For three matrices $A, B$ and $C$ of the same order, if $A$=$B$, then $AC$=$BC$, but converse is not true.

Question

For three matrices $A, B$ and $C$ of the same order, if $A$=$B$, then $AC$=$BC$, but converse is not true.

I guess $A,B,C$ all are square matrices of the same order. $$ A=B $$ multiplying by $C$ from right side, $$ \implies AC=BC $$

But why is the second statement "converse is not true" ?

$AC=BC$, multiplying by $C^{-1}$ from the right $$ ACC^{-1}=BCC^{-1}\implies AI=BI\implies A=B $$ right ?

Not every matrix has an inverse. – Angina Seng Feb 04 '18 at 13:39 — Angina Seng, Feb 04 '18 at 13:39
Yes, and if $C = \mathbf 0$? – Sarvesh Ravichandran Iyer Feb 04 '18 at 13:39 — Sarvesh Ravichandran Iyer, Feb 04 '18 at 13:39

Emilio Novati · Accepted Answer · 2018-02-04T14:00:55.977

0

Hint.

Consider: $$A= \begin{bmatrix} 1&0\\0&2 \end{bmatrix} \qquad B= \begin{bmatrix} 1&0\\0&3 \end{bmatrix} \qquad C= \begin{bmatrix} 0&1\\0&0 \end{bmatrix} $$

and note that $C$ is not invertible

edited Feb 04 '18 at 14:00

answered Feb 04 '18 at 13:50

Emilio Novati

62,675

alright. so does that mean that if $C$ is invertible the converse is true ? – Sooraj S Feb 04 '18 at 13:56
1

Yes. If it is invertible your proof is correct – Emilio Novati Feb 04 '18 at 14:00
thanx. i think tht clears my doubt – Sooraj S Feb 04 '18 at 14:14

For three matrices $A, B$ and $C$ of the same order, if $A$=$B$, then $AC$=$BC$, but converse is not true.

1 Answers1