Verify the calculations in (6.15) ,(6.16), and (6.17) .

Matrix multiplication is a binary operation that creates a matrix by multiplying two matrices together. For matrix multiplication to work, the number of columns in the first matrix must equal the number of rows in the second matrix.

For example, if matrix A and B is defined as $\left(3\times 2\right)$and $\left(3\times 2\right)$then, the product of the matrices A and B is meaningless as the columns in the first matrix (here, 2) is not equal to the number of rows in the second matrix (here, 3).

In (6.15), a given matrix is,

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

In (6.16) , a function has been given as follow.

$\mathrm{f}\left(\mathrm{A}\right)=3-2{\mathrm{A}}^2-{\mathrm{A}}^3-5{\mathrm{A}}^4+{\mathrm{A}}^6$

And in (6.17) the given equation is,

${\mathrm{e}}^{\mathrm{kA}}=1+\mathrm{k}\mathrm{A}+\frac{{\mathrm{k}}^2{\mathrm{A}}^2}{2!}+\frac{{\mathrm{k}}^3{\mathrm{A}}^3}{3!}+\frac{{\mathrm{k}}^4{\mathrm{A}}^4}{4!}+\frac{{\mathrm{k}}^5{\mathrm{A}}^5}{5!}+$

The powers ${\mathrm{A}}^2,{\mathrm{A}}^3,{\mathrm{A}}^4$, and so on need to be calculated to verify the fact that the higher powers simply repeat the four results: $\mathrm{A},-\mathrm{I},-\mathrm{A},\mathrm{I},$ , over and over. Also, The value of f (A) and${\mathrm{e}}^{\mathrm{kA}}$ are to be found.

Find the product ${\mathrm{A}}^2$.

$$\begin{gathered} A^2=\left(\begin{array}{cc}1& \sqrt{2}\\ -\sqrt{2}& -1\end{array}\right)\left(\begin{array}{cc}1& \sqrt{2}\\ -\sqrt{2}& -1\end{array}\right) \\ =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right) \\ =-\mathrm{I} \end{gathered}$$

Find the product ${\mathrm{A}}^3$.

$$\begin{gathered} {\mathrm{A}}^3={\mathrm{AA}}^2 \\ =\mathrm{A}(-\mathrm{I}) \\ =-\mathrm{A} \end{gathered}$$

Find the value of ${\mathrm{A}}^4$.

$$\begin{gathered} {\mathrm{A}}^4={\mathrm{A}}^2{\mathrm{A}}^2 \\ =(-\mathrm{I})(-\mathrm{I}) \\ =\mathrm{I} \end{gathered}$$

Find the product ${\mathrm{A}}^5$.

$$\begin{gathered} {\mathrm{A}}^5={\mathrm{A}}^4\mathrm{A} \\ =\mathrm{IA} \\ =\mathrm{A} \end{gathered}$$

Find the product ${\mathrm{A}}^6$.

$$\begin{gathered} {\mathrm{A}}^6={\mathrm{A}}^4{\mathrm{A}}^2 \\ =\mathrm{I}(-\mathrm{I}) \\ =-\mathrm{I} \end{gathered}$$

It is verified that the higher powers of A repeats the four results $\mathrm{A},-\mathrm{I},-\mathrm{A},\mathrm{I}$.

Substitute the values of ${\mathrm{A}}^2,{\mathrm{A}}^3,{\mathrm{A}}^4,$and ${\mathrm{A}}^6$into the function f (A) .

$$\begin{gathered} \mathrm{f}\left(\mathrm{A}\right)=3-2{\mathrm{A}}^2-{\mathrm{A}}^3-5{\mathrm{A}}^4+{\mathrm{A}}^6 \\ =3\mathrm{I}+2\mathrm{I}+\mathrm{A}-5\mathrm{I}-\mathrm{I} \\ =\mathrm{A}-\mathrm{I} \end{gathered}$$

Use the values of A and, I .

$$\begin{gathered} f\left(A\right)=\left(\begin{array}{cc}1& \sqrt{2}\\ -\sqrt{2}& -1\end{array}\right)-\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}1& \sqrt{2}\\ -\sqrt{2}& -2\end{array}\right) \end{gathered}$$

The value of f(A) is $\left(\begin{array}{cc}1& \sqrt{2}\\ -\sqrt{2}& -2\end{array}\right)$.

Substitute the values of ${\mathrm{A}}^2{\mathrm{A}}^3{\mathrm{A}}^4{\mathrm{A}}^5$and ${\mathrm{A}}^6$into the function.

$$\begin{gathered} {\mathrm{e}}^{\mathrm{kA}}=1+\mathrm{kA}+\frac{{\mathrm{k}}^2{\mathrm{A}}^2}{2!}+\frac{{\mathrm{k}}^3{\mathrm{A}}^3}{3!}+\frac{{\mathrm{k}}^4{\mathrm{A}}^4}{4!}+\frac{{\mathrm{k}}^5{\mathrm{A}}^5}{5!}+... \\ =1+\mathrm{kA}+\frac{{\mathrm{k}}^2\left(-\mathrm{I}\right)}{2!}+\frac{{\mathrm{k}}^3\left(-\mathrm{A}\right)}{3!}+\frac{{\mathrm{k}}^4\mathrm{I}}{4!}+\frac{{\mathrm{k}}^5\mathrm{A}}{5!}+... \end{gathered}$$

Evaluate the equation.

role="math" localid="1658985408095" $$\begin{gathered} {\mathrm{e}}^{\mathrm{kA}}=\left(1-\frac{{\mathrm{k}}^2}{2!}+\frac{{\mathrm{k}}^4}{4!}+...\right)\mathrm{I}+\left(\mathrm{k}-\frac{{\mathrm{k}}^3}{3!}+\frac{{\mathrm{k}}^5}{5!}+...\right)\mathrm{A} \\ =\left(\cos \mathrm{k}\right)\mathrm{I}+\left(\sin \mathrm{k}\right)\mathrm{A} \\ =\cos \mathrm{k}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\sin \mathrm{k}\left(\begin{array}{cc}1& \sqrt{2}\\ -\sqrt{2}& -1\end{array}\right) \\ =\left(\begin{array}{cc}\mathrm{cosk}& 0\\ 0& \mathrm{cosk}\end{array}\right)+\left(\begin{array}{cc}\mathrm{sink}& \sqrt{2}\mathrm{sink}\\ -\sqrt{2}\mathrm{sink}& -\mathrm{sink}\end{array}\right) \end{gathered}$$

${\mathrm{e}}^{\mathrm{kA}}=\left(\begin{array}{cc}\cos k+\mathrm{sink}& \sqrt{2}\mathrm{sink}\\ -\sqrt{2}\sin k& \cos k-\sin k\end{array}\right)$

Hence, the value of ${\mathrm{e}}^{\mathrm{kA}}$ is role="math" localid="1658985513112" $\left(\begin{array}{cc}\cos k+\mathrm{sink}& \sqrt{2}\mathrm{sink}\\ -\sqrt{2}\sin k& \cos k-\sin k\end{array}\right)$.