Q. 92 - Precalculus Enhanced with Graphing Utilities

Question

If $a, b$, and $c\ne 0$ are real numbers with $ac = bc$, then $a = b$. Does this same property hold for matrices? In other words, if $A$, $B$, and $C$, are matrices and $AC = BC$, must $A = B$?

Step-by-Step Solution

Answer

The cancelation law holds good in real numbers but it does not hold good in matrices.

Step 1. Consider an example

If $a,b$ and $c\ne 0$ are real numbers with $ac=bc$, then $a=b$. The same property does not hold good for matrices. Thus if $A,B$ and $C$ are matrices and $AC=BC$ then $A\ne B$.

Let,

$$\begin{gathered} A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right] \\ C=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Now, we have to find $AC$

$$\begin{gathered} AC=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{ll}1\times 1+2\times 0& 1\times 0+2\times 0\\ 3\times 1+4\times 0& 3\times 0+4\times 0\end{array}\right] \\ \begin{array}{r}=\left[\begin{array}{ll}1& 0\\ 3& 0\end{array}\right]\dots \dots \left(1\right)\end{array} \end{gathered}$$

Now, let us find $BC$

$$\begin{gathered} BC=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}5\times 1+6\times 0& 5\times 0+6\times 0\\ 7\times 1+8\times 0& 7\times 0+8\times 0\end{array}\right] \\ =\left[\begin{array}{cc}5& 0\\ 7& 0\end{array}\right] .........\left(2\right) \end{gathered}$$

Step 2

From both equation $1$ and $2$, we find that $AC=BC$ but $A\ne B$

Therefore, we could conclude that the cancelation law holds good in real numbers but it does not hold good in matrices.