# Finding values of x for linear dependence

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

• Would you mind using MathJax to format your matrices? I’m not sure I understand what your matrices look like. – Jack Moody Nov 19 '18 at 21:17
• I have edited it – vjhyvjhvjhvjhg Nov 19 '18 at 21:28
• Are you trying to prove that the four matrices are linearly independent, or that each individual matrix has linearly-independent rows/columns? – amd Nov 21 '18 at 1:28

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.

• I am unclear as how the determinant would work for this example? Can you show me? – vjhyvjhvjhvjhg Nov 19 '18 at 23:06
• @connttttt see the updated answer – Jack Moody Nov 19 '18 at 23:13