Find all real numbers $x$ for which the matrices are linearly dependent
$$M_1=\begin{bmatrix} 1&1\\x&0 \end{bmatrix}\quad M_2= \begin{bmatrix} 1&-x\\x-1&3 \end{bmatrix}$$ $$M_3= \begin{bmatrix} 0&-2\\1&x \end{bmatrix}\quad M_4= \begin{bmatrix} 1&-3\\-1&2x \end{bmatrix}$$
Note: As usual, it will be useful if you start your answer by stating what condition you have to verify. Be careful not to divide by zero in your calculation.
To start this question I gave the condition of Linear dependence and set up a gaussian elimination with the last columns as 0's but this is as far as I can go as I have never worked with a question of this sort, so I need help on how to conduct the gaussian on this question
You want to find where the determinant is zero. The determinant of matrix $A$, where $A$ is given by
$$A = \begin{pmatrix}a & b\\ c & d \end{pmatrix}$$
The determinant of $A$ is $ad-bc$. So you want to solve $ad-bc = 0$.
For example, consider $M_{3}$. The determinant is given by
$$\text{det}(M_{3}) = \begin{vmatrix} 0 & -2\\ 1 & x \end{vmatrix} = 0 \cdot x + 1 \cdot -2 =-2.$$
Since this determinant can never be zero, you have no values of $x$ such that the rows are linearly dependent.
Alternatively, you can find which scalar multiples work such that multiplying the first row by the scalar gives the second row.