Let A, B and C be square matrices of the same order. Consider the following statements:
(A) If \(AB = AC\), then \(B = C\).
(B) If A is invertible and \(AB = AC\), then \(B = C\).
(C) If A is invertible and \(BA = CA\), then \(B = C\).
(D) If A is invertible and \(AB = CA\), then \(B = C\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (C) only
When A is invertible, an equation can be multiplied by \(A^{-1}\) on the same side as A:
\(AB = AC \Rightarrow A^{-1}(AB) = A^{-1}(AC) \Rightarrow B = C\)
\(BA = CA \Rightarrow (BA)A^{-1} = (CA)A^{-1} \Rightarrow B = C\)
So A can be cancelled only when it is invertible and stands on the same side in both products.
Checking the statements
(A) If \(AB = AC\), then \(B = C\). — Incorrect. \(A = \begin{bmatrix}1&0\\0&0\end{bmatrix}\), \(B = \begin{bmatrix}0&0\\0&1\end{bmatrix}\), \(C = O\) give \(AB = AC = O\), but \(B \ne C\).
(B) If A is invertible and \(AB = AC\), then \(B = C\). — Correct. Multiplying by \(A^{-1}\) on the left gives \(B = C\).
(C) If A is invertible and \(BA = CA\), then \(B = C\). — Correct. Multiplying by \(A^{-1}\) on the right gives \(B = C\).
(D) If A is invertible and \(AB = CA\), then \(B = C\). — Incorrect. \(A = \begin{bmatrix}1&1\\0&1\end{bmatrix}\), \(B = \begin{bmatrix}1&0\\0&0\end{bmatrix}\), \(C = \begin{bmatrix}1&-1\\0&0\end{bmatrix}\) give \(AB = CA = \begin{bmatrix}1&0\\0&0\end{bmatrix}\), but \(B \ne C\).