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[h] IB Mathematics AI HL Flashcards- Definition of a matrix
[q] Matrices
This may be one of the first truly new areas of maths you have encountered so far in this course, but they aren’t as daunting as they may look. There is no great secret to their meaning — they are simply just grids/‘arrays’ of numbers that have all sorts of meanings:
For example, it could represent:
– the price per kg of carrots and leeks in Albania and Greece:
\[
\text{Albania} \quad \text{Greece} \\
\begin{matrix}
\text{Carrots} & 1.64 & 2.13 \\
\text{Leeks} & 1.42 & 1.27
\end{matrix}
\]
→ instead of
\[
\begin{pmatrix} 1.64 & 2.13 \\ 1.42 & 1.27 \end{pmatrix}
\]
[q]
This is a 2×2 matrix.
It could also represent:
– the probability of the weather:
\[
\begin{pmatrix} 0.6 & \text{Sun} \\ 0.4 & \text{Rain} \end{pmatrix}
\]
and so on. More on these in 4.19 and 3.15.
However, most of the matrices you’ll see in unit will just be presented abstractly, as you learn how to use them.
[a]
TERMINOLOGY — If you take the following matrix, we say that it has 2 rows and 3 columns, which gives us an order of 2×3. In general, we say a matrix has m rows (the horizontal), and n columns (vertical), for an order of m×n. A specific element of the matrix is simply one of the numbers in the array and is denoted with \(a_{ij}\); the element in the \(i^{th}\) row & \(j^{th}\) col. For example, from above, \(a_{2,1} = 7\) or \(a_{2,1} = 1\). When we have 2×2, 3×3, etc., it is called a square matrix.
[q]
ADD/SUBTRACT — To add or subtract matrices, we simply add or subtract the elements that are in the same position:
\[
\text{E.G. 1} \quad A + B = \begin{pmatrix} 1 & 3 \\ 7 & 2 \\ 0 \end{pmatrix} + \begin{pmatrix} 4 & 0 \\ 0 & 7 \\ -1 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & 7 \\ -1 & 4 \end{pmatrix} + A – B
\]
[a]
EQUALITY — For 2 matrices to be equal, all elements must be equal.
DETERMINANT — As ad-bc keeps cropping up, we give it a name: the determinant
ADD/SUBTRACT — To add or subtract matrices, we simply add or subtract the elements that are in the same position:
– E.G. 1 Calculate \( A + B \), if:
\[
A = \begin{pmatrix} 1 & 3 \\ 7 & 2 \\ 0 & -3 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 4 & 0 \\ 0 & 7 \\ -1 & 4 \end{pmatrix}
\]
Then:
\[
A + B = \begin{pmatrix} 1+4 & 3+0 \\ 7+0 & 2+7 \\ 0-1 & -3+4 \end{pmatrix} = \begin{pmatrix} 5 & 3 \\ 7 & 9 \\ -1 & 1 \end{pmatrix}
\]
[q]
MATRIX MULTIPLICATION — This is where matrices start to get a little complicated. We don’t just multiply elements in a similar way to addition. Instead, we multiply a whole row by a column to get each final element. You need an example:
– E.G. 5 Take A from above, and D:
\[
D = \begin{pmatrix} 2 & 3 \\ 6 & 0 \\ 1 & -4 \end{pmatrix}
\]
and find \( AD \):
\[
AD = \begin{pmatrix} 1 & 3 \\ 7 & 2 \\ 0 & -3 \end{pmatrix} \times \begin{pmatrix} 2 & 3 \\ 6 & 0 \\ 1 & -4 \end{pmatrix} = \begin{pmatrix} (1 \times 2) + (3 \times 6) & (1 \times 3) + (3 \times 0) \\ (7 \times 2) + (2 \times 6) & (7 \times 3) + (2 \times -4) \\ (0 \times 2) + (-3 \times 6) & (0 \times 3) + (-3 \times 0) \end{pmatrix} = \begin{pmatrix} 20 & 3 \\ 54 & 9 \\ 63 & -8 \end{pmatrix}
\]
– Note: Here we see that we can get the rule: \( AB \neq BA \), and it may not even be the same order.
[a]
PROPERTIES — We also have:
\[
(A+B) = (A+C)
\]
INVERSE — For regular numbers, we say
For the process, Continuing the transcription:
E.G. 6: Calculate \( DA \):
If:
\[
D = \begin{pmatrix} 2 & 3 \\ 6 & 0 \\ 1 & -4 \end{pmatrix}, \quad A = \begin{pmatrix} 1 & 3 \\ 7 & 2 \\ 0 & -3 \end{pmatrix}
\]
then:
\[
DA = \begin{pmatrix} 2 & 3 \\ 6 & 0 \\ 1 & -4 \end{pmatrix} \times \begin{pmatrix} 1 & 3 \\ 7 & 2 \\ 0 & -3 \end{pmatrix} = \begin{pmatrix} (2 \times 1) + (3 \times 7) & (2 \times 3) + (3 \times 2) \\ (6 \times 1) + (0 \times 7) & (6 \times 3) + (0 \times 2) \\ (1 \times 1) + (-4 \times 7) & (1 \times 3) + (-4 \times 2) \end{pmatrix} = \begin{pmatrix} 23 & 12 \\ 6 & 18 \\ -27 & -8 \end{pmatrix}
\]
> Note here how a 3×2 multiplied by a 2×3 gives us a 3×3.
[q]
PROPERTIES — We also have:
– \((AB)C = A(BC)\) (Associative property)
– \(A(B + C) = AB + AC\) (Distributive property)
– \(AI = IA = A\) where \(I\) is the identity matrix \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) for 2×2 matrices.
[a]
INVERSE — For regular numbers, we say \(\frac{1}{3}\) is the inverse of 3 because to solve \(3x = 12\), we do \(\frac{12}{3} = 4\). Now we need an inverse, \(A^{-1}\), for matrices. The equivalent of “1” for matrices is the identity, \(I\). So, we need to find \(A^{-1}\) that gives \(AA^{-1} = I\). If we call \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) and \(A^{-1} = \begin{pmatrix} w & x \\ y & z \end{pmatrix}\), then we can figure it out:
\[
AA^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \times \begin{pmatrix} w & x \\ y & z \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
[q]
This gives us:
\[
A^{-1} = \frac{1}{\text{detA}} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
\]
DETERMINANT — As ad – bc keeps cropping up, we give it a name: the determinant. So,
\[
\text{det} A = ad – bc
\]
– E.G. 7 If \( A = \begin{pmatrix} 2 & 1 \\ 4 & 7 \end{pmatrix} \), find \( A^{-1} \):
\[
A^{-1} = \frac{1}{(2 \times 7) – (1 \times 4)} \begin{pmatrix} 7 & -1 \\ -4 & 2 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} 7 & -1 \\ -4 & 2 \end{pmatrix} = \begin{pmatrix} 0.7 & -0.1 \\ -0.4 & 0.2 \end{pmatrix}
\]
[a]
GDC — For 3×3 matrices and beyond, you’ll use a GDC (Graphing Display Calculator). Just create the matrix and raise it to the power of -1 to find the inverse.
– TI-nspire:
[MENU] → (7: MATRIX) → (1: CREATE) → (1: MATRIX) → Enter no. of rows & cols → ENTER or use the \(\boxed{B}\) button.
– TI-84:
[2ND] → [x⁻¹] → EDIT → [A] → Enter no. of rows & cols → [2ND] → [MODE] → [2ND] → [x⁻¹] → Names → Select [A].
Then the final step on both calculators is just raising them to the power of -1, to find the inverse.
[q]
SYSTEMS OF EQUATIONS
If you briefly consider the following matrix multiplication:
\[
\begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10 \\ 16 \end{pmatrix}
\]
then this looks very familiar — like a system of equations. So we can exploit this:
If you have:
\[
2x + 3y = 10 \\
4x + 5y = 16
\]
it can be rewritten as:
\[
\begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10 \\ 16 \end{pmatrix}
\]
[a]
In itself, that isn’t particularly helpful. But then we realize that we solved something similar in E.G. 8, using inverses.
So we get \( x = 1, y = 4 \), without a traditional method. And it extends beyond 2×2 matrices.
– E.G. 10 Solve the following system using matrices:
\[
3x – y + z = -1 \\
x + 3y – 2z = 5 \\
5x + 4y + z = 5
\]
Matrix form:
\[
\begin{pmatrix} 3 & -1 & 1 \\ 1 & 3 & -2 \\ 5 & 4 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 \\ 5 \\ 5 \end{pmatrix}
\]
Solve using GDC (graphing calculator).
> Note: There is a separate method, by row reduction, but it isn’t needed as well.
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