Hey guys,
Sorry if this was the wrong area to post this, but I was given the cambridge senior maths chapter 18 (matrices) because my methods tutor (i'm in year 11) said it would give us a bigger overall picture on matrices. I'm having a little bit of trouble with exercise 18C. I know how to do the basic multiplying, but don't understand a few of the trickier questions, and I'd be super grateful if you guys could assist me with the following!!
Question 2: http://puu.sh/A9Yqb/283fa45c3e.png
Question 4 and Question 5: http://puu.sh/A9YrO/0bea0789de.png
Question 12: http://puu.sh/A9Yts/f21438a27d.png
Even if you guys can help me with even one of the questions it'd help a lot!!
Cheers!
Question 2 is testing your understanding of the matrix multiplication process, that is:
A product of 2 matrices is defined only if the number of columns in the first matrix is equal to the number of rows in the second matrix.
The matrices have the following dimensions:
\(\mathbf{X}:\ 2\times1,\ \mathbf{Y}:\ 2\times1,\ \mathbf{A}:\ 2\times2,\ \mathbf{B}:\ 2\times2,\ \mathbf{C}:\ 2\times2,\ \mathbf{I}:\ 2\times2\).
Note that the dimensions are listed as "number of rows"\(\times\)"number of columns"
Therefore the following matrix products are defined (of the ones listed):
\(\mathbf{A}\mathbf{Y},\ \mathbf{C}\mathbf{I}\).
For question 4, consider the matrix product \(\mathbf{AB}\) for two arbitrary \(2\times2\) matrices:
If \(\mathbf{AB}=\mathbf{O}\) then \(ae+bg=0,\ af+bh=0,\ ce+dg=0,\ cf+dh=0\).
All we need to determine is if there is any possible combination of these such that not all of \(a,\ b,\ c\text{ and }d\) are 0 and not all of \(e,\ f,\ g\text{ and }h\) are 0.
We have \(ae=-bg\Rightarrow\frac{a}{b}=-\frac{g}{e}\) and \(af=-bh\Rightarrow\frac{a}{b}=-\frac{h}{f}\). Therefore, \(\frac{g}{e}=\frac{h}{f}\).
We also have \(ce=-dg\Rightarrow\frac{c}{d}=-\frac{g}{e}\) and \(cf=-dh\Rightarrow\frac{c}{d}=-\frac{h}{f}\).
Combining all four of these gives us \(\frac{a}{b}=\frac{c}{d}=-\frac{g}{e}=-\frac{h}{f}\).
If we can find an example where this is true, for which none of \(a,\ b,\ c,\ d,\ e,\ f\text{ and }g\) are 0, then we have disproven the statement.
Consider the simple case where \(a=1,\ b=1,\ c=1,\ d=1,\ g=1,\ e=-1,\ h=1\text{ and }f=-1\). In this case, we get:
Thus, the statement is disproved (i.e. the null factor law is not true for matrices - although the product \(\mathbf{OX}=\mathbf{O}\), this doesn't necessarily mean that the product \(\mathbf{AB}=\mathbf{O}\) implies that \(\mathbf{A}=\mathbf{O}\) or \(\mathbf{B}=\mathbf{O}\) (or both).
Note that the way question 5 is asked gives a big hint as to the answer to question 4.
Firstly, if \(\mathbf{A}^2\) is defined, then \(\mathbf{A}\) must be square. Also, if the dimensions of \(\mathbf{A}\) are \(m\times m\), then the dimensions of \(\mathbf{A}^2\) and therefore \(\mathbf{O}\) are also \(m\times m\).
If \(\mathbf{A}\times\mathbf{A}=\mathbf{O}\), then \(\mathbf{A}^{-1}\mathbf{A}\mathbf{A}=\mathbf{A}^{-1}\mathbf{O}\), which gives \(\mathbf{I}\mathbf{A}=\mathbf{A}^{-1}\mathbf{O}\), finally giving \(\mathbf{A}=\mathbf{A}^{-1}\mathbf{O}\). Note that this seems to indicate that \(\mathbf{A}=\mathbf{O}\), however, this is ignoring the case where the inverse doesn't exist (i.e. when \(\text{det}\left(\mathbf{A}\right)=0\).)
Let's consider the \(2\times2\) case, to keep the arithmetic as simple as possible.
Now we have \(a^2+bc=0,\ ab+bd=0,\ ac+cd=0,\ bc+d^2=0\text{ and }ad=bc\), which gives \(a^2=-bc,\ ab=-bd,\ ac=-cd,\ bc=-d^2\text{ and }ad=bc\).
Consider the simple case where \(c=0\). This gives: \(a^2=-b\times0,\ ab=-bd,\ a\times0=-0\times d,\ b\times0=-d^2\text{ and }ad=b\times0\), which gives \(a^2=0,\ ab=-bd,\ 0=0,\ 0=-d^2\text{ and }ad=0\), finally giving \(a=0, c=0, d=0\). This will be true for ANY real value of \(b\), so simply choose a non-zero \(b\) and you have a matrix \(mathbf{A}\) for which this property is true (note that there are other solutions to)
For question 12a, note that the top row of the \(2\times2\) matrix contains the time, in minutes, that it takes John to drink a milkshake and eat a banana split, respectively and its bottom row contains the cost of a milkshake and a banana split respectively.
The matrix product gives us
The intermediate multiplication step gives us an insight into what the product represents. We have, in the top row, the time it takes John to drink a milkshake multiplied by 1, added to the time it takes John to eat a banana split multiplied by 2 and, in the bottom row, the cost of a milkshake multiplied 1, plus the cost of a banana split multiplied by 2.
Hence, the product tells us the time it would take John to drink a milkshake and eat 2 banana splits (row 1) and the cost of one milkshake and two banana splits (row 2).
The interpretation for part b, follows from that of part a:
Assuming that the speed that John's friends eat and drink at is exactly the same as John, we can interpret the information in this matrix as follows:
The time it takes John to drink a milkshake and eat 2 banana splits (row 1, column 1)
The cost of John's milkshake and 2 banana splits (row 2, column 1)
The time it takes one of John's friends to drink 2 milkshakes and eat a banana split (row 1, column 2)
The cost of John's first friend's banana split and 2 milkshakes (row 2, column 2)
The time it takes the second of John's friends to eat a banana split (row 1, column 3)
The cost of John's second friend's banana split (row 2, column 3)
Given that we weren't actually given any information about how fast John's friends eat and drink, you may interpret this matrix slightly differently (i.e. maybe John ordered different things on different days), but the information about his friends indicates that this is the logical way to answer it.
An additional note: The knowledge of these matrix ideas is well beyond what is required, even at Unit 3&4 Methods. A basic understanding of how matrices are multiplied is sufficient for what can be assessed on the VCAA exams.