Question

Find a $\mathbf{2}\mathbf{\times }\mathbf{2}$matrixsuch that the system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$has

role="math" localid="1659603098595" $\overrightarrow{\mathbf{x}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{2}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{+}\mathbf{3}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\\ \mathbf{3}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{+}\mathbf{4}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\end{array}\mathbf{\right)}$

as one of its solutions.

Short answer

The matrix$A=\left[\begin{array}{cc}-11& -6\\ 12& -18\end{array}\right]$

Step-by-step solution

Simplify the equation x→(t)=(2e2t+3e3t3e2t+4e3t)

Consider a$2\times 2$matrix A such that the system$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$hasrole="math" localid="1659603143898" $\overrightarrow{\mathrm{x}}\left(\mathrm{t}\right)=\left(\begin{array}{c}2{\mathrm{e}}^{2\mathrm{t}}+3{\mathrm{e}}^{3\mathrm{t}}\\ 3{\mathrm{e}}^{2\mathrm{t}}+4{\mathrm{e}}^{3\mathrm{t}}\end{array}\right)$.

Simplify the equation $\overrightarrow{\mathrm{x}}\left(\mathrm{t}\right)=\left(\begin{array}{c}2{\mathrm{e}}^{2\mathrm{t}}+3{\mathrm{e}}^{3\mathrm{t}}\\ 3{\mathrm{e}}^{2\mathrm{t}}+4{\mathrm{e}}^{3\mathrm{t}}\end{array}\right)$as follows.

role="math" localid="1659603367251" $$\begin{gathered} \overrightarrow{\mathrm{x}}\left(\mathrm{t}\right)=\left(\begin{array}{c}2{\mathrm{e}}^{2\mathrm{t}}+3{\mathrm{e}}^{3\mathrm{t}}\\ 3{\mathrm{e}}^{2\mathrm{t}}+4{\mathrm{e}}^{3\mathrm{t}}\end{array}\right) \\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{c}2e^{2t}\\ 3e^{2t}\end{array}\right]+\left[\begin{array}{c}3e^{3t}\\ 4e^{3t}\end{array}\right] \\ \overrightarrow{x}\left(t\right)=e^{2t}\left[\begin{array}{c}2\\ 3\end{array}\right]+e^{3t}\left[\begin{array}{c}3\\ 4\end{array}\right] \end{gathered}$$

Determine the values of c→1,c→2andv→1,v→2

Compare the equations $\overrightarrow{x}\left(t\right)=e^{2t}\left[\begin{array}{c}2\\ 3\end{array}\right]+e^{3t}\left[\begin{array}{c}3\\ 4\end{array}\right]$and $\overrightarrow{x}\left(t\right)=c_1{e}^{{\lambda }_{1}t}{\overrightarrow{v}}_1+c_2{e}^{{\lambda }_{2}t}{\overrightarrow{v}}_2$as follows

$$\begin{gathered} e^{2t}\left[\begin{array}{c}2\\ 3\end{array}\right]+e^{3t}\left[\begin{array}{c}3\\ 4\end{array}\right]=c_1{e}^{{\lambda }_{1}t}{\overrightarrow{v}}_1+c_2{e}^{{\lambda }_{2}t}{\overrightarrow{v}}_2 \\ {c}_1=1,c_2=1 \\ {\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 3\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}3\\ 4\end{array}\right] \\ {\lambda }_1=2,{\lambda }_2=3 \end{gathered}$$

Determine the values of and

As$B=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]$, substitute the values 2 for${\lambda }_1$and 3 for${\lambda }_2$in the equation$B=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]$as follows.

$B=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]$

$B=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]$

Asrole="math" localid="1659605575540" $S=\left[\begin{array}{cc}{v}_1& v_2\end{array}\right]$, substitute the values$\left[\begin{array}{c}2\\ 3\end{array}\right]$for$v_1$and$\left[\begin{array}{c}3\\ 4\end{array}\right]$for$v_2$in the equation$S=\left[\begin{array}{cc}{v}_1& v_2\end{array}\right]$as follows.

$$\begin{gathered} S=\left[\begin{array}{cc}{v}_1& v_2\end{array}\right] \\ S=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right] \end{gathered}$$

 Step 3: Determine the matrix A

By the definition of diagonal matrix B, the matrix A is defined as follows.

$$\begin{gathered} S^{-1}AS=B \\ A=SB{S}^{-1} \end{gathered}$$

Asrole="math" localid="1659606078279" $S=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]$, the matrix $S^{-1}$is defined as follows.

role="math" localid="1659606135230" $$\begin{gathered} S^{-1}=\frac{1}{-1}\left[\begin{array}{cc}4& -3\\ -3& 2\end{array}\right] \\ {S}^{-1}=\left[\begin{array}{cc}4& -3\\ -3& 2\end{array}\right] \end{gathered}$$

Substitute the values $\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]$for S ,$\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]$ for B and$\left[\begin{array}{cc}4& -3\\ -3& 2\end{array}\right]$for $S^{-1}$in the equation$A=SB{S}^{-1}$ as follows.

$A=SB{S}^{-1}$

role="math" localid="1659606360364" $A=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right]$

Simplify the equation $A=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right]$as follows

role="math" localid="1659606926953" $$\begin{gathered} A=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right] \\ A=\left[\begin{array}{cc}4& 9\\ 6& 12\end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right] \\ A=\left[\begin{array}{cc}16-27& 12-18\\ -24+36& 18-36\end{array}\right] \\ A=\left[\begin{array}{cc}-11& -6\\ 12& -18\end{array}\right] \end{gathered}$$

Hence, for a $2\times 2$matrixA such that the system $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$has $\overrightarrow{x}\left(t\right)=\left[\begin{array}{c}2e^{2t}+3e^{3t}\\ 3e^{2t}+4e^{3t}\end{array}\right]$then .

$A=\left[\begin{array}{cc}-11& -6\\ 12& -18\end{array}\right]$