Rewrite the following system of linear equations as an augmented matrix.
\[\begin{aligned} a - b + 2c - d &= -1 \\ 2a + b - 2c - 2d &= -2 \\ -a + 2b - 4c + d &= 1 \\ -3c &= -3 \\ \end{aligned}\]\[\left( \begin{array}{cccc|c} 1 & -1 & 2 & -1 & -1 \\ 2 & 1 & -2 & -2 & -2 \\ -1 & 2 & -4 & 1 & 1 \\ 0 & 0 & -3 & 0 & -3 \end{array} \right)\]
Using Gauss-Jordan elimination, find a reduced row echelon matrix which is row equivalent to your answer in part (a). You have to show your work here for full credit: Be sure to show me the intermediate matrices you get and which elementary row operations you use at each step.
\[\begin{aligned} \left( \begin{array}{cccc|c} 1 & -1 & 2 & -1 & -1 \\ 2 & 1 & -2 & -2 & -2 \\ -1 & 2 & -4 & 1 & 1 \\ 0 & 0 & -3 & 0 & -3 \end{array} \right) \\ \stackrel{(2)=(2)-2(1) \\ (3)=(3)+(1) \\ (4) = -\frac{1}{3}(4)}{\sim} \left( \begin{array}{cccc|c} 1 & -1 & 2 & -1 & -1 \\ 0 & 3 & -6 & 0 & 0 \\ 0 & 1 & -2 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \end{array} \right) \\ \stackrel{(2) = \frac{1}{3}(2)}{\sim} \left( \begin{array}{cccc|c} 1 & -1 & 2 & -1 & -1 \\ 0 & 1 & -2 & 0 & 0 \\ 0 & 1 & -2 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \end{array} \right) \\ \stackrel{(3) = (3)-(2)}{\sim} \left( \begin{array}{cccc|c} 1 & -1 & 2 & -1 & -1 \\ 0 & 1 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \end{array} \right) \\ \stackrel{(3) \leftrightarrow (4)}{\sim} \left( \begin{array}{cccc|c} 1 & -1 & 2 & -1 & -1 \\ 0 & 1 & -2 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right) \\ \stackrel{(2) = (2)+2(3) \\ (1) = (1) - 2(3)}{\sim} \left( \begin{array}{cccc|c} 1 & -1 & 0 & -1 & -3 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right) \\ \stackrel{(1) = (1) +(2)}{\sim} \left( \begin{array}{cccc|c} 1 & 0 & 0 & -1 & -1 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right) \end{aligned}\]
Using your solution to part (b), describe the solution to the system of linear equations from part (a).
As there is no pivot in the column for \(d\), that means \(d\) is free. The remaining three columns have pivots, whose equations correspond to,
\[\begin{aligned} a-d &= -1 \\ b = 2 \\ c = 1 \end{aligned}\]Solving these equations for the pivot variables yields a canonical solution \(a = d-1\), \(b=2\), and \(c=1\) (with \(d\) free).
For parts (a) – (d), determine whether the system of linear equations each augmented matrix represents has one unique solution, infinitely many solutions, or no solution.
The third row is inconsistent, so no solution.
There is a pivot in every column, so one solution.
The second column is missing a pivot, so infinitely many solutions.
There is a pivot in every column, and all rows are consistent (this is a homogeneous system), so one solution, 0.
Suppose you want to find \(a\), \(b\), and \(c\) so that the parabola \(y = ax^2 + bx + c\) is guaranteed to pass through the points \((1,2)\), \((2,-3)\), and \((-1,2)\). Write down (but do not solve) a system of linear equations whose solutions will give values for \(a\), \(b\), and \(c\).
Plugging in each \((x,y)\) point to the parabola equation yields the three linear equations,
\[\begin{aligned} 2 &= a + b + c \\ -3 &= 4a + 2b + c \\ 2 &= a - b + c. \end{aligned}\]
How many solutions does your system in part (a) have? Why? It might help to think about what you know about parabolas.
None of the points given are vertically stacked, so there is at least one parabola through these points. Furthermore, as there is exactly one parabola which passes through any three points, there is exactly one solution to the system.
\[\begin{pmatrix} 5 \\ -4 \\ 0 \\ 9 \\ 7 \\ 11 \end{pmatrix}\]
\[\begin{pmatrix} -2 \\ 0 \\ -4 \\ -8 \\ -12 \end{pmatrix}\]
Compute the matrix product \(AB\) of the matrices \(A\) and \(B\) below.
\[A = \begin{pmatrix} 1 & 2 & -3 \\ -2 & 3 & 1 \end{pmatrix}\] \[B = \begin{pmatrix} 1 & -1 \\ 3 & -4 \\ 4 & -2 \end{pmatrix}\]\[AB = \begin{pmatrix} -5 & -3 \\ 11 & -12 \end{pmatrix}\]