Question-78
Consider the set of vectors given below:
\[ \displaystyle B = \{( x,1,1,1) ,( 1,x,1,1) ,( 1,1,x,1) ,( 1,1,1,x)\} \]
Choose all values of \(\displaystyle x\) for which \(\displaystyle B\) is not a basis for \(\displaystyle \mathbb{R}^{4}\).
We have:
\[ \begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ 1 & 1 & 1 & x \end{bmatrix} \]
If \(\displaystyle x=1\), this is clearly a matrix of rank \(\displaystyle 1\). So let us assume that \(\displaystyle x\neq 1\). We now proceed with row reduction:
Step-1
\[ \begin{equation*} \begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ 1 & 1 & 1 & x \end{bmatrix}\xrightarrow{R_{4}\rightarrow R_{4} +R_{3} +R_{2} +R_{1}}\begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ x+3 & x+3 & x+3 & x+3 \end{bmatrix} \end{equation*} \]
Step-2
If \(x = -3\), the last row is zero. The rank of the matrix is \(3\). This shows that \(B\) cannot be a basis for \(\mathbb{R}^{4}\). Let us now assume that \(x \neq -3\). This allows us to divide the last row by \(x + 3\).
\[ \begin{equation*} \begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ x+3 & x+3 & x+3 & x+3 \end{bmatrix}\xrightarrow{R_{4}\rightarrow \frac{1}{x+3} R_{4}}\begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ 1 & 1 & 1 & 1 \end{bmatrix} \end{equation*} \]
Step-3
We can use the last row to eliminate a lot of entries in the first three rows:
\[ \begin{equation*} \begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ 1 & 1 & 1 & 1 \end{bmatrix}\rightarrow \begin{bmatrix} x-1 & 0 & 0 & 0\\ 0 & x-1 & 0 & 0\\ 0 & 0 & x-1 & 0\\ 1 & 1 & 1 & 1 \end{bmatrix} \end{equation*} \]
Step-4
Since \(\displaystyle x\neq 1\), we can divide by \(\displaystyle x-1\) in each of the first three rows:
\[ \begin{equation*} \begin{bmatrix} x-1 & 0 & 0 & 0\\ 0 & x-1 & 0 & 0\\ 0 & 0 & x-1 & 0\\ 1 & 1 & 1 & 1 \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1 \end{bmatrix} \end{equation*} \]
After a few more operations, we get the identity matrix. This shows that \(\displaystyle A\) has rank \(\displaystyle 4\) when \(\displaystyle x\neq 1\) and \(x \neq -3\). So \(B\) is a basis for \(\mathbb{R}^{4}\) so long as \(x \not \in \{-3, 1\}\).