Solution - Matrix core operations

$$rref\left(\left[\begin{array}{cc}-4& 3\\ 2& 1\end{array}\right]\right)$$

Apply Gauss-Jordan elimination column by column to create pivot columns in RREF form.

$$rref\left(\left[\begin{array}{cc}-4& 3\\ 2& 1\end{array}\right]\right)$$

For each pivot column: normalize the pivot to 1, then eliminate entries above and below that pivot.

$$rref\left(\left[\begin{array}{cc}-4& 3\\ 2& 1\end{array}\right]\right)$$

R1 <- -1/4R1

$$\left[\begin{array}{cc}1& -0.75\\ 2& 1\end{array}\right]$$

R2 <- R2 - 2R1

$$\left[\begin{array}{cc}1& -0.75\\ 0& 2.5\end{array}\right]$$

R2 <- 2/5R2

$$\left[\begin{array}{cc}1& -0.75\\ 0& 1\end{array}\right]$$

R1 <- R1 + 3/4R2

$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

Follow each row operation in order and watch how pivot columns are formed.

$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

The matrix is now in reduced row-echelon form with normalized pivots.

$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

Study tip: inspect each pivot column and confirm it has one leading 1 with zeros elsewhere.

$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

Present the final matrix or scalar outcome in canonical form for stable routing and review.