Question 16
Consider the system of linear equations
$$\begin{aligned} x+y+z&=2\\ 2x-y+3z&=1\\ 5x-y+\lambda z&=p \end{aligned}$$
where $\lambda,p\in R$.
Use a Gaussian technique to determine the values of $\lambda$ and $p$ for which the system has infinitely many solutions.
[3 marks] b)Use your result from Question 16a) to determine the solutions of the system. Express your answer in the form of a vector equation of a line.
[2 marks] c)Describe the geometric interpretation of the system when $\lambda=7$ and $p\neq4$.
[1 mark]Three planes and one row of zeros
Row-reduce until row 3 holds only $\lambda$ and $p$. Then watch what the planes do when $\lambda=7$: one shared line when $p=4$, a triangular prism when $p\neq4$.
The planes are drawn to scale. Only plane 3 depends on $p$, so only plane 3 moves. The prism uses $p=10$.
Reduce the matrix, then make row 3 a row of zeros
Apply the row operations yourself. When row 3 holds only $\lambda$ and $p$, choose values that make it a row of zeros. Then slide $k$ along the solutions and check them in all three equations.
Row 1 has a leading 1 in the $x$ column. Use it to clear the $x$ entries in rows 2 and 3.
Row 2 starts with 2, so you subtracted $2R_1$. Its $x$ entry is now 0.
Row 3 starts with 5, so you subtracted $5R_1$. Its $x$ entry is now 0.
The $x$ column is clear below row 1. Now row 3 needs a 0 in the $y$ column. Which multiple of row 2 does that?
Clear the $x$ entries in rows 2 and 3 first. Row 1 does that job.
That row already has a 0 in the $x$ column. Doing it again would undo your work.
The matrix is already in row-echelon form. Use the sliders to choose $\lambda$ and $p$.
That gives $\left[\,0\ \ -3\ \ \lambda-6\ \,|\ \,p-7\,\right]$. The $y$ entry is still $-3$, so row 3 is not in echelon form and a conclusion such as $\lambda=6$, $p=7$ is unsupported. The $y$ entries are $-3$ and $-6$, and $\dfrac{-6}{-3}=2$. Try again with that multiple.
Row 3 is now $\left[\,0\ \ 0\ \ \lambda-7\ \,|\ \,p-4\,\right]$. It says $(\lambda-7)z=p-4$. Now choose $\lambda$ and $p$ to make it a row of zeros.
The $z$ entry of row 3 is , not 0, so row 3 is not a row of zeros. Change $\lambda$.
With $\lambda=7$, row 3 reads . That is impossible, so this system has no solution. This is the case in part c). For infinitely many solutions, change $p$.
Row 3 is a row of zeros, so it says $0=0$. Rows 1 and 2 are two equations in three unknowns, so there are infinitely many solutions. That needs $\lambda=7$ and $p=4$. Now slide $k$.
Every value of $k$ gives a point that satisfies all three equations. The solutions form the line $\boldsymbol{r}=\begin{bmatrix}1\\1\\0\end{bmatrix}+k\begin{bmatrix}-4\\1\\3\end{bmatrix}$.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Write the augmented matrix, then clear the $x$ entries below row 1.
$$\begin{aligned}&\left[\begin{array}{ccc|c}1&1&1&2\\2&-1&3&1\\5&-1&\lambda&p\end{array}\right]\\[6pt]&\left[\begin{array}{ccc|c}1&1&1&2\\0&-3&1&-3\\0&-6&\lambda-5&p-10\end{array}\right]\\&R_2'=R_2-2R_1,\quad R_3'=R_3-5R_1\end{aligned}$$
Row 1 has a leading 1 in the $x$ column. Row 2 starts with 2, so subtract $2R_1$; row 3 starts with 5, so subtract $5R_1$.
Marker: correctly applies row operations to eliminate $x$ from rows 2 and 3
Clear the $y$ entry in row 3 using row 2.
$$\begin{aligned}&\left[\begin{array}{ccc|c}1&1&1&2\\0&-3&1&-3\\0&0&\lambda-7&p-4\end{array}\right]\\&R_3''=R_3'-2R_2'\end{aligned}$$
Infinitely many solutions require row 3 to be a row of zeros.
The $y$ entries in rows 2 and 3 are $-3$ and $-6$. Since $\dfrac{-6}{-3}=2$, subtract $2R_2'$ from $R_3'$. Use row 2, not row 1, so the $x$ entry of row 3 stays 0.
Marker: correctly reduces row 3 to $\left[\,0\ \ 0\ \ \lambda-7\ \,|\ \,p-4\,\right]$
Set both entries of row 3 equal to zero.
$$\lambda-7=0\ \text{ and }\ p-4=0$$
Row 3 then says $0=0$, leaving two equations in three unknowns. One variable is free, so there are infinitely many solutions.
Marker: correctly determines the values of $\lambda$ and $p$
With $\lambda=7$ and $p=4$, let $z=3k$ and back-substitute into rows 2 and 1.
$$\begin{aligned}&z=3k\quad(k\in R)\\\text{Row 2: }&{-3y}+3k=-3\ \Rightarrow\ y=1+k\\\text{Row 1: }&x=2-(1+k)-3k=1-4k\end{aligned}$$
Two equations in three unknowns leave one variable free. Solving row 2 for $y$ divides by $-3$, so $z=3k$ (rather than $z=k$) avoids fractions.
Marker: correctly introduces a parameter and expresses $x$, $y$ and $z$ in terms of it
Separate the constant parts from the parts multiplying $k$.
Any vector equation of the same line is accepted: the point must satisfy all three equations and the direction must be parallel to $-4\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+3\hat{\boldsymbol{k}}$. Follow-through applies in part b) from an incorrect reduction in part a).
Marker: correctly expresses the solutions as a vector equation of a line
Substitute $\lambda=7$ into row 3, then check the normals from the coefficients: no two are scalar multiples, so no two planes are parallel.
$$0x+0y+0z=p-4\neq0$$
In part c), “no solution” alone does not gain the mark.
Marker: correctly describes the geometric interpretation
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- Using the wrong multiple to clear $y$. Using $R_3'-R_2'$ instead of $R_3'-2R_2'$ gives $\left[\,0\ \ -3\ \ \lambda-6\ \,|\ \,p-7\,\right]$. The $y$ entry is still $-3$, so row 3 is not in echelon form and any conclusion such as $\lambda=6$, $p=7$ is unsupported.
- A sign slip on the $\lambda$ entry. Writing $(\lambda-5)+2$ instead of $(\lambda-5)-2(1)$ gives $\lambda-3$ and so $\lambda=3$.
- A right-hand-side slip in $R_3-5R_1$. Writing $p-5$ instead of $p-5(2)=p-10$ leads to $p-5+6=p+1$ and so $p=-1$. A sign slip in $R_2'$ (writing $+3$ instead of $1-2(2)=-3$) gives $p-10-2(3)=p-16$ and so $p=16$.
- A back-substitution sign error in part b). Writing $x=2-y+z$ gives $x=2-(1+k)+3k=1+2k$. Substituting $(1+2k,\ 1+k,\ 3k)$ into equation 2 gives $1+12k$, not 1, so the “line” is not the solution set. Always check your answer in all three original equations.
- An incomplete answer to part c). Writing only “no solution”, or saying “two planes are parallel”, earns no mark. Your answer must state that no two planes are parallel and that the three planes meet pairwise in three parallel lines (a triangular prism).
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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