What operation does \(A^T\) denote?

What is the size of the identity matrix \(I_4\)?

Can a \(2\times3\) matrix be added to a \(2\times2\) matrix? Explain.

If \(A\) is \(3\times2\) and \(B\) is \(2\times5\), what is the size of \(AB\)?

Compute \(\begin{pmatrix}1&-2\\3&0\end{pmatrix}+\begin{pmatrix}4&5\\-1&7\end{pmatrix}\).

Compute \(-3\begin{pmatrix}2&-1\\0&4\end{pmatrix}\).

Find \(A^T\) for \(A=\begin{pmatrix}1&2&3\\4&5&6\end{pmatrix}\).

For \(A=\begin{pmatrix}1&2\\3&4\end{pmatrix}\) and \(B=\begin{pmatrix}0&5\\6&-1\end{pmatrix}\), compute \(2A-B\).

Compute \(\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}5\\6\end{pmatrix}\).

Compute \(AB\) for \(A=\begin{pmatrix}1&0&2\\-1&3&1\end{pmatrix}\) and \(B=\begin{pmatrix}2&1\\0&4\\5&-2\end{pmatrix}\).

Show with a calculation that matrix multiplication is not commutative for \(A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\), \(B=\begin{pmatrix}1&0\\1&1\end{pmatrix}\).

Explain why \(AI_3=A\) is valid when \(A\) is \(2\times3\), but \(AI_2\) is not defined.

Find all real \(a\) for which \(A=\begin{pmatrix}2&a\\a+1&5\end{pmatrix}\) is symmetric.

Let \(A\) be \(2\times3\), \(B\) be \(3\times2\), and \(C\) be \(2\times2\). Which of \(AB+C\), \(BA+C\), and \(AC\) are defined?

For what values of \(t\) is \(A=\begin{pmatrix}1&t\\3&2\end{pmatrix}\) such that \(A+A^T\) has equal off-diagonal entries?

Find the condition on \(a,b,c\) for \(\begin{pmatrix}a&b\\0&c\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}5\\2\end{pmatrix}\).

For which sizes of \(A\) can both \(AI_4\) and \(I_3A\) be defined? Determine the size of \(A\).

Diagnose the error: a learner computes \(\begin{pmatrix}1&2\end{pmatrix}\begin{pmatrix}3&4\end{pmatrix}=\begin{pmatrix}3&8\end{pmatrix}\) by multiplying entries position by position.

Prove using entry reasoning that \((A+B)^T=A^T+B^T\) for matrices of the same size.

A physics transformation sends \(\begin{pmatrix}x\\y\end{pmatrix}\) to \(\begin{pmatrix}2x-y\\x+3y\end{pmatrix}\). Find its matrix and explain why the order of entries matters.