Problem of the week #9 - May 28th, 2012

[Jameson]

Thank you to Chris L T521 for submitting this problem!

Consider a $2\times 2$ matrix

\[A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}.\]

If $\det A = ad-bc \neq 0$, show that

\[A^{-1}=\frac{1}{\det A} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}.\]


Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) Sudharaka

Solution:

Here is the way Sudharaka correctly demonstrated that the given matrix inverse for A satisfied the conditions necessary to in fact be its inverse.

Spoiler

\(\mbox{Let, }A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}\mbox{ and }B=\frac{1}{\det A} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\mbox{ where }\det A=ad-bc\neq 0\,.\)

\begin{eqnarray}

AB &=& \frac{1}{\det A}\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\\

&=&\frac{1}{\det A}\begin{bmatrix}ad-bc & 0 \\ 0 & ad-bc\end{bmatrix}\\

&=&\frac{1}{ad-bc}\begin{bmatrix}ad-bc & 0 \\ 0 & ad-bc\end{bmatrix}\\

&=&\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\\

\therefore AB &=&\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}~~~~~~~~~~(1)

\end{eqnarray}

Similarly,

\begin{eqnarray}

BA &=& \frac{1}{\det A}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\begin{bmatrix}a & b \\ c & d\end{bmatrix}\\

&=&\frac{1}{\det A}\begin{bmatrix}ad-bc & 0 \\ 0 & ad-bc\end{bmatrix}\\

&=&\frac{1}{ad-bc}\begin{bmatrix}ad-bc & 0 \\ 0 & ad-bc\end{bmatrix}\\

&=&\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\\

\therefore BA &=&\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}~~~~~~~~~~(1)

\end{eqnarray}

By (1) and (2);

\[AB=BA=I\mbox{ where }I\mbox{ is the identity matrix of order 2}\]

\[\therefore B=A^{-1}=\frac{1}{\det A} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\]

Q.E.D