Let B be a \(4 \times 3\) matrix in reduced echelon form.

a) If B has three nonzero rows, then determine the form of B. According to Fig. 1.5 of Section 1.2,since the matrix is in reduced echelon form with three nonzero rows, then one of the four rows must: contain all zero entries. Hence, the form of B is

\(B=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{bmatrix}\)

b)Suppose that a system of 4 linear equations in 2 unknowns has augmented matrix A, where A is a \(4 \times 3\) matrix row equivalent to B.

Demonstrate that the system of equations is inconsistent.

Let the system of four linear in 2 unknowns \(x_1\ and\ x_2\), then

\(a_{11}x_1+a_{12}x_2=a\)

\(a_{21}x_1+a_{22}x_2=b\)

\(a_{31}x_1+a_{32}x_2=c\)

\(a_{41}x_1+a_{42}x_2=d\)

The augmented matrix of the system is

\(A=\begin{bmatrix}a_{11}&a_{12}&a\\ a_{21}&a_{22}&b\\ a_{31}&a_{32}&c\\ a_{41}&a_{42}&d\end{bmatrix}\)

Since, A is a \(4 \times 3\) matrix row equivalent to B then is of the form \(A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{bmatrix}\)

Since, the system has 2 unknowns, then from the third reduced echelon form \(A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{bmatrix}\) the equation obtained is 0 = 1 which is not possible.

Hence , the system has not solution.