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Calculate these matrix products:
1. $\begin{pmatrix}3&-3\\2&0\\1&4\end{pmatrix} \begin{pmatrix}2&-1&5\\0&3&-2\end{pmatrix}$
Tanish from Pate's Grammar School in the UK, Ci Hui from Queensland Academy for Science Mathematics and Technology (QASMT) in Australia, Mizuki and Yuhan from St George's British International School, Rome in Italy and Beren from the United Kingdom correctly found this matrix product. This is Ci Hui's work:
2. $\begin{pmatrix}2&-1\\3&5\end{pmatrix}\begin{pmatrix}5&-3\\-1&0\end{pmatrix}$
Tanish, Mizuki, Yuhan, Ci Hui and Beren correctly found this matrix product. Here is Beren's work:
3. Let ${\bf P}=\begin{pmatrix}2&3&-1\end{pmatrix}$ and let ${\bf Q}=\begin{pmatrix}-1\\0\\5\end{pmatrix}$. Find the products ${\bf P}{\bf Q}$ and ${\bf Q}{\bf P}$.
Tanish, Mizuki, Yuhan, Beren and Ci Hui worked out both products. Here is Mizuki's work:
4. Let ${\bf A} = \begin{pmatrix}3&-1&0\\-2&5&1\end{pmatrix}$ and ${\bf B}=\begin{pmatrix}3&2 \\ 0 & -1\end{pmatrix}$. State which of the products ${\bf A}{\bf B}$ and ${\bf B}{\bf A}$ can be calculated and find this product.
Tanish, Mizuki, Yuhan, Beren and Ci Hui all concluded that only one product could be calculated. This is Tanish's work:
For the rest of this question we will consider square $2 \times 2$ matrices, so matrices of the form $\begin{pmatrix}a&b\\c&d\end{pmatrix}$. Here are some questions to think about:
1. With numbers it does not matter which order we use to multiply, so $2 \times 3 = 3 \times 2$. Is the same true of matrices, i.e. do we have ${\bf A}{\bf B} = {\bf B}{\bf A}$ in all cases?
Beren explained why this may not be true of matrices:
This is not true. With matrices the answer will change as the multiplication is different.
Ci Hui used an example to show that this is not true for all matrices:
Mizuki and Yuhan also found examples to show that it is not true for all matrices. Mizuki and Yuhan's examples are of square matrices. This is Yuhan's work, which also includes an example of two matrices $\bf A$ and $\bf B$ where ${\bf A}{\bf B}$ is equal to ${\bf B}{\bf A}.$
Tanish used algebra to show that, in general, ${\bf A}{\bf B} \neq {\bf B}{\bf A}:$
2. If ${\bf A}{\bf B}={\bf 0}$, must we have at least one of ${\bf A}$ or ${\bf B}$ equal to ${\bf 0}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$?
Beren explained how ${\bf A}{\bf B}$ could be ${\bf 0}$ even if ${\bf A}$ and ${\bf B}$ are not equal to ${\bf 0}:
This is not true because we are multiplying them so each position could be perhaps $1-1$ which will total to $0.$
Mizuki and Ci Hui both found pairs of matrices where neither matrix is the zero matrix, but the product is the zero matrix. This is Mizuki's example:
This is Ci Hui's example:
Tanish used algebra to describe two families of matrices where ${\bf A}$ and ${\bf B} are not ${\bf 0},$ but ${\bf A}{\bf B}={\bf 0}:$
3. Consider the matrix ${\bf M}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. What is the matrix ${\bf M}^{2023}$?
Tanish, Yuhan, Ci Hui and Mizuki worked this out by calculating the first few powers of ${\bf M}.$ This is Yuhan's work:
4. The $2 \times 2$ matrix ${\bf P}$ satisfies ${\bf P}{\bf X} = {\bf X}{\bf P}$ for all possible $2 \times 2$ matrices ${\bf X}$. Find the possible matrices that ${\bf P}$ could be.
Mizuki found one matrix that ${\bf P}$ could be:
Ci Hui and Tanish noticed that ${\bf P}$ could be the identity matrix, and Tanish found a family of matrices that ${\bf P}$ could be:
In fact, there are even more matrices that satisfy ${\bf P}{\bf X} = {\bf X}{\bf P}$ for all possible $2 \times 2$ matrices ${\bf X}.$ Can you find any more?