Inconsistent Matrices
I'm teaching myself Linear Algebra and am not sure how to approach this problem:
Let A be a 4×4 matrix, and let b and c be two vectors in R4. We are told that the system Ax = b is inconsistent. What can you say about the number of solutions of the system Ax = c?
Bretscher, Otto (2013-02-21). Linear Algebra with Applications (2-Download) (5th Edition) (Page 35). Pearson HE, Inc.. Kindle Edition.
I start with a generic 4x4 matrix, a vector x, and another generic vector b:
$$ \left[ \begin{array}{cccc} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{array} \right] \left[ \begin{array}{c} x_{0} \\ x_{1} \\ x_{2} \\ x_{3} \end{array} \right] = \left[ \begin{array}{c} b_{0} \\ b_{1} \\ b_{2} \\ b_{3} \end{array} \right] $$
And the definition of inconsistent in terms of systems of equations. In the case of A, A would be inconsistent if one of its rows looked like this:
$$ \left[ \begin{array}{cccc} 0 & 0 & 0 & q \end{array} \right] $$
Where q does not equal zero.
My first hangup is--what does it even mean for this equation to be inconsistent? Given the above definition of inconsistent, only matrices with more than one column can be inconsistent. If this is the case, then A must be inconsistent, b must be undefined, and therefore c must be undefined as well.
But the solutions at the end of the book say Ax = c has infinitely many solutions or no solutions.
What am I missing?
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$\begingroup$Perhaps things will go faster with a simpler example. Consider the inconsistent equations $$ \begin{align} x + y &= 1 \\ x + y &= 0 \end{align} $$ The linear system is $$ \begin{align} % \mathbf{A} x &= b\\ % \left[ \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right] \left[ \begin{array}{c} x \\ y \\ \end{array} \right] &= \left[ \begin{array}{r} 1 \\ 0 \\ \end{array} \right] % \end{align} $$ There are no exact solutions for this problem.
But you are asking about $$ \begin{align} % \mathbf{A} x &= c\\ % \left[ \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right] \left[ \begin{array}{c} x \\ y \\ \end{array} \right] &= \left[ \begin{array}{r} c_{1} \\ c_{2} \\ \end{array} \right] % \end{align} $$
A solution exists when $$ \left[ \begin{array}{r} c_{1} \\ c_{2} \\ \end{array} \right] = \alpha \left[ \begin{array}{r} 1 \\ 1 \\ \end{array} \right] $$ Obivous solutions are $$ \left[ \begin{array}{c} x \\ y \\ \end{array} \right] = \alpha \left[ \begin{array}{c} 1 \\ 0 \\ \end{array} \right] % \qquad % \left[ \begin{array}{c} x \\ y \\ \end{array} \right] = \alpha \left[ \begin{array}{c} 0 \\ 1 \\ \end{array} \right] % \qquad % \left[ \begin{array}{c} x \\ y \\ \end{array} \right] = \frac{\alpha}{2} \left[ \begin{array}{c} 1 \\ 1 \\ \end{array} \right] % $$ Linear systems either have no solution (no existence), a unique solution (existence and uniqueness), or an infinitude of solutions (existence, no uniqueness). We must have a infinite number of solutions. They are $$ \left[ \begin{array}{c} x \\ y \\ \end{array} \right] = \left[ \begin{array}{c} x \\ \alpha - x \\ \end{array} \right] $$ This is the set $$ (x,y)_{soln} = \left\{ (x,y) \in \mathbb{C}^{2} \colon y = \alpha - x \right\} $$
Summary: The inconsistent system will either present no solutions or an infinite number of solutions. $\endgroup$ $\begingroup$
The definition you quote is for a system of equations to be inconsistent. At the end you are talking about individual matrices being inconsistent. You can't make that leap. The reduced row echelon form of the augmented matrix could have a row that looks like the row you display. In this case the system of equations (not the matrix) would be inconsistent. The column matrices are not undefined but the system of equations represented by the matrix equation has no solution. Now if the last row of the reduced row echelon form of the augmented matrix, with the vector $c$ in place of the $b$, is of the same type you showed: $[0 \, 0 \, 0 \, q]$ with $q$ not zero, then there would be no solution. If it turned out to be $[0 \, 0 \, 0 \, 0]$ there would be infinitely many solutions.
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