = Rank (A) (B) Rank (AB) = Rank (B) (C) Rank (AB) > Rank (A) (D) Rank (AB) > Rank (B)

Tardigrade Admin · Accepted Answer

Since, A is invertible ⇒ A-1 exists. Now, Rank (B)= operatornameRank(A-1(A B)) ≤ operatornameRank(A B) [∵ Rank(P Q) ≤ operatornameRank Q] But, Rank (A B) ≤ operatornameRank(B) ∴ Rank (A B)= operatornameRank(B)

Q. If $A$ is invertible matrix and $B$ is any matrix, then

Rank $(A B) =$ Rank $(A)$

Rank $(A B) =$ Rank $(B)$

Rank $(A B) >$ Rank $(A)$

Rank $(A B) >$ Rank $(B)$

Since, $A$ is invertible $\Rightarrow A^{- 1}$ exists.
Now, Rank $(B) = Rank (A^{- 1} (A B))$
$\leq Rank (A B) [∵ R ank (PQ) \leq Rank Q]$
But, Rank $(A B) \leq Rank (B)$
$∴$ Rank $(A B) = Rank (B)$

Q. If A is invertible matrix and B is any matrix, then

Rank (AB)= Rank (A)

Rank (AB)= Rank (B)

Rank (AB)> Rank (A)

Rank (AB)> Rank (B)