Skip to main content

Standard Checks

Print preview

Paper size
Additional printing options
Print headers
Print footers

Handout 1.1 Key

Section 1.1.1 Linear Systems

Subsection 1.1.1.1 Solving linear systems: be able to set up a system, solve using row reduction (by hand), and state and use the result.

Checkpoint 1.1.1.
Find all solutions to this system of equations.
\begin{equation*} \begin{array}{rrrcr} x & +y & +8z & = & 42. \\ x & +2y & +10z & = & 53. \\ 7x & +2y & +47z & = & 244. \end{array} \end{equation*}
🔗
Solution.
\begin{align*} \begin{bmatrix} 1 & 1 & 8 & 42 \\ 1 & 2 & 10 & 53 \\ 7 & 2 & 47 & 244 \\ \end{bmatrix} & \sim & \begin{array}{l} \\ R_2 \leftarrow -1R_1+R_2 \\ R_3 \leftarrow -7R_1+R_3 \\ \end{array}\\ \begin{bmatrix} 1 & 1 & 8 & 42 \\ 0 & 1 & 2 & 11 \\ 0 & -5 & -9 & -50 \\ \end{bmatrix} & \sim & \begin{array}{l} \\ \\ R_3 \leftarrow 5R_2+R_3 \\ \end{array}\\ \begin{bmatrix} 1 & 1 & 8 & 42 \\ 0 & 1 & 2 & 11 \\ 0 & 0 & 1 & 5 \\ \end{bmatrix} & \sim & \begin{array}{l} R_1 \leftarrow -1R_2+R_1 \\ \\ \\ \end{array}\\ \begin{bmatrix} 1 & 0 & 6 & 31 \\ 0 & 1 & 2 & 11 \\ 0 & 0 & 1 & 5 \\ \end{bmatrix} & \sim & \begin{array}{l} R_1 \leftarrow -6R_3+R_1 \\ R_2 \leftarrow -2R_3+R_2 \\ \\ \end{array}\\ \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 5 \\ \end{bmatrix} \end{align*}
Thus the solutions are \(\begin{bmatrix} 1 \\ 1 \\ 5 \end{bmatrix}\)
🔗
🔗
🔗
Checkpoint 1.1.2.
Find all solutions to this system of equations.
\begin{equation*} \begin{array}{rrrcr} x & -7y & -23z & = & -5. \\ 5x & -34y & -112z & = & -24. \\ 6x & -40y & -132z & = & -28. \end{array} \end{equation*}
🔗
Solution.
\begin{align*} \begin{bmatrix} 1 & -7 & -23 & -5 \\ 5 & -34 & -112 & -24 \\ 6 & -40 & -132 & -28 \\ \end{bmatrix} & \sim & \begin{array}{l} \\ R_2 \leftarrow -5R_1+R_2 \\ R_3 \leftarrow -6R_1+R_3 \\ \end{array}\\ \begin{bmatrix} 1 & -7 & -23 & -5 \\ 0 & 1 & 3 & 1 \\ 0 & 2 & 6 & 2 \\ \end{bmatrix} & \sim & \begin{array}{l} \\ \\ R_3 \leftarrow -2R_2+R_3 \\ \end{array}\\ \begin{bmatrix} 1 & -7 & -23 & -5 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} & \sim & \begin{array}{l} R_1 \leftarrow 7R_2+R_1 \\ \\ \\ \end{array}\\ \begin{bmatrix} 1 & 0 & -2 & 2 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \end{align*}
The solutions are \(\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix} x_3\)
🔗
🔗
🔗
Checkpoint 1.1.3.
Find all solutions to this system of equations.
\begin{equation*} \begin{array}{rrrcr} x & +2y & +10z & = & 23. \\ 2x & +3y & +13z & = & 37. \\ 2x & +y & +-4z & = & 12. \end{array} \end{equation*}
🔗
Solution.
\begin{align*} \begin{bmatrix} 1 & 2 & 10 & 23 \\ 2 & 3 & 13 & 37 \\ 2 & 1 & -4 & 12 \\ \end{bmatrix} & \sim & \begin{array}{l} \\ R_2 \leftarrow -2R_1+R_2 \\ R_3 \leftarrow -2R_1+R_3 \\ \end{array}\\ \begin{bmatrix} 1 & 2 & 10 & 23 \\ 0 & -1 & -7 & -9 \\ 0 & -3 & -24 & -34 \\ \end{bmatrix} & \sim & \begin{array}{l} \\ \\ R_3 \leftarrow -3R_2+R_3 \\ \end{array}\\ \begin{bmatrix} 1 & 2 & 10 & 23 \\ 0 & -1 & -7 & -9 \\ 0 & 0 & -3 & -7 \\ \end{bmatrix} & \sim & \begin{array}{l} R_1 \leftarrow 2R_2+R_1 \\ R_2 \leftarrow -1 R_2 \\ \\ \end{array}\\ \begin{bmatrix} 1 & 0 & -4 & 5 \\ 0 & 1 & 7 & 9 \\ 0 & 0 & -3 & -7 \\ \end{bmatrix} & \sim & \begin{array}{l} R_1 \leftarrow -\frac{4}{3}R_3+R_1 \\ R_2 \leftarrow \frac{7}{3}R_3+R_2 \\ R_3 \leftarrow -\frac{1}{3} R_3 \\ \end{array}\\ \begin{bmatrix} 1 & 0 & 0 & \frac{43}{3} \\ 0 & 1 & 0 & -\frac{22}{3} \\ 0 & 0 & 1 & \frac{7}{3} \\ \end{bmatrix} \end{align*}
Thus the solutions are \(\begin{bmatrix} \frac{43}{3} \\ -\frac{22}{3} \\ \frac{7}{3} \end{bmatrix}\)
🔗
🔗
🔗
🔗

Subsection 1.1.1.2 Homogeneous solutions: be able to use the solution to a homogeneous solution to solve non-homogeneous problems

Checkpoint 1.1.4.
Solutions to a homogeneous system are given by \(\begin{bmatrix} 2 \\ 4 \\ 7 \end{bmatrix} x_3 \text{.}\) One solution to the related non-homogeneous system is \(\begin{bmatrix} 2 \\ 0 \\ -6 \end{bmatrix} \text{.}\) Write the general form for solutions to the non-homogeneous system.
🔗
Solution.
\begin{equation*} \begin{bmatrix} 2 \\ 4 \\ 7 \end{bmatrix} x_3 + \begin{bmatrix} 2 \\ 0 \\ -6 \end{bmatrix} \end{equation*}
🔗
🔗
🔗
Checkpoint 1.1.5.
The solutions to the following system are: \(x=9-y+4z\) with \(y\) and \(z\) free. Write the general form of the solutions to the associated homogeneous system.
\begin{equation*} \begin{array}{rrrcr} x & +y & -4z & = & 9. \\ -7x & -7y & +28z & = & -63. \\ -3x & -3y & +12z & = & -27. \end{array} \end{equation*}
🔗
Solution.
\begin{equation*} \begin{bmatrix} 9 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}y + \begin{bmatrix} 4 \\ 0 \\ 1 \end{bmatrix}z \end{equation*}
🔗
🔗
🔗
🔗
🔗

Section 1.1.2 Linear Combinations

Subsection 1.1.2.1 Linear combinations and spans: be able to write a linear combination of an arbitrary set of vectors; be able to determine if something is in the span of a set of vectors

These can fulfill the \(\R^n\)/\(\C^n\) portion of this standard.
🔗
Checkpoint 1.1.1.
Write a vector that is in the span of \(\vec{b}_1 = [3,0,9]\) and \(\vec{b}_2 = [1,2,6]\text{.}\)
🔗
Solution.
\begin{equation*} 1[3,0,9]+2[1,2,6] = [5,4,21] \end{equation*}
🔗
🔗
🔗
Checkpoint 1.1.2.
Is \(\vec{c} = [30,-6,81]\) in the span of \(\vec{b}_1 = [3,0,9]\) and \(\vec{b}_2 = [1,2,6]\text{?}\)
🔗
Solution.
If it is in the span there is a solution to the following system
\begin{equation*} [30,-6,81] = a_1 [3,0,9] + a_2 [1,2,6]\text{.} \end{equation*}
We can solve this by row reducing
\begin{equation*} \begin{bmatrix} 3 & 1 & 30 \\ 0 & 2 & -6 \\ 9 & 6 & 81 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 11 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \end{bmatrix} \end{equation*}
Thus it is in the span, because
\begin{equation*} [30,-6,81] = 11 [3,0,9] -3 [1,2,6] \end{equation*}
🔗
🔗
🔗
Checkpoint 1.1.3.
Is \(\vec{d} = [1,4,2]\) in the span of \(\vec{b}_1 = [3,0,9]\text{,}\) \(\vec{b}_2 = [1,2,6]\text{,}\) and \(\vec{b}_3 = [-1,10,12]\text{?}\)
🔗
Solution.
If it is in the span there is a solution to the following system
\begin{equation*} [1,4,2] = a_1 [3,0,9] + a_2 [1,2,6] + a_3 [-1,10,12]\text{.} \end{equation*}
We can solve this by row reducing
\begin{equation*} \begin{bmatrix} 3 & 1 & -1 & 1 \\ 0 & 2 & 10 & 4 \\ 9 & 6 & 12 & 2 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & 5 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{equation*}
Thus it is not in the span, because there is no solution.
🔗
🔗
🔗
🔗
🔗
🔗
PrevTopNext