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  • On using elementary column operations C2arrow C2 arrow 2C1 in the following matrix equation [1 -3 2 4] we have:

On using elementary column operations C2→ C2 — 2C1 in the following matrix equation

\left[\begin{array}{cc} 1 & -3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right] we have:

A.\left[\begin{array}{cc}1 & -5 \\ 0 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]
B. \left[\begin{array}{cc}1 & -5 \\ 0 & 4\end{array}\right]=\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\left[\begin{array}{cc}3 & -5 \\ -0 & 5\end{array}\right]
C.\left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -2 & 4\end{array}\right]
D. \left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]

Answers (1)

\left[\begin{array}{cc} 1 & -3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right]

For column transformation, we operate the post matrix.

As,

\left[\begin{array}{cc} 1 & -3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right]

By Applying C2→ C2 — 2C1,

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

Clearly, it matches with option (D).
Hence we can say that,

∴ Option (D) is the correct answer.

Posted by

infoexpert22

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