Mfrhtyj

Consider the matrix below.

(a) Apply the elementary row operation R2 ← R2 − 3R1.
begin{pmatrix}
1 & 1 \
3 & 2 \
0 & 3
end{pmatrix}
Which I think would become:
begin{pmatrix}
1 & 1 \
0 & -1 \
0 & 3
end{pmatrix}

(b) Express the application of the row operation from part (a) as the multiplication of an elementary matrix with A.

This part I don't know how to do.

Your part (a) looks good.

The elementary matrix corresponding to this will be the identity matrix with a $-3$ in the $(2,1)$ entry since you're adding $-3$ times the first row to the second row. So the elementary matrix is $$E=begin{bmatrix}1&0&0 \ -3&1&0 \ 0&0&1end{bmatrix}$$ Then this corresponds to the row operation in $A$ by multiplying $$EA=begin{bmatrix}1&1 \ 0&-1 \ 0&3end{bmatrix}$$

When you multiply two matrices, the rows of the result are linear combinations of the rows of the right-hand matrix, with the coefficients coming from the corresponding row of the left-hand matrix.

Here, the first and third rows of the given matrix are unchanged, so the elementary matrix will have the form $$begin{bmatrix}1&0&0 \ *&*&* \ 0&0&1end{bmatrix}.$$ The second row of the result is supposed to be $R_2-3R_1$, from which we can read that the second row of the elementary matrix must be $(-3,1,0)$.

One way to see why this is the case is to write the right-hand matrix of the matrix product $AB$ in block form, so that it looks like you’re multiplying a vector by a matrix: $$Abegin{bmatrix}B_1\B_2\vdots\B_nend{bmatrix}.$$ The $i$th row of the product is then formally just the dot product of the $i$th row of $A$ with this vector: $a_{i1}B_1+a_{i2}B_2+dots+a_{in}B_n$, a linear combination of the rows of $B$.