# 2018 Nov Paper 1 - Q4

I’m going to assume, like A-level:

• A matrix method is expected here
• People will have calculators that can invert and multiply matrices

Write the system as:

\begin{bmatrix}2 & 4 & -1 \\ 1 & 2 & a \\ 5 & 12 & 0\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}10 \\ 5 \\ 2a\end{bmatrix}

(a)

The system of equations does not have a unique solution when:

\begin{align*}\begin{vmatrix}2 & 4 & -1 \\ 1 & 2 & a \\ 5 & 12 & 0\end{vmatrix} & = 2(2 \times 0 - 12a) - 4(1 \times 0 - 5a) - 1(12 - 10) \\ & = -24a + 20a - 2 \\ & = -4a - 2 \\ & = 0\end{align*}

satisfied iff a = -\dfrac 1 2.

(b)

When a = 2, the system has a unique solution.

We have:

\displaystyle \begin{bmatrix}2 & 4 & -1 \\ 1 & 2 & 2 \\ 5 & 12 & 0\end{bmatrix}^{-1} = \frac 1 {10} \begin{bmatrix}24 & 12 & -10 \\ -10 & -5 & 5 \\ -2 & 4 & 0\end{bmatrix}

So:

\displaystyle \begin{align*}\begin{bmatrix}x \\ y \\ z\end{bmatrix} &= \frac 1 {10} \begin{bmatrix}24 & 12 & -10 \\ -10 & -5 & 5 \\ -2 & 4 & 0\end{bmatrix} \begin{bmatrix}10 \\ 5 \\ 4\end{bmatrix} \\ & = \begin{bmatrix}26 \\ -\frac {21} 2 \\ 0 \end{bmatrix}\end{align*}

So x = 26, y = -\dfrac {21} 2, z = 0 is the solution to the system.

2 Likes

(To everyone doing IB)

For Part (a) we can just multiply the 2nd equation by 2 to get:

2x + 4y + 2az = 10

Hence subtracting this equation from the 1st:
-1 - 2a = 0
a = -1/2

For Part (b) we can use row echelon form or row reduction and we get the answer above

Ah that’s interesting.

So no determinants and inversion is done by row reduction? Will keep this in mind for future questions.

Yes in the IB syllabus we don’t teach matrices except for questions like this
Also this is Paper 1 so no calculators allowed