The following augmented matrices correspond to a linear system which has already been reduced by Gaussian elimination.
For each part, describe all of the solutions to the system described by the augmented matrix. (You may express your answer in any way that you feel comfortable (as \(n\)-tuples, vectors, variable assignments…), but if you use variable names, make sure to let me know what they mean).
\[\left( \begin{array}{ccc|c} 1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 7 \end{array} \right)\]
\[\left( \begin{array}{ccc|c} 1 & 0 & 0 & -3 \\ 0 & 1 & -3 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right)\]
\[\left( \begin{array}{ccccc|c} 1 & -6 & 0 & 0 & 3 & -2 \\ 0 & 0 & 1 & 0 & 4 & 7 \\ 0 & 0 & 0 & 1 & 5 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right)\]
\[\left( \begin{array}{ccc|c} 1 & 0 & 0 & -3 \\ 0 & 1 & -3 & 2 \\ 0 & 0 & 0 & 3 \end{array} \right)\]
There are many different ways to express the solutions. We’ll use vectors here.
- \[\begin{pmatrix}-3 \\ 0 \\ 7\end{pmatrix}\]
- \[\begin{pmatrix}-3 \\ 2+3s \\ s\end{pmatrix}, s \in \mathbb{R}\]
- \[\begin{pmatrix}-2+6s-3t\\s\\7-4t\\-5t\\t\end{pmatrix}, s \in \mathbb R, t \in \mathbb R\]
- No solutions. The last row makes the system inconsistent.
Put the following linear system into an augmented matrix, reduce it using Gaussian elimination, and express the solution.
\[\begin{aligned} 2a + 2b + 2c &= 0 \\ -2a + 5b + 2c &= 1 \\ 8a + b + 4c &= -1 \end{aligned}\]The system becomes the matrix,
\[\begin{pmatrix} 2 & 2 & 2 & 0 \\ -2 & 5 & 2 & 1 \\ 8 & 1 & 4 & -1 \end{pmatrix}\]Which has RREF (REF is ok too for this kind of problem),
\[\begin{pmatrix} 1 & 0 & 3/7 & -1/7 \\ 0 & 1 & 4/7 & 1/7 \\ 0 & 0 & 0 & 0 \end{pmatrix}\]So the solutions would be,
\[\begin{pmatrix}-1/7-3/7s \\ 1/7-4/7s \\ s\end{pmatrix}, s \in \mathbb R\]
In each of the following parts, each asterisk \(*\) corresponds to any real number (they might all be different). If possible, answer whether the system represented by the augmented matrix has (I) no solutions, (II) one unique solution, (III) infinitely many solutions, or that there is (IV) not enough information to tell.
\[\left( \begin{array}{ccc|c} 1 & * & * & * \\ 0 & 1 & * & 0 \\ 0 & 0 & 1 & * \end{array} \right)\]
\[\left( \begin{array}{ccc|c} 1 & * & * & * \\ 0 & 1 & * & * \\ 0 & 0 & 0 & * \end{array} \right)\]
\[\left( \begin{array}{ccc|c} 1 & * & * & * \\ 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 \end{array} \right)\]
\[\left( \begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & * & * & * \end{array} \right)\]
(II). The matrix is in RREF and consistent, and there are as many pivots as columns on the LHS.
(IV). The (*) in the lower right could be a 0 (solutions) or not (no solutions) and we don’t know.
(III). The matrix is in RREF and consistent, and there are 2 pivots and 3 columns. So, there is 1 free variable and there are infinitely many solutions.
(I). The first two rows are inconsistent, saying \(x=0\) and \(x=1\). This is impossible, as 1 and 0 can’t be the same number.
Under what conditions on the real numbers \(a\) and \(b\) will the following linear system have no solutions, one solution, and infinitely many solutions?
\[\begin{aligned} 2x - 3y &= a \\ 4x - 6y &= b \end{aligned}\]In REF, the matrix for the solution above is,
\[\begin{pmatrix} 2 & -3 & a \\ 0 & 0 & b-2a \end{pmatrix}\]If \(b-2a \ne 0\), the system isn’t consistent and there are no solutions. If \(b-2a=0\) then the system is consistent, and there is 1 pivot and 2 variables, so there is 1 free variable and thus infinitely many solutions.
The equation for a parabola is \(ax^2 + bx + c = y\). Plug each of the three points in \(xy\)-coordinates \((0, 3)\), \((1,2)\), and \((2, 4)\) into the parabola equation one at a time to obtain a linear system of three equations.
As an augmented matrix, the system is,
\[\begin{pmatrix} 0 & 0 & 1 & 3 \\ 1 & 1 & 1 & 2 \\ 4 & 2 & 1 & 4 \end{pmatrix}\]
How many solutions are there to your system of linear equations from part (a)? You may use what you know about parabolas to answer this question.
For any three points which pass the vertical line test, there is exactly one parabola which pass through those three points. So, there will be exactly one system to the solution. Here it is.
Another way to discover this is by solving your system above. You’ll find one answer.