To solve the problem of finding the appropriate identity matrices [tex]\( I_m \)[/tex] and [tex]\( I_n \)[/tex] such that [tex]\( I_m A = A \)[/tex] and [tex]\( A I_n = A \)[/tex], let's consider the dimensions of the given matrix [tex]\( A \)[/tex].
The matrix [tex]\( A \)[/tex] is given as:
[tex]\[
A = \begin{pmatrix}
6 & 1 & -2 \\
4 & -3 & 4
\end{pmatrix}
\][/tex]
[tex]\( A \)[/tex] is a [tex]\( 2 \times 3 \)[/tex] matrix, meaning it has 2 rows and 3 columns.
### Finding [tex]\( I_m \)[/tex]:
For the product [tex]\( I_m A = A \)[/tex], the identity matrix [tex]\( I_m \)[/tex] must have dimensions such that it can multiply directly with [tex]\( A \)[/tex] on the left. This means that [tex]\( I_m \)[/tex] must be a [tex]\( 2 \times 2 \)[/tex] identity matrix because [tex]\( I_m A \)[/tex] will involve multiplying a [tex]\( 2 \times 2 \)[/tex] matrix with a [tex]\( 2 \times 3 \)[/tex] matrix, resulting in a [tex]\( 2 \times 3 \)[/tex] matrix (the same dimensions as [tex]\( A \)[/tex]).
The [tex]\( 2 \times 2 \)[/tex] identity matrix [tex]\( I_m \)[/tex] is:
[tex]\[
I_m = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\][/tex]
Thus, the appropriate identity matrix [tex]\( I_m \)[/tex] is:
[tex]\[
I_m = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\][/tex]
So, [tex]\( I_m = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)[/tex].
