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Find the basis for the nullspace, the row space, and the column space of the given matrix.$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 3 & 2 & 5 & 2 \\ -1 & 2 & 1 & 3 \\ 1 & 1 & 2 & 2 \end{bmatrix}$$

My Try $$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 3 & 2 & 5 & 2 \\ -1 & 2 & 1 & 3 \\ 1 & 1 & 2 & 2 \end{bmatrix}_{R_2\rightarrow R_2-3R_1\\R_3\rightarrow R_3+R_1\\R_4\rightarrow R_4-R_1}$$ $$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & -4 & -4 & -1 \\ 0 & 4 & 4 & 4 \\ 0 & -1 & -1 & 1 \end{bmatrix}_{R_2\rightarrow \frac{R_2}{-1}\\R_3\rightarrow \frac{R_3}{4}}$$ $$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & -1 & -1 & 1 \end{bmatrix}_{R_3\rightarrow 4R_3-R_2\\R_4\rightarrow4R_4+R_2}$$ $$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ I got solution as inconsistent. How to proceed further?

I also referred to $1^{st}$ question of This

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  • $\begingroup$ The last row can be reduced to a zero row by $R_4\rightarrow R_4-\dfrac{R_3}{3}.$ $\endgroup$ – poyea Jul 31 '18 at 21:46
  • $\begingroup$ There is no such thing as “the” basis for any of these spaces. At best you can find a basis for each. $\endgroup$ – amd Aug 1 '18 at 0:08
  • $\begingroup$ “Solution” to what? There’s no system of linear equations here, just a matrix. $\endgroup$ – amd Aug 1 '18 at 0:09
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We don't say that the RREF is not consistent in that case.

We need another step to obtain

$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix}\to \begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

therefore a basis for the nullspace is given by solving $Ax=0$ that is $$(1,1,-1,0)$$

For the column space, a basis is formed by the columns of the original matrix containing the pivots in the RREF.

For the row space, as a basis we can select the first three rows in the RREF.

can you see why?

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  • $\begingroup$ So it doesn't matter even if it is inconsistent. We only need to check for the column with pivotal element right? $\endgroup$ – user572932 Jul 31 '18 at 21:48
  • $\begingroup$ @philip You are making confusion with the system $Ax=b$ when in the RREF we have a zero row and a value $\neq 0$ in the corresponding position for the vector $b'$. $\endgroup$ – gimusi Jul 31 '18 at 21:50
  • $\begingroup$ @philip Here we are working directly on the matrix to find by RREF the independent columns and rows. Pivot positions are the key to understand what it is going on. We see that colmns 1,2 and 4 and the first 3 rows are are linearly independent. $\endgroup$ – gimusi Jul 31 '18 at 21:57
  • $\begingroup$ @philip Yes by pivotal elements we find the independent rows and columns and by the RREF we can easily solve $Ax=0$ which has always solution and then we can find also the basis of the nullspace. $\endgroup$ – gimusi Jul 31 '18 at 22:17

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