Question 16
Consider this system of equations that corresponds to three planes.
Use a Gaussian technique to determine the value of $\lambda$ for which this system of equations has infinitely many solutions.
[4 marks] b)Use the result from Question 16a) to determine the infinitely many solutions. Express your answer in the form of a vector equation of a line.
[3 marks]The left side of row 3 cancels out
The row operations clear the whole left side of row 3. Then the planes are seen end-on, and plane 3 slides as $\lambda$ changes until all three share a line.
In the end-on view each plane looks like a line. The gaps between the planes are stretched vertically so you can see them, and the orange dot is the common line seen end-on.
Slide the third plane until all three share a line
Change $\lambda$ to move plane 3. Then slide the point $P$ along the line where planes 1 and 2 meet, and watch which equations it satisfies. Drag the picture to turn it, or use the button to look straight along the planes.
The small grey arrows show the directions of the axes. The orange line appears only when the three planes share a line.
This is the question’s value. The bottom row reads $0=0$, so one variable is free. Wherever you slide $P$ it satisfies all three equations, so the three planes share one line and every point on it is a solution.
The bottom row reads , which can never be true, so there are no solutions. Each pair of planes still meets in a line, but the three lines are parallel and never meet. The planes form a triangular prism.
The left side of equation 3 is the left side of equation 1 minus the left side of equation 2. So at every point where planes 1 and 2 meet, $8y-3z=1-3=-2$. Plane 3 contains that line only when $\lambda=-2$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow Method 1 of the QCAA sample response.
Before you can build a matrix, each equation needs $x$, $y$ and $z$ on the left in that order, with the constant on the right. Change the sign of every term that crosses the equals sign.
$$\begin{aligned}x+5y-2z&=1\\x-3y+z&=3\\8y-3z&=\lambda\end{aligned}$$
This mark can be implied by your later working, so a correct augmented matrix earns it too.
Marker: correctly rearranges the three equations
The coefficients of $x$, $y$ and $z$ go in the first three columns, with a $0$ for the missing $x$ in equation 3. The constants go behind the partition line.
$$\left[\begin{array}{ccc|c}1&5&-2&1\\1&-3&1&3\\0&8&-3&\lambda\end{array}\right]$$
If you rearranged an equation incorrectly, you can still earn this mark and the later ones by working correctly from your own matrix.
Marker: establishes an augmented matrix
Clear the $1$ under the leading entry of row 1. Then add the new row 2 to row 3.
$R_2'=R_2-R_1$
$$\left[\begin{array}{ccc|c}1&5&-2&1\\0&-8&3&2\\0&8&-3&\lambda\end{array}\right]$$
$R_3'=R_3+R_2$
$$\left[\begin{array}{ccc|c}1&5&-2&1\\0&-8&3&2\\0&0&0&\lambda+2\end{array}\right]$$
The whole left side of row 3 has become zeros. After the first operation, row 2 and row 3 have opposite signs on the left. Spotting that and using it to reach $\lambda=-2$ is also accepted. This mark can be implied by your later working.
Marker: establishes a row of zeros in the row containing $\lambda$
The bottom row says $0x+0y+0z=\lambda+2$. If $\lambda+2$ is not zero, the row says $0$ equals a non-zero number, so there are no solutions. For infinitely many solutions the whole row must be zeros.
$$\lambda+2=0$$
Marker: determines a value of $\lambda$
With $\lambda=-2$ the bottom row is all zeros, so one variable is free. Let $z=k$, where $k\in\mathbb{R}$, and solve row 2 for $y$.
$$\left[\begin{array}{ccc|c}1&5&-2&1\\0&-8&3&2\\0&0&0&0\end{array}\right]$$
$$\begin{aligned}-8y+3k&=2\\8y-3k&=-2\\y&=\frac{3k-2}{8}\end{aligned}$$
An equivalent parameter is accepted, such as $z=t$. Stay away from $\lambda$, because it already has a job in this question. Equivalent expressions are accepted, such as $y=\frac{3k}{8}-\frac14$.
Marker: expresses $y$ in terms of a parameter
Substitute $z=k$ and your $y$ into row 1, which says $x+5y-2z=1$.
$$\begin{aligned}&x+5\left(\frac{3k-2}{8}\right)-2k=1\\&x+\frac{15k-10}{8}-\frac{16k}{8}=1\\&x=\frac{k}{8}+\frac94\end{aligned}$$
Equivalent expressions are accepted, such as $x=\frac{k+18}{8}$.
Marker: expresses $x$ in terms of a parameter
Stack $x$, $y$ and $z$ in a column and split each entry into the part without $k$ and the part with $k$. The parts without $k$ give a point on the line, and the coefficients of $k$ give its direction.
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\dfrac94+\dfrac18k\\[6pt]-\dfrac14+\dfrac38k\\[6pt]0+k\end{pmatrix}$$
where $k\in\mathbb{R}$.
Equivalent answers are accepted. Any multiple of the direction gives the same line, so these are fine too, with $l\in\mathbb{R}$:
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\3\\8\end{pmatrix}l+\frac14\begin{pmatrix}9\\-1\\0\end{pmatrix}$$
$$\begin{aligned}\boldsymbol{r}&=\left(\frac94\hat{\boldsymbol{\imath}}-\frac14\hat{\boldsymbol{\jmath}}\right)\\&\quad+k\left(\frac18\hat{\boldsymbol{\imath}}+\frac38\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}\right)\end{aligned}$$
Unsimplified fractions are accepted too.
Marker: determines the infinite solutions expressed in the form of a vector equation of a line
The marking guide gives two other routes for part b), and each earns the same three marks. In Method 2 you keep reducing: divide row 2 by $-8$, then use $R_1-5R_2$ to clear the $5$, and read $x$ and $y$ straight off the matrix. In Method 3 you go back to the matrix before the row operations, with $\lambda=-2$, and let $y=k$ instead. That gives $z=\frac{8k}{3}+\frac23$ and $x=\frac{k}{3}+\frac73$, so
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\dfrac73\\[6pt]0\\[6pt]\dfrac23\end{pmatrix}+k\begin{pmatrix}\dfrac13\\[6pt]1\\[6pt]\dfrac83\end{pmatrix}$$
It looks different, but it is the same line. Its direction is a multiple of $(1,3,8)$, and the point $\left(\frac73,0,\frac23\right)$ is on the Method 1 line, where $k=\frac23$.
Each phrase of the question gave you something
Nothing in the question tells you which row operations to use or warns you that a row is about to disappear. You tidy the equations, build the matrix, choose the operations and decide what a row of zeros means, then carry $\lambda=-2$ into part b) and turn a free variable into a vector equation of a line. Joining those steps yourself is what takes it past simple familiar. It is still complex familiar because the wording names both routines you have practised: a Gaussian technique and a vector equation of a line.
If I had to guess where you might lose marks, it would be the rearranging, before any matrix appears. $x+z=3y+3$ has to become $x-3y+z=3$, and $8y-\lambda=3z$ has to become $8y-3z=\lambda$. One sign slip there and row 3 no longer cancels. The 2022 subject report’s general advice asks teachers to remind students to “take care when simplifying algebraic terms, especially when expanding brackets with negative signs”. In part b) that applies when you expand $5\left(\frac{3k-2}{8}\right)$. The last mark needs the vector form, so finish with a point plus a multiple of a direction after you find $x$ and $y$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2022 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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