# Method of Matrix Exponential

## Solved Problems

### Example 1.

Find the general solution of the system, using the matrix exponential:

$\frac{{dx}}{{dt}} = 2x + 3y,\; \frac{{dy}}{{dt}} = 3x + 2y.$

Solution.

We solve this system by following the algorithm described above. Calculate the eigenvalues of the matrix $$A:$$

$\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}} {2 - \lambda }&3\\ 3&{2 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow {\left( {2 - \lambda } \right)^2} - 9 = 0,\;\; \Rightarrow 4 - 4\lambda + {\lambda ^2} - 9 = 0,\;\; \Rightarrow {\lambda ^2} - 4\lambda - 5 = 0,\;\; \Rightarrow {\lambda _1} = 5,\;{\lambda _2} = - 1.$

We find the corresponding eigenvectors for each of the eigenvalues. For the number $${\lambda _1} = 5$$ we have:

$\left[ {\begin{array}{*{20}{c}} {2 - 5}&3\\ 3&{2 - 5} \end{array}} \right[\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} { - 3}&3\\ 3&{ - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow 3{V_{11}} - 3{V_{21}} = 0,\;\; \Rightarrow {V_{11}} - {V_{21}} = 0.$

By setting $${V_{21}} = t,$$ we find the eigenvector $${\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}:$$

${V_{21}} = t,\;\; \Rightarrow {V_{11}} = {V_{21}} = t,\;\; \Rightarrow {\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} t\\ t \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right].$

Similarly, we find the eigenvector $${\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T}$$ associated with the eigenvalue $${\lambda _2} = -1:$$

$\left[ {\begin{array}{*{20}{c}} {2 - \left( { - 1} \right)}&3\\ 3&{2 - \left( { - 1} \right)} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = 0,\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} 3&3\\ 3&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = 0,\;\; \Rightarrow 3{V_{12}} + 3{V_{22}} = 0,\;\; \Rightarrow {V_{12}} + {V_{22}} = 0.$

Let $${V_{22}} = t.$$ Then $${V_{12}} = -{V_{22}}= -t.$$ Hence,

${\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - t}\\ t \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} { - 1}\\ 1 \end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}} { - 1}\\ 1 \end{array}} \right].$

We compose the matrix $$H$$ of the found eigenvectors $${\mathbf{V}_1}$$ and $${\mathbf{V}_2}:$$

$H = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 1&1 \end{array}} \right].$

Next, we compute the inverse matrix $${H^{ - 1}}:$$

$\Delta \left( H \right) = \left| {\begin{array}{*{20}{c}} 1&{ - 1}\\ 1&1 \end{array}} \right| = 1 + 1 = 2,$
${H^{ - 1}} = \frac{1}{{\Delta \left( H \right)}}{\left[ {\begin{array}{*{20}{c}} {{H_{11}}}&{{H_{12}}}\\ {{H_{21}}}&{{H_{22}}} \end{array}} \right]^T} = \frac{1}{2}{\left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 1&1 \end{array}} \right]^T} = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 1}&1 \end{array}} \right].$

As in this example the eigenvalues are simple roots of the characteristic equation, we can immediately write down the Jordan form, which will have a simple diagonal form:

$J = \left[ {\begin{array}{*{20}{c}} {{\lambda _1}}&0\\ 0&{{\lambda _2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} \color{blue}5&0\\ 0&\color{red}{ - 1} \end{array}} \right].$

We verify this by using the formula of the transition from the original matrix $$A$$ to the Jordan normal form $$J:$$

$J = {H^{ - 1}}AH = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&3\\ 3&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 1&1 \end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} 5&5\\ 1&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 1&1 \end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {5 + 5}&{ - 5 + 5}\\ {1 - 1}&{ - 1 - 1} \end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {10}&0\\ 0&{ - 2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&0\\ 0&{ - 1} \end{array}} \right] = J.$

Now we form the matrix $${e^{tJ}}$$ (it can also be called the matrix exponential):

${e^{tJ}} = \left[ {\begin{array}{*{20}{c}} {{e^{5t}}}&0\\ 0&{{e^{ - t}}} \end{array}} \right].$

Compute the matrix exponential $${e^{tA}}:$$

${e^{tA}} = H{e^{tJ}}{H^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{e^{5t}}}&0\\ 0&{{e^{ - t}}} \end{array}} \right] \cdot \frac{1}{2}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 1}&1 \end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {{e^{5t}}}&{ - {e^{ - t}}}\\ {{e^{5t}}}&{{e^{ - t}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 1}&1 \end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {{e^{5t}} + {e^{ - t}}}&{{e^{5t}} - {e^{ - t}}}\\ {{e^{5t}} - {e^{ - t}}}&{{e^{5t}} + {e^{ - t}}} \end{array}} \right].$

The general solution of the system can be written as

$\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = {e^{tA}}\left[ {\begin{array}{*{20}{c}} {{C_1}}\\ {{C_2}} \end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {{e^{5t}} + {e^{ - t}}}&{{e^{5t}} - {e^{ - t}}}\\ {{e^{5t}} - {e^{ - t}}}&{{e^{5t}} + {e^{ - t}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{C_1}}\\ {{C_2}} \end{array}} \right],$

where $${C_1},{C_2}$$ are arbitrary numbers.

This answer can also be expressed in another form:

$\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {{C_1}{e^{5t}} + {C_1}{e^{ - t}} + {C_2}{e^{5t}} - {C_2}{e^{ - t}}}\\ {{C_1}{e^{5t}} - {C_1}{e^{ - t}} + {C_2}{e^{5t}} + {C_2}{e^{ - t}}} \end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {{e^{5t}}\left( {{C_1} + {C_2}} \right) + {e^{ - t}}\left( {{C_1} - {C_2}} \right)}\\ {{e^{5t}}\left( {{C_1} + {C_2}} \right) - {e^{ - t}}\left( {{C_1} - {C_2}} \right)} \end{array}} \right] = \frac{1}{2}\left( {{C_1} + {C_2}} \right){e^{5t}}\left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right] + \frac{1}{2}\left( {{C_1} - {C_2}} \right){e^{ - t}}\left[ {\begin{array}{*{20}{c}} 1\\ { - 1} \end{array}} \right] = {B_1}{e^{5t}}\left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right] + {B_2}{e^{ - t}}\left[ {\begin{array}{*{20}{c}} 1\\ { - 1} \end{array}} \right],$

where $${B_1}$$ and $${B_2}$$ denote arbitrary constants associated with $${C_1},{C_2}.$$

### Example 2.

Solve the system of equations by the method of matrix exponential:

$\frac{{dx}}{{dt}} = 4x,\; \frac{{dy}}{{dt}} = x + 4y.$

Solution.

We solve the auxiliary equation and find the eigenvalues:

$\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}} {4 - \lambda }&0\\ 1&{4 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow {\left( {4 - \lambda } \right)^2} = 0,\;\; \Rightarrow {\lambda _1} = 4.$

So, we have one eigenvalue $${\lambda _1} = 4$$ of multiplicity $$2.$$ Determine the eigenvector $${\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}:$$

$\left[ {\begin{array}{*{20}{c}} {4 - 4}&0\\ 1&{4 - 4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} 0&0\\ 1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow 1 \cdot {V_{11}} + 0 \cdot {V_{21}} = 0.$

It follows that the coordinate $${V_{11}} = 0,$$ and the coordinate $${V_{21}}$$ can be any number. We choose for simplicity $${V_{21}} = 1.$$ Hence, the eigenvector $${\mathbf{V}_1}$$ is equal: $${\mathbf{V}_1} = {\left( {0,1} \right)^T}.$$

The second linearly independent vector is defined as the generalized eigenvector $${\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T},$$ connected to $${\mathbf{V}_1}.$$ It can be found from the equation

$\left( {A - {\lambda _1}I} \right){\mathbf{V}_2} = {\mathbf{V}_1},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} 0&0\\ 1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right],\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {0 \cdot {V_{12}} + 0 \cdot {V_{22}} = 0}\\ {1 \cdot {V_{12}} + 0 \cdot {V_{22}} = 1} \end{array}} \right..$

Here the coordinate $${{V_{22}}}$$ can be any number. We choose $${{V_{22}}} = 0.$$ Then we obtain $${{V_{11}}} = 1.$$ Thus, the generalized eigenvector is $${\mathbf{V}_2} = {\left( {1,0} \right)^T}.$$

Now, using the basis vectors, we form the matrix $$H$$ - the transition matrix from $$A$$ to the Jordan canonical form $$J:$$

$H = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right].$

Calculate the inverse matrix $${H^{ - 1}}:$$

$\Delta \left( H \right) = \left| {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right| = 0 - 1 = - 1,\;\; {H^{ - 1}} = \frac{1}{{\Delta \left( H \right)}}{\left[ {\begin{array}{*{20}{c}} {{H_{11}}}&{{H_{12}}}\\ {{H_{21}}}&{{H_{22}}} \end{array}} \right]^T}.$

Here $${H_{ij}}$$ denote the cofactors of the elements of the matrix $$H.$$ In the result of the calculations we find:

${H^{ - 1}} = \frac{1}{{\left( { - 1} \right)}}{\left[ {\begin{array}{*{20}{c}} 0&{ - 1}\\ { - 1}&0 \end{array}} \right]^T} = \left( { - 1} \right)\left[ {\begin{array}{*{20}{c}} 0&{ - 1}\\ { - 1}&0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right].$

Interestingly, in this case the inverse matrix $${H^{ - 1}}$$ coincides with the initial matrix $$H.$$ Such an effect is possible, if the square of the original matrix is the identity matrix:

${H^2} = {\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]^2} = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0 + 1}&{0 + 0}\\ {0 + 0}&{1 + 0} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = I.$

The Jordan form $$J$$ of the matrix $$A$$ is

$J = {H^{ - 1}}AH = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4&0\\ 1&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0 + 1}&{0 + 4}\\ {4 + 0}&{0 + 0} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&4\\ 4&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0 + 4}&{1 + 0}\\ {0 + 0}&{4 + 0} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} \color{blue}4&\color{blue}1\\ \color{blue}0&\color{blue}4 \end{array}} \right].$

We see that the Jordan form $$J$$ consists of a single block of size $$2.$$

Compose the matrix $${e^{tJ}}:$$

${e^{tJ}} = \left[ {\begin{array}{*{20}{c}} {{e^{4t}}}&{t{e^{4t}}}\\ 0&{{e^{4t}}} \end{array}} \right] = {e^{4t}}\left[ {\begin{array}{*{20}{c}} 1&t\\ 0&1 \end{array}} \right].$

Calculate the matrix exponential $${e^{tA}}:$$

${e^{tA}} = H{e^{tJ}}{H^{ - 1}} = {e^{4t}}\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&t\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right] = {e^{4t}}\left[ {\begin{array}{*{20}{c}} {0 + 0}&{0 + 1}\\ {1 + 0}&{t + 0} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right] = {e^{4t}}\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&t \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right] = {e^{4t}}\left[ {\begin{array}{*{20}{c}} {0 + 1}&{0 + 0}\\ {0 + t}&{1 + 0} \end{array}} \right] = {e^{4t}}\left[ {\begin{array}{*{20}{c}} 1&0\\ t&1 \end{array}} \right].$

The general solution is written as

$\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = {e^{tA}}\mathbf{C} = {e^{4t}}\left[ {\begin{array}{*{20}{c}} 1&0\\ t&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{C_1}}\\ {{C_2}} \end{array}} \right],$

where $${C_1}, {C_2}$$ are arbitrary constants.

### Example 3.

Solve the system of equations using the matrix exponential:

$\frac{{dx}}{{dt}} = x + y,\;\ \frac{{dy}}{{dt}} = - x + y.$

Solution.

In this case, the coefficient matrix $$A$$ is

$A = \left[ {\begin{array}{*{20}{c}} 1&1\\ { - 1}&1 \end{array}} \right].$

Calculate its eigenvalues:

$\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}} {1 - \lambda }&1\\ { - 1}&{1 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow {\left( {1 - \lambda } \right)^2} + 1 = 0,\;\; \Rightarrow {\left( {\lambda - 1} \right)^2} = - 1,\;\; \Rightarrow \lambda - 1 = \pm i,\;\; \Rightarrow {\lambda _{1,2}} = 1 \pm i.$

Hence, the matrix $$A$$ has a pair of complex conjugate eigenvalues. For each eigenvalue, we find the corresponding eigenvector (it can have complex coordinates).

Let $${\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}$$ be an eigenvector, associated with the eigenvalue $${\lambda _1} = 1 + i.$$ The coordinates of this vector satisfy the following matrix-vector equation:

$\left[ {\begin{array}{*{20}{c}} {1 - \left( {1 + i} \right)}&1\\ { - 1}&{1 - \left( {1 + i} \right)} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\;\Rightarrow \left[ {\begin{array}{*{20}{c}} { - i}&1\\ { - 1}&{ - i} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} { - i{V_{11}} + {V_{21}} = 0}\\ { - {V_{11}} - i{V_{21}} = 0} \end{array}} \right.,\;\; \Rightarrow \left. {\left\{ {\begin{array}{*{20}{l}} {{V_{11}} + i{V_{21}} = 0}\\ {i{V_{11}} - {V_{21}} = 0} \end{array}} \right.\;} \right|\begin{array}{*{20}{l}} {}\\ \small{{R_2} - i{R_1}}\normalsize \end{array},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{V_{11}} + i{V_{21}} = 0}\\ {0 = 0} \end{array}} \right.,\;\; \Rightarrow {V_{11}} + i{V_{21}} = 0.$

We set $${V_{21}} = t.$$ Then $${V_{11}} = -it.$$ Hence, the eigenvector $${\mathbf{V}_1}$$ is given by

${\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - it}\\ t \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} { - i}\\ 1 \end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}} { - i}\\ 1 \end{array}} \right].$

Similarly, we find the eigenvector $${\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T},$$ associated with the number $${\lambda _2} = 1 - i:$$

$\left[ {\begin{array}{*{20}{c}} {1 - \left( {1 - i} \right)}&1\\ { - 1}&{1 - \left( {1 - i} \right)} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} { i}&1\\ { - 1}&{i} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} { i{V_{12}} + {V_{22}} = 0}\\ { -{V_{12}} + i{V_{22}} = 0} \end{array}} \right.,\;\; \Rightarrow \left. {\left\{ {\begin{array}{*{20}{l}} {{V_{12}} - i{V_{22}} = 0}\\ {i{V_{12}} + {V_{22}} = 0} \end{array}} \right.\;} \right|\begin{array}{*{20}{l}} {}\\ \small{{R_2} - i{R_1}}\normalsize \end{array},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{V_{12}} - i{V_{22}} = 0}\\ {0 = 0} \end{array}} \right.,\;\; \Rightarrow {V_{12}} - i{V_{22}} = 0.$

Here we set $${V_{22}} = t.$$ Therefore, $${V_{12}} = it.$$ The vector $${\mathbf{V}_2}$$ is equal to

${\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}} {{V_{12}}}\\ {{V_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {it}\\ t \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} {i}\\ 1 \end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}} {i}\\ 1 \end{array}} \right].$

We form the matrix $$H$$ of the found eigenvectors $${\mathbf{V}_1}$$ and $${\mathbf{V}_2}:$$

$H = \left[ {\begin{array}{*{20}{c}} { - i}&i\\ 1&1 \end{array}} \right].$

Calculate the inverse matrix $${H^{ - 1}}$$ by the formula

${H^{ - 1}} = \frac{1}{{\Delta \left( H \right)}}{\left[ {\begin{array}{*{20}{c}} {{H_{11}}}&{{H_{12}}}\\ {{H_{21}}}&{{H_{22}}} \end{array}} \right]^T},$

where $${\Delta \left( H \right)}$$ is the determinant of the matrix $$H,$$ and $${H_{ij}}$$ are cofactors of the elements of the matrix $$H.$$ As a result, we obtain:

$\Delta \left( H \right) = \left| {\begin{array}{*{20}{c}} { - i}&i\\ 1&1 \end{array}} \right| = - i - i = - 2i,$
${H^{ - 1}} = \frac{1}{{\left( { - 2i} \right)}}{\left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ { - i}&{ - i} \end{array}} \right]^T} = \frac{1}{{\left( { - 2i} \right)}}\left[ {\begin{array}{*{20}{c}} 1&{ - i}\\ { - 1}&{ - i} \end{array}} \right] = \frac{1}{{2i}}\left[ {\begin{array}{*{20}{c}} { - 1}&i\\ 1&i \end{array}} \right].$

Now we find the Jordan form $$J$$ by the formula

$J = {H^{ - 1}}AH.$

By performing calculations, we find:

$J = \frac{1}{{2i}}\left[ {\begin{array}{*{20}{c}} { - 1}&i\\ 1&i \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 1}&1 \end{array}} \right] \left[ {\begin{array}{*{20}{c}} { - i}&i\\ 1&1 \end{array}} \right] = \frac{1}{{2i}}\left[ {\begin{array}{*{20}{c}} { - 1 - i}&{ - 1 + i}\\ {1 - i}&{1 + i} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} { - i}&i\\ 1&1 \end{array}} \right] = \frac{1}{{2i}}\left[ {\begin{array}{*{20}{c}} {2i - 2}&0\\ 0&{2i + 2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{i - 1}}{i}}&0\\ 0&{\frac{{i + 1}}{i}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} \color{blue}{1 + i}&0\\ 0&\color{red}{1 - i} \end{array}} \right].$

Generally speaking, we can immediately write the Jordan form $$J,$$ which in this case is diagonal (as the eigenvalues $${\lambda _1},$$ $${\lambda _2}$$ have multiplicity $$1$$). We will assume that the computation is done to test the Jordan form $$J,$$ and the matrices $$H$$ and $${H^{ - 1}}$$ are needed to define the matrix exponential.

Now we form the matrix $${e^{tJ}}:$$

${e^{tJ}} = \left[ {\begin{array}{*{20}{c}} {{e^{\left( {1 + i} \right)t}}}&0\\ 0&{{e^{\left( {1 - i} \right)t}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{e^t}{e^{it}}}&0\\ 0&{{e^t}{e^{ - it}}} \end{array}} \right] = {e^t}\left[ {\begin{array}{*{20}{c}} {{e^{it}}}&0\\ 0&{{e^{ - it}}} \end{array}} \right].$

Compute the matrix exponential $${e^{tA}}:$$

${e^{tA}} = H{e^{tJ}}{H^{ - 1}} = \frac{{{e^t}}}{{2i}}\left[ {\begin{array}{*{20}{c}} { - i}&i\\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{e^{it}}}&0\\ 0&{{e^{ - it}}} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} { - 1}&i\\ 1&i \end{array}} \right].$

The exponential functions $${{e^{it}}},$$ $${{e^{-it}}},$$ can be expanded by Euler's formula:

${e^{it}} = \cos t + i\sin t,\;\;{e^{ - it}} = \cos t - i\sin t.$

We get the following result:

${e^{tA}} = \frac{{{e^t}}}{{2i}}\left[ {\begin{array}{*{20}{c}} { - i}&i\\ 1&1 \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {\cos t + i\sin t}&0\\ 0&{\cos t - i\sin t} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} { - 1}&i\\ 1&i \end{array}} \right] = \frac{{{e^t}}}{{2i}} \left[ {\begin{array}{*{20}{c}} { - i\cos t + \sin t}&{i\cos t + \sin t}\\ {\cos t + i\sin t}&{\cos t - i\sin t} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} { - 1}&i\\ 1&i \end{array}} \right] = \frac{{{e^t}}}{{2i}}\left[ {\begin{array}{*{20}{c}} {2i\cos t}&{2i\sin t}\\ { - 2i\sin t}&{2i\cos t} \end{array}} \right] = {e^t}\left[ {\begin{array}{*{20}{c}} {\cos t}&{\sin t}\\ { - \sin t}&{\cos t} \end{array}} \right].$

The general solution of the system of differential equations is given by

$\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = {e^{tA}}\mathbf{C} = {e^t}\left[ {\begin{array}{*{20}{c}} {\cos t}&{\sin t}\\ { - \sin t}&{\cos t} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{C_1}}\\ {{C_2}} \end{array}} \right],$

where $$C = {\left( {{C_1},{C_2}} \right)^T}$$ is an arbitrary vector.