Question

In Problems 1and 2 fill in the blanks.

The vector X =k(4/5)is a solution of

$X\text{'}=\left[\begin{array}{cc}1& 4\\ 2& -1\end{array}\right]X-\left[\begin{array}{c}8\\ 1\end{array}\right]$

for k = .

Short answer

The vector $X=k\left[\begin{array}{c}4\\ 5\end{array}\right]$is a solution of $X\text{'}=\left[\begin{array}{cc}1& 4\\ 2& -1\end{array}\right]X-\left[\begin{array}{c}8\\ 1\end{array}\right]$ fork = 1/3.

Step-by-step solution

Define matrix exponential

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

$A^0=I,A^1=A,A^2=A.A,A^3=A^2.A,.....,A^k=A.\underset{ktimes}{A}.....$

where I is the unit matrix of order n.

Therefore, the infinite matrix power series is $I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$.

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,$e^{tA}=\sum _{k=0}^{\infty }\frac{t^k}{k!}{A}^k$ .

Find the value of k.

It is given that,

$X=k\left[\begin{array}{c}4\\ 5\end{array}\right]$

……

$X\text{'}=\left[\begin{array}{cc}1& 4\\ 2& -1\end{array}\right]X-\left[\begin{array}{c}8\\ 1\end{array}\right]$ ……

Differentiating (1) with respect to x,

$X\text{'}=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Substitute the differentiated value in (2), then it gives

$$\begin{gathered} \left[\begin{array}{c}0\\ 0\end{array}\right]=k\left[\begin{array}{cc}1& 4\\ 2& -1\end{array}\right]\left(\begin{array}{c}4\\ 5\end{array}\right)-\left(\begin{array}{c}8\\ 1\end{array}\right) \\ =\left[\begin{array}{cc}1& 4\\ 2& -1\end{array}\right]\left(\begin{array}{c}4k\\ 5k\end{array}\right)-\left(\begin{array}{c}8\\ 1\end{array}\right) \\ =\left[\begin{array}{cc}4k+& 20k\\ 8k-& 5k\end{array}\right]-\left(\begin{array}{c}8\\ 1\end{array}\right) \\ \left(\begin{array}{c}8\\ 1\end{array}\right)=\left[\begin{array}{c}24k\\ 3k\end{array}\right] \end{gathered}$$

Therefore, the value of k is found as:

role="math" localid="1663952367550" $$\begin{gathered} 8=24k \\ =\left[\begin{array}{c}1\\ 3\end{array}\right] \\ 1=3k \\ =\left[\begin{array}{c}1\\ 3\end{array}\right] \end{gathered}$$

Hence the value of (1/3) .