Question-52
Consider \(\displaystyle A=\begin{bmatrix} 1 & 3 & 2\\ a & 6 & 2\\ 0 & 9 & 5 \end{bmatrix}\).
For what value of \(\displaystyle a\) will a row interchange be required during Gaussian elimination?
\(2\)
The first row needn’t budge. We already have a zero in the last entry of the first column. To get a zero in the second element of the first column, we have to perform \(\displaystyle R_{2}\rightarrow R_{2} -aR_{1}\), which results in:
\[ \begin{equation*} \begin{bmatrix} 1 & 3 & 2\\ 0 & 6-3a & 2-2a\\ 0 & 9 & 5 \end{bmatrix} \end{equation*} \]
A row interchange would be required if the second row is of the form \(\displaystyle \begin{bmatrix} 0 & 0 & \# \end{bmatrix}\), which happens when \(\displaystyle 6-3a=0\). From this, we get \(\displaystyle a=2\).