Question

The following matrix can be converted to a Identity matrix using elementary row transformations $\left[\begin{array}{ccc}1& 2& 2\\ 2& 5& 5\\ -1& 1& 1\end{array}\right]$

• A. True
• B. False

Answer 1

The correct option is B

False

$R_1\to R_1+R_3$ $\left[\begin{array}{ccc}1-1& 2+1& 2+1\\ 2& 5& 5\\ -1& 1& 1\end{array}\right]$

$\left[\begin{array}{ccc}0& 3& 3\\ 2& 5& 5\\ -1& 1& 1\end{array}\right]$

$R_2\to R_2-2R_3$ $\left[\begin{array}{ccc}0& 3& 3\\ 2-2& 5+2& 5+2\\ -1& 1& 1\end{array}\right]$

$\left[\begin{array}{ccc}0& 3& 3\\ 0& 7& 7\\ -1& 1& 1\end{array}\right]$

$R_1\to R_2-3/7R_2$ $\left[\begin{array}{ccc}0& 3-\frac{3}{7}\left(7\right)& 3-\frac{3}{7}\left(7\right)\\ 1& 7& 7\\ -1& 1& 1\end{array}\right]$

$\left[\begin{array}{ccc}0& 0& 0\\ 0& 7& 7\\ -1& 1& 1\end{array}\right]$

$R_2\to R_2/7$ $\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 1\\ -1& 1& 1\end{array}\right]$

$R_3\to R_3-R_2$ $\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 1\\ -1& 0& 0\end{array}\right]$

$R_3\to (-1)R_{3}and{R}_1\leftrightarrow R_3$

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

Now irrespective of what we do we cannot make it a identity matrix. Without disturbing other elements. A more detailed explanation can be understood using rank which we will see ahead.