Show that the given linear system has this specific solution if $\det \neq 0$

Question

Show that the linear system $\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix}= \begin{pmatrix} b_1\\ b_2 \end{pmatrix} \,\,\,\,\,\, \text{ with }\,\,\,\, A= \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}$

has the unique solution $\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$ with $x_1 = \frac{\det \begin{pmatrix} b_1 & a_{12}\\ b_2 & a_{22} \end{pmatrix}}{\det A} \,\,\,\,\,\,; \,\,\,\,\, x_2 = \frac{\det \begin{pmatrix} a_{11} & b_1\\ a_{21} & b_2 \end{pmatrix}}{\det A}$, if $\det A \neq 0.$

I have an idea and I think it's almost correct but something is missing or wrong..

Let $\vec{x} = \begin{pmatrix} x_1\\ x_2 \end{pmatrix}$ and let $\vec{b} = \begin{pmatrix} b_1\\ b_2 \end{pmatrix}$. For the linear system $A\vec{x} = \vec{b}$ there exists the unique solution $\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$, if $\det A \neq 0$ because we can use the inverse of matrix $A$:

$$A\vec{x}=\vec{b} \Leftrightarrow A^{-1}A\vec{x} = A^{-1} \vec{b} \Leftrightarrow A^{-1} \vec{b} = \vec{x}$$

Know that $A^{-1} = \frac{1}{\det A} \begin{bmatrix} a_{22} & -a_{12}\\ -a_{21} & a_{11} \end{bmatrix}$

Then we have $A^{-1} \vec{b} = \vec{x} \Leftrightarrow \frac{1}{\det A} \begin{bmatrix} a_{22} & -a_{12}\\ -a_{21} & a_{11} \end{bmatrix} \begin{bmatrix} b_1\\ b_2 \end{bmatrix} = \begin{bmatrix} x_1\\ x_2 \end{bmatrix}$

But how is this supposed to be equal to the solution given in the task?

I did a mistake somewhere? :(

Answer 1

You are almost there; simply write out what the two determinants $$\det \begin{pmatrix} b_1 & a_{12}\\ b_2 & a_{22} \end{pmatrix}\qquad\text{ and }\qquad\det \begin{pmatrix} a_{11} & b_1\\ a_{21} & b_2 \end{pmatrix}$$ are. You will find that these are precisely the coefficients of the vector $$\begin{bmatrix} a_{22} & -a_{12}\\ -a_{21} & a_{11} \end{bmatrix} \begin{bmatrix} b_1\\ b_2 \end{bmatrix}.$$