# Linear Algebra - Linear transformation is inconsistent?

So I have this matrix

$$M = \left[ \begin{array}{ccc} 1&7&9&3\\ 2&15&19&8\\ 7&52&66&27\\ 3&4&10&-24\\ \end{array} \right]$$

and I need to show that the linear transformation is inconsistent. So basically

$$M \left[ \begin{array}{ccc} x\\ y\\ z\\ t\\ \end{array} \right]$$

$$= \left[ \begin{array}{ccc} a\\ b\\ c\\ d\\ \end{array} \right]$$

that means it's inconsistent

this is the reduced form.

$$\left[ \begin{array}{ccc} 1&0&2&0\\ 0&1&1&0\\ 0&0&0&1\\ 0&0&0&0\\ \end{array} \right]$$

So how do I show that the linear transformation is inconsistent. Thanks!

• What exactly does it mean to say that a linear transformation is "inconsistent"? What is the definition? – David Nov 30 '15 at 3:44
• Well it the book, it says Let M be an m x n matrix and if m < n, then for each vector b in R^m that linear system Ax=b is either inconsistent or has infinitely many solution. – Shawn Edward Nov 30 '15 at 3:45
• That is what's meant by a linear system being inconsistent but you are asking about a linear transformation being inconsistent. – David Nov 30 '15 at 3:47
• "Show that the linear transformation Tm: R^4 -> R ^4 by multiplication of column vectors on the left by M is not surjective by exhibiting a column vector not in the image." Thats from the question, So if M[x,y,z,t] = [a.b,c,d] then it inconsistent – Shawn Edward Nov 30 '15 at 3:51
• No ... If $M[x,y,z,t]=[a,b,c,d]$ has no solutions, then it is inconsistent. – Christopher Carl Heckman Nov 30 '15 at 3:57

Reducing $M\left[\begin{array}{c}x\\y\\z\\t\end{array}\right]=\left[\begin{array}{c}a\\b\\c\\d\end{array}\right]$ we get $\left[\begin{array}{cccc|c} 1&7&9&3&a\\ 2&15&19&8&b\\ 7&52&66&27&c\\ 3&4&10&-24&d\\ \end{array}\right]$ $\rightarrow \left[\begin{array}{cccc|c} 1&7&9&3&a\\ 0&1&1&2&b-2a\\ 0&3&3&6&c-7a\\ 0&-17&-17&-33&d-3a\\ \end{array}\right]$ $\rightarrow \left[\begin{array}{cccc|c} 1&7&9&3&a\\ 0&1&1&2&b-2a\\ 0&0&0&0&c-a-3b\\ 0&0&0&1&d-37a+17b\\ \end{array}\right]$.
This is consistent only if $c-a-3b=0$. If we pick any vector where this doesn't hold, e.g. $\left[\begin{array}{c}a\\b\\c\\d\end{array}\right]=\left[\begin{array}{c}0\\0\\1\\0\end{array}\right]$, then that vector is not in the image of left-multiplication by $M$, which (as per your comment) is what your question actually asked you to show.