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  • On using elementary row operation R1→ R1 — 3R2 in the following matrix equation: [4 2 3 3 ] = [1 2 0 3 ]

On using elementary row operation R1→ R1 — 3R2 in the following matrix equation:

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

\begin{array}{l} \text { A. }\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & -7 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right] \\ \\B.{\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} -1 & -3 \\ 1 & 1 \end{array}\right]} \\ \\C.{\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ 1 & -7 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right]} \\\\ D.{\left[\begin{array}{cc} 4 & 2 \\ -5 & -7 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ -3 & -3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right]} \end{array}

Answers (1)

Elementary row transformation is applied on the first matrix of RHS.

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

By Applying R1→ R1 — 3R2 we get -

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

\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & -7 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right] \\

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

∴ Option (A) is the correct answer.

Posted by

infoexpert22

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