Question

In Problems 19-20verify the foregoing result for the given matrix.

$A=\left(\begin{array}{cc}2& 1\\ -3& 6\end{array}\right)$

Short answer

The foregoing result for the given matrix is $\left(\begin{array}{cc}1& 1\\ 3& 1\end{array}\right)\left(\begin{array}{cc}5& 0\\ 0& 3\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right)=\left(\begin{array}{cc}2& 1\\ -3& 6\end{array}\right)=A$. The result is verified.

Step-by-step solution

Define matrix exponential.

Consider a square matrix A of size n*n. This matrix can contain either complex numbers or real numbers. The matrix can be calculated as

where I is the unit matrix of order n.

Therefore the infinite matrix power series is

Now, the matrix exponential is defined as the sum of the infinite matrix power series. It is denoted by the expression e^At. It is given by the formula,

Find the eigenvalues.

We are given,

$A=\left(\begin{array}{cc}2& 1\\ -3& 6\end{array}\right)$

Solve the matrix.

$$\begin{gathered} det(A-\lambda I)=0\to \left(\begin{array}{cc}2-\lambda & 1\\ -3& 6-\lambda \end{array}\right)=0 \\ (2-\lambda )(6-\lambda )+3=0 \\ 12-2\lambda -6\lambda +{\lambda }^2+3=0 \\ {\lambda }^2-8\lambda +15=0 \\ (\lambda -5)(\lambda -3)=0 \end{gathered}$$

Hence the eigenvalues are ${\lambda }_1=5,{\lambda }_2=3$

Find the eigenvectors.

For${\lambda }_1=5$

$$\begin{gathered} (A-5I|0)=\left(\begin{array}{ccc}2-5& 1& 0\\ -3& 6-5& 0\end{array}\right) \\ =\left(\begin{array}{ccc}-3& 1& 0\\ -3& 1& 0\end{array}\right) \end{gathered}$$

Apply row operation,

$$\begin{gathered} R_2-R_1\to R_2 \\ =\left(\begin{array}{ccc}-3& 1& 0\\ 0& 0& 0\end{array}\right) \\ -3k_1+k_2=0\to k_1=\frac{1}{3}{k}_2 \end{gathered}$$

If we choose k₂=3, it gives k₁ =1.

It gives the eigenvector,

$K_1=\left(\begin{array}{c}1\\ 3\end{array}\right)$

For ${\lambda }_2=3$

$$\begin{gathered} (A-3I|0)=\left(\begin{array}{ccc}2-3& 1& 0\\ -3& 6-3& 0\end{array}\right) \\ =\left(\begin{array}{ccc}-1& 1& 0\\ -3& 3& 0\end{array}\right) \end{gathered}$$

Apply row operation,

$$\begin{gathered} R_2-3R_1\to R_2: \\ =\left(\begin{array}{ccc}-1& 1& 0\\ 0& 0& 0\end{array}\right) \\ -k_1+k_2=0\to k_1=k_2 \end{gathered}$$

If we choose k₂ =1, it gives k₁=1.

It gives the eigenvector,

$k_2=\left(\begin{array}{c}1\\ 1\end{array}\right)$

Hence the eigenvectors are $k_1=\left(\begin{array}{c}1\\ 3\end{array}\right)and{k}_2=\left(\begin{array}{c}1\\ 1\end{array}\right)$.

Verification of the foregoing result of the given matrix.

P denotes a matrix whose columns are eigenvectors K₁and K₂so

$$\begin{gathered} P=\left(\begin{array}{cc}1& 1\\ 3& 1\end{array}\right) \\ det\left(P\right)=\left|\begin{array}{cc}1& 1\\ 3& 1\end{array}\right|=1-3=-2 \end{gathered}$$

Now let’s find the inverse matrix of P.

$P^{-1}=\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right)$

We know that,

role="math" localid="1665069574023" $D=\left(\begin{array}{cc}5& 0\\ 0& 3\end{array}\right)$

The inverse of the matrix PDP is,

$$\begin{gathered} PD{P}^{-1}=\left(\begin{array}{cc}1& 1\\ 3& 1\end{array}\right)\left(\begin{array}{cc}5& 0\\ 0& 3\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right) \\ =\left(\begin{array}{cc}5& 3\\ 15& 3\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right) \\ =\left(\begin{array}{cc}2& 1\\ -3& 6\end{array}\right) \\ =A \end{gathered}$$

Hence it is verified that A=PDP^-1.