AI HL · AHL 5
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Systems of equations using the inverse matrix
AHL 5.5
a^-1 · b
Screenshot and video to come
By first forming a matrix equation, solve \(3x + 5y = 4\) and \(x + 2y = 1\).
Have a go first. Then open the working one line at a time.
\(\begin{pmatrix}3 & 5 \\ 1 & 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}4 \\ 1\end{pmatrix}\)
Coefficients, unknowns, right-hand sides.
\(\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2 & -5 \\ -1 & 3\end{pmatrix}\begin{pmatrix}4 \\ 1\end{pmatrix} = \begin{pmatrix}3 \\ -1\end{pmatrix}\)
The determinant is 1.
\(x = 3\), \(y = -1\)
Check in the second equation: \(3 + 2(-1) = 1\).
A bakery sells loaves, rolls and pastries. On Monday it sells 20 loaves, 35 rolls and 12 pastries for €128. On Tuesday it sells 25, 30 and 18 for €156.50. On Wednesday it sells 15, 50 and 20 for €142.50. Find the price of each item.
Have a go first. Then open the working one line at a time.
\(\begin{pmatrix}20 & 35 & 12 \\ 25 & 30 & 18 \\ 15 & 50 & 20\end{pmatrix}\begin{pmatrix}l \\ r \\ p\end{pmatrix} = \begin{pmatrix}128 \\ 156.5 \\ 142.5\end{pmatrix}\)
One row for each day.
a^-1 · b
Store the 3 × 3 matrix as a and the column as b.
Loaf €3.50, roll €0.80, pastry €2.50
Check Monday: \(20(3.5) + 35(0.8) + 12(2.5) = 128\).
Find the values of \(k\) for which the equations \(kx + 2y = 3\) and \(8x + ky = 5\) do not have a unique solution.
Have a go first. Then open the working one line at a time.
\(\begin{pmatrix}k & 2 \\ 8 & k\end{pmatrix}\)
The coefficients of \(x\) and \(y\).
\(k^2 - 16 = 0\)
No inverse means no unique solution.
\(k = 4\) or \(k = -4\)
For these values the two equations describe parallel lines.
Find the values of \(k\) for which the equations \(2x + 3y = 5\) and \(4x + 6y = k\) have (a) infinitely many solutions, (b) no solutions.
Have a go first. Then open the working one line at a time.
\(\det\begin{pmatrix}2 & 3 \\ 4 & 6\end{pmatrix} = 12 - 12 = 0\)
It is zero for every \(k\), so there is never exactly one solution.
\(4x + 6y = 2(2x + 3y)\)
The left of the second equation is twice the left of the first.
\(k = 2(5) = 10\)
When the right-hand side is also doubled, the equations are the same line.
\(k \ne 10\)
The lines are parallel and different, so they never meet.