AI HL · AHL 5
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Determinants
AHL 5.3
Menu > Matrix & Vector > Determinant · det(a)
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Drag A and B. The columns of the matrix are the position vectors of A and B. Say what the determinant will be before you read it.
Select a point (Tab) and use the arrow keys to move it.
Find the determinant of \(\begin{pmatrix}4 & -3 \\ 2 & 5\end{pmatrix}\).
Have a go first. Then open the working one line at a time.
\(4(5) - (-3)(2)\)
Leading diagonal first, then subtract the other diagonal.
\(20 - (-6) = 26\)
Subtracting a negative adds.
Find the determinant of \(\begin{pmatrix}1 & 6 \\ 3 & 2\end{pmatrix}\), and say what its sign tells you.
Have a go first. Then open the working one line at a time.
\(1(2) - 6(3) = -16\)
Leading diagonal, less the other diagonal.
Negative
The second column, \(\begin{pmatrix}6 \\ 2\end{pmatrix}\), is clockwise from the first, \(\begin{pmatrix}1 \\ 3\end{pmatrix}\). The area is 16.
Find the values of \(k\) for which \(\begin{pmatrix}k & 6 \\ 2 & k-1\end{pmatrix}\) has determinant zero.
Have a go first. Then open the working one line at a time.
\(k(k - 1) - 12 = 0\)
The leading diagonal is \(k(k-1)\). The other is \(6 \times 2\).
\(k^2 - k - 12 = 0\), so \((k - 4)(k + 3) = 0\)
By factorising, or on the calculator.
\(k = 4\) or \(k = -3\)
For these values the two columns lie along the same line.
Find the area of triangle \(OAB\) where \(A\) is \((5, 2)\) and \(B\) is \((1, 4)\).
Have a go first. Then open the working one line at a time.
\(\begin{pmatrix}5 & 1 \\ 2 & 4\end{pmatrix}\)
The position vectors of \(A\) and \(B\).
\(5(4) - 1(2) = 18\)
The parallelogram has area 18.
\(\frac{1}{2}(18) = 9\)
The triangle is half the parallelogram.
Find the area of the triangle with vertices \(P(1, 1)\), \(Q(6, 3)\) and \(R(2, 7)\).
Have a go first. Then open the working one line at a time.
\(\overrightarrow{PQ} = \begin{pmatrix}5 \\ 2\end{pmatrix}\), \(\overrightarrow{PR} = \begin{pmatrix}1 \\ 6\end{pmatrix}\)
Both start at \(P\).
\(5(6) - 1(2) = 28\)
The two vectors are the columns.
\(\frac{1}{2}|28| = 14\)
Take the size: an area is never negative.
Find the determinant of \(\begin{pmatrix}1 & 2 & 0 \\ 3 & -1 & 4 \\ 2 & 0 & 5\end{pmatrix}\).
Have a go first. Then open the working one line at a time.
a := the 3 × 3 matrix
Use the matrix template.
det(a) = −19
You do not need to do a 3 × 3 determinant by hand.