91Ó°ÊÓ

Eigenvectors are a nonzero vector that is mapped by a given linear transformation of a vector space onto a vector that is the product of a scalar multiplied by the original vector.

The objective is to find $2×2$ matrix A such that $\left(\begin{array}{c}3\\ 1\end{array}\right)$ and $\left(\begin{array}{c}1\\ 2\end{array}\right)$ are Eigen vectors of A with respect to eigen values 5 and 10 respectively.

Let, a matrix$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

If x is a Eigen vector and$\mathrm{λ}$is an Eigen value of matrix A then the following equation should be hold.

$Ax=\mathrm{λ}x$

Substitute$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],x_1=\left(\begin{array}{c}3\\ 1\end{array}\right)$and${\mathrm{λ}}_1=5$in$A{x}_1={\mathrm{λ}}_1{x}_1$

$$\begin{gathered} A{x}_1={\mathrm{λ}}_1{x}_1\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left(\begin{array}{c}3\\ 1\end{array}\right) \\ =5\left(\begin{array}{c}3\\ 1\end{array}\right)\left(\begin{array}{c}3a+b\\ 3c+d\end{array}\right) \\ =\left(\begin{array}{c}15\\ 5\end{array}\right) \end{gathered}$$

Now equating on both sides, then

$\begin{array}{ll}3a+b=15& .....\left(1\right)\\ 3c+d=5& ......\left(2\right)\end{array}$

Substitute $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],x_2=\left(\begin{array}{c}1\\ 2\end{array}\right)$ and ${\mathrm{λ}}_2=10\mathrm{in}{\mathrm{Ax}}_2={\mathrm{λ}}_2{\mathrm{x}}_2$

$$\begin{gathered} A{x}_2={\mathrm{λ}}_2{x}_2\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left(\begin{array}{c}1\\ 2\end{array}\right) \\ =10\left(\begin{array}{c}1\\ 2\end{array}\right)\left(\begin{array}{c}a+2b\\ c+2d\end{array}\right) \\ =\left(\begin{array}{c}10\\ 20\end{array}\right) \end{gathered}$$

Now equating on both sides, then

$\begin{array}{l}a+2b=10.……………\left(3\right)\\ c+2d=20.…………….\left(4\right)\end{array}$

Now solve the equations (1) and (3) to solve the values of a and b

$\begin{array}{l}3a+b=15..….\left(1\right)\\ a+2b=10.…..\left(3\right)\end{array}$

Multiply the equation (3) with 3 and subtract from (1), then

$\begin{array}{l}3a+b=153\\ a+6b=30\\ ---------\\ 5b=-15\\ b=3\end{array}$

Substitutein (1), then

$\begin{array}{l}3a+3=15\\ ⇒3a=12\\ ⇒a=4\end{array}$

Now solve the equations (2) and (4), then

$\begin{array}{l}3c+d=5\\ c+2d=2\end{array}$

Multiply the equation (4) with 3 and subtract from (1), then

$$\begin{gathered} \begin{array}{c}3c+d=5\\ 3c+6d=0\end{array} \\ ------- \\ 5d=-55 \\ d=11 \end{gathered}$$

Now substitute d=11 in equation (2), then

$\begin{array}{l}3c+11=5\\ ⇒3c=-6\\ ⇒c=-2\end{array}$

Now substitute the values ofandin matrix, then

$\begin{array}{l}A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\\ ⇒A=\left[\begin{array}{cc}4& 3\\ -2& 11\end{array}\right]\end{array}$

Therefore, required matrix A is$\left[\begin{array}{cc}4& 3\\ -2& 11\end{array}\right]$ .