Using the Inverse matrix to solve a linear system.

Question

I have the following question:

$$x - 2y = 1$$ $$2x + 3y = 4$$

So let's say:

$$A = \left[ \begin{matrix} 1 & -2 \\ 2 & 3 \end{matrix} \right]$$

So according to some formula to $2 \times 2$ matrices:

$$A^{-1} = \frac{1}{7} \left [ \begin{matrix} 3 & 2 \\ -2 & 1 \end{matrix} \right ]$$

$$= \left [ \begin{matrix} \frac{3}{7} & \frac{2}{7} \\ \frac{-2}{7} & \frac{1}{7} \end{matrix} \right ]$$

So since

$$Ax = b$$

$$A^{-1} * Ax = A^{-1}b$$

$$x = A^{-1}b$$

So

$$A^{-1}b = \left [ \begin{matrix} \frac{3}{7} & \frac{2}{7} \\ \frac{-2}{7} & \frac{1}{7} \end{matrix} \right ] * \left [ \begin{matrix} 1 \\ 4 \end{matrix} \right ] = x = \left [ \begin{matrix} \frac{11}{7} \\ \frac{2}{7} \end{matrix} \right ] $$

so $x_1 = \frac{11}{7}$ and $y = \frac{2}{7}$

My question: When you solve for a column, is x the way to denote a whole column of variables? If B is a 2x1 matrix and $A^{-1}$ is a 2x2 matrix, does this mean that the x will assign two variables?

Also, does my solution look right?

Answer 1

Regarding your notation, you might like to define things explicitly to avoid misunderstanding.

We let x$=\begin{bmatrix} x\\ y\end{bmatrix}$, hence we know exactly what you are working with. You might like to choose another notation say $v=\begin{bmatrix} x\\ y\end{bmatrix}$ so that verbal communication becomes easier as well.

As to check the solution, it is clear that the determinant is non-zero, hence the solution is unique if exists. We just have to verify your solution.

$$\frac{11-2(2)}{7}=1$$ $$\frac{2(11)+3(2)}{7}=4$$

Your solution is correct.

Answer 2

Your solution looks right.

$\mathbf{x}$ is the vector of variables, i.e. $\mathbf{x}=(x,y)^\top$ in your case. The "dimensions" of $\mathbf{x}$ and $\mathbf{b}$ are necessary from the matrix equation. Note, that you have a $2\times 2$-matrix multiplied from the right by a column vector, necessarily of dimension $2\times 1$ for matrix multiplication to be well defined in this vector case. If you spell out the actual meaning of the equation, you will see the connection to linear systems of equations right away:

$$ A\mathbf{x}=\mathbf{b}\Leftrightarrow\begin{pmatrix}a &b\\c &d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}b_1\\b_2\end{pmatrix}\Leftrightarrow\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}=\begin{pmatrix}b_1\\b_2\end{pmatrix} $$

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Note that you can always check your solution by checking the original equations again.

Answer 3

You are using the variable $x$ to mean two different things which is confusing. When you are solving $$ AX=b$$ your $X$ is a vector not a real number .

That is $$X= \left [ \begin{matrix} \frac{11}{7} \\ \frac{2}{7} \end{matrix} \right ] = \left [ \begin{matrix} x \\ y \end{matrix}\right ] $$

Thus your $x=11/7$ and $y=2/7$