Question

Given

$\mathbf{A}\mathbf{=}\left(\begin{array}{cc}1& 2\\ 3& 6\end{array}\right)\mathbf{,}\mathbf{}\mathbf{C}\mathbf{=}\left(\begin{array}{cc}7& 6\\ 2& 3\end{array}\right)\mathbf{,}\mathbf{}\mathbf{D}\mathbf{=}\left(\begin{array}{cc}-3& 2\\ 7& 5\end{array}\right)$

Show that,$\mathbf{AC}\mathbf{=}\mathbf{AD}\mathbf{,}\mathbf{but}\mathbf{}\mathbf{C}\mathbf{\ne }\mathbf{D}\mathbf{}\mathbf{AND}\mathbf{}\mathbf{A}\mathbf{\ne }\mathbf{0}$.

Short answer

Find the products AC and AD of the given matrices to prove that AC=AD, but $\mathrm{C}\ne \mathrm{D}\mathrm{and}\mathrm{A}\ne 0.$.

Step-by-step solution

Definition of Matrix multiplication:

The element in row iand column j of the product matrix ABis equal to row iof Atimes column j of B. In index notation${\left(\mathrm{AB}\right)}_{\mathbf{ij}}\mathbf{}\mathbf{=}\underset{\mathbf{k}}{\mathbf{\sum }{\mathbf{A}}_{\mathbf{ik}}{\mathbf{B}}_{\mathbf{kj}}}$.

For example, if $$\begin{gathered} \mathrm{A}=\left(\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right)\mathrm{and}\mathrm{B}=\left(\begin{array}{cc}\mathrm{e}& \mathrm{f}\\ \mathrm{c}& \mathrm{d}\end{array}\right)\mathrm{then}, \\ \mathrm{AB}=\left(\begin{array}{cc}\mathrm{ae}+\mathrm{bg}& \mathrm{af}+\mathrm{bh}\\ \mathrm{ce}+\mathrm{dg}& \mathrm{cf}+\mathrm{dh}\end{array}\right) \end{gathered}$$and then,

 Multiply Matrices:

The give matrices are

$\mathrm{A}=\left(\begin{array}{cc}1& 2\\ 3& 6\end{array}\right),\mathrm{C}=\left(\begin{array}{cc}7& 6\\ 2& 3\end{array}\right)\mathrm{and}\mathrm{D}=\left(\begin{array}{cc}-3& 2\\ 7& 5\end{array}\right)$

It needs to be proved that $\mathrm{AC}=\mathrm{AD},\mathrm{but}\mathrm{C}\ne \mathrm{D}\mathrm{and},\mathrm{A}\ne 0$

$$\begin{gathered} \mathrm{AC}=\left(\begin{array}{cc}1& 2\\ 3& 6\end{array}\right)\left(\begin{array}{cc}7& 6\\ 2& 3\end{array}\right) \\ =\left(\begin{array}{cc}7+4& 6+6\\ 21+12& 18+18\end{array}\right) \\ =\left(\begin{array}{cc}11& 12\\ 33& 36\end{array}\right) \\ \mathrm{And} \\ \mathrm{AD}=\left(\begin{array}{cc}1& 2\\ 3& 6\end{array}\right)\left(\begin{array}{cc}-3& 2\\ 7& 5\end{array}\right) \\ =\left(\begin{array}{cc}-3+14& 2+10\\ -9+42& 6+30\end{array}\right) \\ =\left(\begin{array}{cc}11& 12\\ 33& 36\end{array}\right) \end{gathered}$$

$\mathrm{Thus},\mathrm{AC}=\mathrm{AD},\mathrm{while}\mathrm{C}\ne \mathrm{D}\mathrm{and}\mathrm{A}\ne 0.$