91Ó°ÊÓ

Matrices: A rectangular array is made up of a set of numbers arranged in rows and columns. The numbers are referred to as the matrix's elements or entries.

Multiply both matrices

$$\begin{gathered} \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{cc}{x}_2& x_1\\ y_2& y_1\end{array}\right)=\left(\begin{array}{cc}{x}_2& x_1\\ y_2& y_1\end{array}\right)\left(\begin{array}{cc}6& 0\\ 0& 1\end{array}\right) \\ \left(\begin{array}{cc}5x_2-2y_2& 5x_1-2y_1\\ -2x_2+2y_2& -2x_1+2y_1\end{array}\right)=\left(\begin{array}{c}6x_2\\ 6y_2\end{array}\frac{x_1}{y_1}\right).......\left(1\right) \end{gathered}$$

Equate both matrices

$$\begin{gathered} 5x_2-2y_2=6x_2 \\ 5x_1-2y_1=x_1 \\ -2x_2+2y_2=6y_2 \\ -2x_1+2y_1=y_1 \end{gathered}$$

However, it is convenient to pick numerical value of $x_1$and $y_1$to make unit vector $r_1$and $r_2$ :

Here,

$$\begin{gathered} r_1=\left(x_1,y_1\right) \\ {r}_2=\left(x_2,y_2\right) \end{gathered}$$

By solve for unit vector and find the value of ${\text{x}}_1,{\text{x}}_2,{\text{y}}_1,{\text{y}}_2$:

$$\begin{gathered} x_1=\frac{1}{\sqrt{5}} \\ {x}_2=\frac{-2}{\sqrt{5}} \\ {y}_1=\frac{2}{\sqrt{5}} \\ {y}_2=\frac{1}{\sqrt{5}} \end{gathered}$$

Substitute the value of $x_1,x_2,y_1,y_2$in equation (1):

$$\begin{gathered} \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{cc}{x}_2& x_1\\ y_2& y_1\end{array}\right)=\left(\begin{array}{cc}{x}_2& x_1\\ y_2& y_1\end{array}\right)\left(\begin{array}{cc}6& 0\\ 0& 1\end{array}\right) \\ \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{cc}\frac{-2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\end{array}\right)=\left(\begin{array}{cc}\frac{-2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\end{array}\right)\left(\begin{array}{cc}6& 0\\ 0& 1\end{array}\right) \end{gathered}$$

Represent these matrices by letters

$MC=CD$

Here,

$$\begin{gathered} M=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right) \\ C=\left(\begin{array}{cc}\frac{-2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\end{array}\right) \\ D=\left(\begin{array}{cc}6& 0\\ 0& 1\end{array}\right) \end{gathered}$$

Because determinate of $C$ is not zero, then $C$ has an inverse matrix

$C^{-1}MC=D$

and here,

$C{C}^{-1}=\text{unit matrix}$

Multiply both matrices

$$\begin{gathered} \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{cc}{x}_1& x_2\\ y_1& y_2\end{array}\right)=\left(\begin{array}{cc}{x}_1& x_2\\ y_1& y_2\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 6\end{array}\right) \\ \left(\begin{array}{c}5x_1-2y_1\\ -2x_1+2y_1\end{array}\begin{array}{c}5x_2-2y_2\\ -2x_2+2y_2\end{array}\right)=\left(\begin{array}{cc}{x}_1& x_2\\ y_1& 6y_2\end{array}\right).....\left(2\right) \end{gathered}$$

Equate both matrices

$$\begin{gathered} 5x_1-2y_1=x_1 \\ 5x_2-2y_2=6x_2 \\ -2x_1+2y_1=y_1 \\ -2x_2+2y_2=6y_2 \end{gathered}$$

By solve for unit vector and find the value of $x_1,x_2,y_1,y_2$:

$$\begin{gathered} x_1=\frac{1}{\sqrt{5}} \\ {x}_2=\frac{-2}{\sqrt{5}} \\ {y}_1=\frac{2}{\sqrt{5}} \\ {y}_2=\frac{1}{\sqrt{5}} \end{gathered}$$

Substitute the value of ${\text{x}}_1,{\text{x}}_2,{\text{y}}_1,{\text{y}}_2$in equation (1):

$$\begin{gathered} \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{cc}{x}_1& x_2\\ y_1& y_2\end{array}\right)=\left(\begin{array}{cc}{x}_1& x_2\\ y_1& y_2\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 6\end{array}\right) \\ \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{-2}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{-2}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 6\end{array}\right) \end{gathered}$$

Represent these matrices by letters:

$MC=CD$

Here,

$$\begin{gathered} M=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right) \\ C=\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{-2}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right) \\ D=\left(\begin{array}{cc}6& 0\\ 0& 1\end{array}\right) \end{gathered}$$

In this case also determinate of $C$ is not zero, then $C$ has an inverse matrix

$C^{-1}MC=D$

and here,

$C{C}^{-1}=\text{unit matrix}$

In both cases we have obtained two different values of $C$.