Answer
a. True
b. False
c. True
d. False
e. False
f. False
g. True
h. True
i. True
j. False
k. True
l. False
m. False
n. True
o. False
p. True
Work Step by Step
a. $AB^T$ will be an m by m matrix and $BA^T$ will be an n by n matrix.
b. B would have 2 columns.
c. If $a_{ii}$ is the entry in row i and column i of matrix A, the elements of row i in matrix B will be multiplied by $a_{ii}$
d. This is only true if B is invertible, which is not always the case
e. The product of 2 nonzero matrices can be the 0 matrix.
Consider the product of $\begin{bmatrix}
1&0\\
0&0\\
\end{bmatrix}$ and $\begin{bmatrix}
0&0\\
0&1\\
\end{bmatrix}$
f. AB=BA is not always true, so -AB+BA is not always the zero matrix
g. An elementary matrix is one that can be obtained by performing 1 elementary row operation on the identity matrix. This can be to scale a row or interchange two rows in which case there will be n nonzero entries or add a multiple of one row to another in which case there will be n+1 nonzero entries.
h. The transpose of an elementary matrix is an elementary matrix.
i. An elementary matrix is formed by performing row operations on the identity matrix, which is square.
j. Not every square matrix is invertible, in which case it would be row equivalent to the identity matrix
k. A has a pivot in each column and row so is invertible. Every invertible matrix is row reducible to the identity matrix, which can be viewed as left multiplying by elementary matrices.
l. A and B need to be square matrices
m. They must be in reverse order on the right hand side
n. $AB=BA$
$A^{-1}AB=A^{-1}BA$
$B=A^{-1}BA$
$BA^{-1}=A^{-1}BAA^{-1}$
$BA^{-1}=A^{-1}B$
o. $(rA)^{-1}=1/rA^{-1}$
p. If the equation has a unique solution, the A must be invertible by the IMT.
