The following matrix can be converted to a Identity matrix using elementary row transformations $⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 2 2 & 5 & 5 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$

True

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False

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Solution

The correct option is B
False

$R_{1} \to R_{1} + R_{3}$ $⎡ ⎢ ⎣ \begin{matrix} 1 - 1 & 2 + 1 & 2 + 1 2 & 5 & 5 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$

$⎡ ⎢ ⎣ \begin{matrix} 0 & 3 & 3 2 & 5 & 5 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$

$R_{2} \to R_{2} - 2 R_{3}$ $⎡ ⎢ ⎣ \begin{matrix} 0 & 3 & 3 2 - 2 & 5 + 2 & 5 + 2 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$

$⎡ ⎢ ⎣ \begin{matrix} 0 & 3 & 3 0 & 7 & 7 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$

$R_{1} \to R_{2} - 3 / 7 R_{2}$ $⎡ ⎢ ⎢ ⎣ \begin{matrix} 0 & 3 - \frac{3}{7} (7) & 3 - \frac{3}{7} (7) 1 & 7 & 7 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦$

$⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 0 & 7 & 7 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$

$R_{2} \to R_{2} / 7$ $⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 0 & 1 & 1 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$

$R_{3} \to R_{3} - R_{2}$ $⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 0 & 1 & 1 - 1 & 0 & 0 \end{matrix} ⎤ ⎥ ⎦$

$R_{3} \to (- 1) R_{3} a n d R_{1} \leftrightarrow R_{3}$

$⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎦$

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.

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The following matrix can be converted to a Identity matrix using elementary row transformations $⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 2 2 & 5 & 5 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$

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The following matrix can be converted to a Identity matrix using elementary row transformations ⎡⎢⎣122255−111⎤⎥⎦

The following matrix can be converted to a Identity matrix using elementary row transformations $⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 2 2 & 5 & 5 - 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$