# System of linear equations and column space of matrix:

If I have a system of equations: $$a_1x+b_1y+c_1z = d_1$$ $$a_2x+b_2y+c_2z = d_2$$ $$a_3x+b_3y+c_3z = d_3$$ where the coefficients $$a_i, b_i, c_i$$ and constants $$d_i$$ are real, then I know that a zero determinant on the coefficient matrix tells us that we have an inconsistent system, and there are either infinitely many solutions or zero solutions. I read that if the column vector $$(d_1,d_2,d_3)^T$$ lies outside of the "column space" of our coefficient matrix then we have zero solutions (makes sense as the matrix cannot map to this vector). My question is, e.g., for the matrix $$\begin{bmatrix}5 & 2&3\\1 & 2&1\\3&2&2\end{bmatrix}$$ how can we check if it maps to e.g. $$\begin{bmatrix}4\\2\\1\end{bmatrix}$$ using this column space property? (i.e. what are the actual steps involved?) Also, can anyone clarify what exactly is meant by the column space?

• $\det(A)=0$ need not imply that $[A|b]$ is inconsistent. – Matthew Pilling Jan 14 at 18:50
• Also, an inconsistent system does not have infinitely many solutions, it has none. – neuroguy123 Jan 14 at 19:05
• @neuroguy123 ,you're right, I always get mixed up between consistency and whether or not it has a single distinct solution – Poo2uhaha Jan 14 at 19:06
• Oh I know. In fact I made a mistake in my comment. $det(A) \ne 0$ tells us that a system is consistent. The corollary does not necessary tell us about inconsistency, as @MatthewPilling pointed out, just that the columns are not linearly independent and this is for square matrices. – neuroguy123 Jan 14 at 19:12

The column space of a matrix is just the set of all linear combinations of the columns of $$A$$. It's the subspace spanned by those columns.
So, if $$d^T$$ is contained in this subspace, then the system is consistent for that specific $$d^T$$, otherwise it is not.
All of this is just another way of asking if row reducing the augmented matrix $$[a^T b^T c^T d^T]$$ is consistent. It sounds like maybe you have not got to subspaces yet, but they are not that difficult of a concept if you already understand spans.
• What confuses me is that when you carry out matrix multiplication on a vector i.e. $\mathbf{Ax}$, we take each element of our vector and multiply it by a corresponding element in the first row of $\mathbf{A}$ instead of the first column, so why is it the space spanned by the columns instead of the rows (a diagram, in particular, may help me understand)? – Poo2uhaha Jan 14 at 19:09
• I can respond in more detail later, but think of $x$ as the weights of the columns of $A$. This is generally how it is taught in a first course as well. – neuroguy123 Jan 14 at 19:15