IBDP Maths AHL 1.16 Solutions of systems of linear equations AA HL Paper 2- Exam Style Questions- New Syllabus

(a)
Use \(z\) as the free variable.

From the second equation:

\(3x-y+3z=12\)

\(y=3x+3z-12\)

Substitute this expression into the first equation.

\(2x+6(3x+3z-12)-8z=13\)

\(2x+18x+18z-72-8z=13\)

\(20x+10z=85\)

\(x=\dfrac{17}{4}-\dfrac{z}{2}\)

Substitute this into \(y=3x+3z-12\).

\(y=3\left(\dfrac{17}{4}-\dfrac{z}{2}\right)+3z-12\)

\(y=\dfrac{3}{4}+\dfrac{3z}{2}\)

Let \(z=t\), where \(t\in\mathbb{R}\). Therefore:

\(x=\dfrac{17}{4}-\dfrac{t}{2},\quad y=\dfrac{3}{4}+\dfrac{3t}{2},\quad z=t\)

✅ Answer: \((x,y,z)=\left(\dfrac{17}{4}-\dfrac{t}{2},\dfrac{3}{4}+\dfrac{3t}{2},t\right)\), where \(t\in\mathbb{R}\)

(b)
For \(a=3\) and \(k=-3\), the third equation is:

\(3x+12y-16z=-3\)

Substitute the parametric expressions from part (a).

\(3\left(\dfrac{17}{4}-\dfrac{t}{2}\right)+12\left(\dfrac{3}{4}+\dfrac{3t}{2}\right)-16t=-3\)

\(\dfrac{51}{4}-\dfrac{3t}{2}+9+18t-16t=-3\)

\(\dfrac{87}{4}+\dfrac{t}{2}=-3\)

\(\dfrac{t}{2}=-\dfrac{99}{4}\)

\(t=-\dfrac{99}{2}\)

Therefore:

\(x=\dfrac{17}{4}-\dfrac{1}{2}\left(-\dfrac{99}{2}\right)=29\)

\(y=\dfrac{3}{4}+\dfrac{3}{2}\left(-\dfrac{99}{2}\right)=-\dfrac{147}{2}\)

\(z=-\dfrac{99}{2}\)

✅ Answer: \((x,y,z)=\left(29,-\dfrac{147}{2},-\dfrac{99}{2}\right)\)

(c)(i)
Substitute the general solution into the third equation.

\(a\left(\dfrac{17}{4}-\dfrac{t}{2}\right)+12\left(\dfrac{3}{4}+\dfrac{3t}{2}\right)-16t=k\)

\(\dfrac{17a}{4}+9+\left(2-\dfrac{a}{2}\right)t=k\)

There is no unique solution when the coefficient of \(t\) is zero.

\(2-\dfrac{a}{2}=0\)

\(a=4\)

✅ Answer: \(a=4\)

(c)(ii)
When \(a=4\), the equation becomes:

\(\dfrac{17(4)}{4}+9=k\)

\(17+9=k\)

\(k=26\)

When \(k=26\), the third equation is satisfied for every value of \(t\), so the system has infinitely many solutions. If \(k\ne26\), the system would have no solution.

✅ Answer: \(k=26\)