IBDP Maths AHL 1.16 Solutions of systems of linear equations AA HL Paper 1- Exam Style Questions- New Syllabus
Consider the following system of equations where \( a, b \in \mathbb{R} \):
\[ \begin{cases} x + 3y + (a – 1)z = 1 \\ 2x + 2y + (a – 2)z = 1 \\ 3x + y + (a – 3)z = b \end{cases} \]a. Show that for any value of \( a \), the planes do not intersect at a unique point. [3]
b. Find the value of \( b \) for which the intersection is a straight line. [4]
▶️ Answer/Explanation
Solution to (a):
METHOD 1 (Determinant Approach)
Calculate the determinant of the coefficient matrix:
\[ \det \begin{pmatrix} 1 & 3 & a-1 \\ 2 & 2 & a-2 \\ 3 & 1 & a-3 \end{pmatrix} = 1(2(a-3)-(a-2)) – 3(2(a-3)-3(a-2)) + (a-1)(2-6) = 0 \]Since the determinant is always 0, there is no unique solution for any \( a \).
[3 marks]
METHOD 2 (Row Operations)
Performing row operations shows the system is always dependent:
\[ \begin{pmatrix} 1 & 3 & a-1 \\ 2 & 2 & a-2 \\ 3 & 1 & a-3 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 3 & a-1 \\ 0 & -4 & -a \\ 0 & 0 & 0 \end{pmatrix} \]Solution to (b):
METHOD 1 (Row Reduction)
For the system to have infinitely many solutions (a line), the augmented matrix must satisfy:
\[ \begin{pmatrix} 1 & 3 & a-1 & | & 1 \\ 0 & -4 & -a & | & -1 \\ 0 & 0 & 0 & | & b-1 \end{pmatrix} \]This requires \( b-1 = 0 \), giving the solution:
\[ \boxed{b = 1} \][4 marks]
METHOD 2 (Consistency Check)
From part (a), we observe the system is consistent only when \( b = 1 \).
Consider the system of equations representing three planes in space:
\[ \begin{cases} x + y + 2z = -2 \\ 3x – y + 14z = 6 \\ x + 2y = -5 \end{cases} \](a) Show that this system has an infinite number of solutions. [2]
(b) Find the parametric equations of the line of intersection of the three planes. [3]
▶️ Answer/Explanation
Solution to (a):
METHOD 1 (Row Reduction)
Performing row operations on the augmented matrix:
\[ \begin{pmatrix} 1 & 1 & 2 & | & -2 \\ 3 & -1 & 14 & | & 6 \\ 1 & 2 & 0 & | & -5 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 1 & 2 & | & -2 \\ 0 & 1 & -2 & | & -3 \\ 0 & 0 & 0 & | & 0 \end{pmatrix} \]The row of zeros indicates infinitely many solutions.
METHOD 2 (Determinant)
The determinant of the coefficient matrix is zero:
\[ \begin{vmatrix} 1 & 1 & 2 \\ 3 & -1 & 14 \\ 1 & 2 & 0 \end{vmatrix} = 0 \]With at least one valid solution, this confirms infinitely many solutions.
[2 marks]
Solution to (b):
METHOD 1 (Parametric Solution)
Let \( y = t \), then:
\[ x = -5 – 2t \] \[ z = \frac{t + 3}{2} \]Thus the parametric equations are:
\[ \begin{cases} x = -5 – 2t \\ y = t \\ z = \frac{3 + t}{2} \end{cases} \]METHOD 2 (Cross Product)
Using the normal vectors of two planes:
\[ \begin{vmatrix} i & j & k \\ 1 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = -4i + 2j + k \]This gives the direction vector \((-4, 2, 1)\), leading to:
\[ \begin{cases} x = 1 – 4t \\ y = -3 + 2t \\ z = t \end{cases} \][3 marks]
Total [5 marks]
The following system of equations represents three planes in space:
\[ \begin{cases} x + 3y + z = -1 \\ x + 2y – 2z = 15 \\ 2x + y – z = 6 \end{cases} \]Find the coordinates of the point of intersection of the three planes.
▶️ Answer/Explanation
Solution:
METHOD 1 (Elimination)
Subtract the first equation from the second:
\[ (x + 2y – 2z) – (x + 3y + z) = 15 – (-1) \\ -y – 3z = 16 \quad \text{(Equation A)} \]Subtract twice the first equation from the third:
\[ (2x + y – z) – 2(x + 3y + z) = 6 – 2(-1) \\ -5y – 3z = 8 \quad \text{(Equation B)} \]Subtract Equation A from Equation B:
\[ (-5y – 3z) – (-y – 3z) = 8 – 16 \\ -4y = -8 \Rightarrow y = 2 \]Substitute y = 2 into Equation A:
\[ -2 – 3z = 16 \Rightarrow z = -6 \]Substitute y = 2 and z = -6 into the first equation:
\[ x + 3(2) + (-6) = -1 \Rightarrow x = -1 \]Thus, the point of intersection is \(\boxed{(-1,\ 2,\ -6)}\).
METHOD 2 (Matrix Row Reduction)
Starting with the augmented matrix:
\[ \begin{pmatrix} 1 & 3 & 1 & | & -1 \\ 1 & 2 & -2 & | & 15 \\ 2 & 1 & -1 & | & 6 \end{pmatrix} \]After row operations, we obtain:
\[ \begin{pmatrix} 1 & 0 & 0 & | & -1 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & -6 \end{pmatrix} \]This confirms the solution \((-1,\ 2,\ -6)\).
[6 marks]
Consider the system of equations:
\[ \begin{cases} x + y + z = 1 \\ 2x + 3y + z = 3 \\ x + 3y – z = \lambda \end{cases} \]where \(\lambda \in \mathbb{R}\).
a. Show that this system does not have a unique solution for any value of \(\lambda\). [4]
b. (i) Determine the value of \(\lambda\) for which the system is consistent. [1]
b. (ii) For this value of \(\lambda\), find the general solution of the system. [4]
▶️ Answer/Explanation
Solution to (a):
Using row operations on the augmented matrix:
\[ \begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 2 & 3 & 1 & | & 3 \\ 1 & 3 & -1 & | & \lambda \end{pmatrix} \]After elimination, we obtain:
\[ y – z = 1 \\ 2y – 2z = \lambda – 1 \]Since the second equation is exactly twice the first, the system is dependent and has no unique solution for any \(\lambda\).
[4 marks]
Solution to b(i):
For consistency, the equations must satisfy the relationship:
\[ 2(y – z) = \lambda – 1 \Rightarrow 2(1) = \lambda – 1 \Rightarrow \lambda = 3 \]Thus, the system is consistent when \(\boxed{\lambda = 3}\).
[1 mark]
Solution to b(ii):
When \(\lambda = 3\), the system has infinitely many solutions. Let \(z = \mu\) (a free parameter), then:
\[ y = 1 + \mu \\ x = -2\mu \]The general solution is:
\[ \begin{cases} x = -2\mu \\ y = 1 + \mu \\ z = \mu \end{cases}, \quad \mu \in \mathbb{R} \][4 marks]