Trent Carpenter

2021-01-31

Let A,B and C be square matrices such that AB=AC , If $A\ne 0$ , then B=C.
Is this True or False?Explain the reasosing behind the answer.

unett

Step 1
Let A, B, and C be square matrices such that AB=AC.
If $A\ne 0$, then B=C.
Determine true or false.
Step 2
Let A, B, and C be square matrices such that AB=AC.
If $A\ne 0$ , then B=C.
This is true only if matrix A is invertible.
If A is invertible then ${A}^{-1}$ exist.
Multiply given equationAB=AC by ${A}^{-1}$ on both sides,
${A}^{-1}AB={A}^{-1}AC$
$\left({A}^{-1}A\right)B=\left({A}^{-1}A\right)C$
$I×B=I×C$
B=C
If A is not invertible then this statement is false.
Counter example:
$A=\left[\begin{array}{cc}2& -3\\ -4& 6\end{array}\right],B=\left[\begin{array}{cc}8& 4\\ 5& 5\end{array}\right],C=\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]$
Here, $det|A|=2×6-\left(-4\right)×\left(-3\right)$
=12-12
=0
That is A is not invertible.
Now, find AB and AC,
$AB=\left[\begin{array}{cc}2& -3\\ -4& 6\end{array}\right]\left[\begin{array}{cc}8& 4\\ 5& 5\end{array}\right]$
$=\left[\begin{array}{cc}16-15& 8-15\\ -32+30& -16+30\end{array}\right]$
$=\left[\begin{array}{cc}1& -7\\ -2& 14\end{array}\right]$
$AC=\left[\begin{array}{cc}2& -3\\ -4& 6\end{array}\right]\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]$
$=\left[\begin{array}{cc}10-9& -4-3\\ -20+18& 8+6\end{array}\right]$
$=\left[\begin{array}{cc}1& -7\\ -2& 14\end{array}\right]$
That is, AB=AC but B is not the same as C.

Jeffrey Jordon