Question-6

Let the reduced row echelon form of a matrix \(A\) be:

\[ \begin{bmatrix}1 & 0 & -1\\ 0 & 1 & 1 \end{bmatrix} \]

The first and the third columns of \(A\) are \(\begin{bmatrix}-1\\ 1 \end{bmatrix}\) and \(\begin{bmatrix}2\\ -1 \end{bmatrix}\) respectively. Find the second column of \(A\) and thereby the complete matrix.

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Let the matrix \(A\) be:

\[ \begin{bmatrix}-1 & a & 2\\ 1 & b & -1 \end{bmatrix} \]

We can start row reduction:

\[ \begin{aligned} \begin{bmatrix}-1 & a & 2\\ 1 & b & -1 \end{bmatrix} & \rightarrow\begin{bmatrix}-1 & a & 2\\ 0 & a+b & 1 \end{bmatrix} \end{aligned} \]

\[ \begin{aligned} \begin{bmatrix}-1 & a & 2\\ 0 & a+b & 1 \end{bmatrix} & \rightarrow\begin{bmatrix}1 & -a & -2\\ 0 & a+b & 1 \end{bmatrix} \end{aligned} \]

At this stage, we have to decide if we can divide the second row by \(a+b\). This can be done if \(a+b\neq0\). This is the case as \(a+b=0\) would mean that the second column is no longer a pivot column.

\[ \begin{bmatrix}1 & -a & -2\\ 0 & a+b & 1 \end{bmatrix}\rightarrow\begin{bmatrix}1 & -a & -2\\ 0 & 1 & \cfrac{1}{a+b} \end{bmatrix} \]

If \(a=0\), the matrix becomes:

\[ \begin{bmatrix}1 & 0 & -2\\ 0 & 1 & \cfrac{1}{b} \end{bmatrix} \]

This is in RREF, but no matter what value we choose for \(b\), we can never make it equal to the RREF given in the question. So \(a\neq0\) and we can proceed with elimination:

\[ \begin{bmatrix}1 & -a & -2\\ 0 & 1 & \cfrac{1}{a+b} \end{bmatrix}\rightarrow\begin{bmatrix}1 & 0 & \cfrac{-a-2b}{a+b}\\ \\0 & 1 & \cfrac{1}{a+b} \end{bmatrix} \]

We can now do a direct comparison:

\[ \begin{aligned} a+b & =1\\ -a-2b & =-1 \end{aligned} \]

From this, we get \(a=1,b=0\). Thus the matrix \(A\) is:

\[ \begin{bmatrix}1 & 1 & 2\\ 1 & 0 & -1 \end{bmatrix} \]