First of - sorry for my English.
My math teacher gave me three tasks to complete, but I can't complete the second one.
Here is the system: $$\begin{cases} x+y-z=3 \\ x-3y+2z=1 \\ 7x-4y+z=7\end{cases}$$
We have to complete them using Cramer's rule. Determinant result is 43 for me, but today when I asked teacher she said there should be 1. I can't figure it out. How it can be 1? It would be great if someone could explain how to get to 1 step by step.
The coefficient matrix for the system $$\begin{cases} x+y-z=3 \\ x-3y+2z=1 \\ 7x-4y+z=7\end{cases}$$ is $$\pmatrix{1 & 1 & -1 \\ 1 & -3 & 2 \\ 7 & -4 & 1}$$
Let's find the determinant:
$$\begin{align}\left|\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -3 & 2 \\ 7 & -4 & 1\end{array}\right| &= \left|\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -4 & 3 \\ 0 & -11 & 8\end{array}\right| \\ &= \frac 13\left|\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -12 & 9 \\ 0 & -11 & 8\end{array}\right| \\ &= \frac 13\left|\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -1 & 1 \\ 0 & -11 & 8\end{array}\right| \\ &= -\frac 13\left|\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & -3\end{array}\right| \\ &= -\frac 13(1)(1)(-3) \\ &= 1\end{align}$$
EDIT: Here's the cofactor expansion method of finding the determinant. I'll expand across the top row:
$$\begin{align}\left|\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -3 & 2 \\ 7 & -4 & 1\end{array}\right| &= (1)\left|\begin{array}{cc} -3 & 2 \\ -4 & 1\end{array}\right|-(1)\left|\begin{array}{cc} 1 & 2 \\ 7 & 1\end{array}\right|+(-1)\left|\begin{array}{cc} 1 & -3 \\ 7 & -4\end{array}\right| \\ &= (1)[-3\cdot 1 - 2\cdot(-4)] -(1)[1\cdot 1 - 2\cdot 7]+(-1)[1\cdot(-4)-(-3\cdot 7)] \\ &= 5+13-17 \\ &=1\end{align}$$