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Consider the matrix, $M = \begin{bmatrix} -4 &1 &1 \\ 2& 0&1 \\ 0&1 &3 \end{bmatrix}$. Are its columns linearly dependent or linearly independent vectors? Justify your answer and use it to find all solutions to the linear system $Mx=0$, where $x$ is a $3 \times 1$ real vector and $0$ is a $3 \times 1$ vector of zeros. Explain your answers.

My solution: To check linear independence $Mx=0 \implies x=0$.

So after doing the matrix algebra I am left with the following system of equations:

  1. $-4x_{1}+x_{2}+x_{3}=0$
  2. $2x_{1}+x_{3}=0$
  3. $x_{2}+3x_{3}=0$

From equation (2): I write $x_{3}=-2x_{1}$ and substitute away $x_{3}$ in equation (3) to get $x_{2} = 6x_{1}$.

Then I substitute away $x_{2}$ and $x_{3}$ in equation (1) and get $-4x_{1}+6x_{1}-2x_{1}=0$. So this just results in $0=0$.

Does this imply that the column vectors are not linearly independent? Also, are we going to have infinite solutions to this linear system as it could be the case that $Rank(M)<n=3$?

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  • $\begingroup$ Do you know how to do Gauss-Jordan elimination? $\endgroup$ – user137731 Sep 14 '15 at 0:46
  • $\begingroup$ No, but my instructor never taught me that and wants me to solve using the linear system of equations. $\endgroup$ – OGC Sep 14 '15 at 0:48
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Gauss-Jordan method: $$\begin{bmatrix} -4 &1 &1 \\ 2& 0&1 \\ 0&1 &3 \end{bmatrix} \sim \begin{bmatrix} 0 &1 &3 \\ 2& 0&1 \\ 0&1 &3 \end{bmatrix} \sim \begin{bmatrix} 2& 0&1 \\ 0 &1 &3 \\ 0&0 &0 \end{bmatrix}$$ From here we can immediately tell this is a rank $2$ matrix.


Triple product method: $$(-4,2,0) \cdot ((1,0,1)\times (1,1,3)) = (-4,2,0)\cdot(-1,-2,1) = 0$$ Therefore this matrix has rank $\lt 3$.


Determinant method: $$\left|\begin{matrix} -4 &1 &1 \\ 2& 0&1 \\ 0&1 &3 \end{matrix}\right| = 0$$

Therefore this matrix has rank $\lt 3$.


Substitution method (following your progress): $$\begin{cases} −4x_1+x_2+x_3=0 \\ 2x_1+x_3=0\\ x_2+3x_3=0\end{cases} \iff \begin{cases} −4x_1+x_2+x_3=0 \\ x_3=−2x_1\\ x_2=6x_1\end{cases} \iff \begin{cases} 0=0 \\ x_3=−2x_1\\ x_2=6x_1\end{cases}$$

Because the last system of equations has $3$ variables and only $2$ nontrivial equations it can't have a unique solution. Because this is a homogeneous system it also can't have no solutions. Therefor it must have infinitely many solutions.


A matrix that has rank $\lt$ the number of columns will have linearly dependent columns. A matrix that has rank $=$ to the number of columns will have linearly independent columns. It's not possible to have the rank of a matrix be larger than the number of columns.

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  • $\begingroup$ So will the linear system $Mx=0$ have infinitely many solutions? $\endgroup$ – OGC Sep 14 '15 at 0:54
  • $\begingroup$ My latest edit should answer that question. $\endgroup$ – user137731 Sep 14 '15 at 0:57
  • $\begingroup$ Cool. And also the column vectors are linearly dependent? It will help me if you mention it in your answer. Thanks. $\endgroup$ – OGC Sep 14 '15 at 1:00
  • $\begingroup$ It seems that the determinant method is the easiest by far. If the determinant of a square matrix is $0$, that is, it is a singular matrix then it is not full rank. So the rank will always be less than the number of columns. $\endgroup$ – OGC Sep 14 '15 at 1:13
  • $\begingroup$ That is correct. I'd actually say the triple product is easiest, but the determinant is pretty easy too. $\endgroup$ – user137731 Sep 14 '15 at 1:15

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