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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES1 Mark
Question
Choose the correct answer from the given four options.

On using elementary row operation R1 → R1 – 3R2 in the following matrix equation $\begin{bmatrix}4&2\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix},$ we have:

  1. $\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$

  2. $\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}-1&-3\\1&1\end{bmatrix}$

  3. $\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\1&-7\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$

  4. $\begin{bmatrix}4&2\\-5&-7\end{bmatrix}=\begin{bmatrix}1&2\\-3&-3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$

✓

Answer

  1. $\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$

​Solution:​​​​​​

We have, $\begin{bmatrix}4&2\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$

Using elementary row operation R1 → R1 - 3R2

$\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$

Since, on using elementary row operation on X = AB, we apply these operation simultaneously on X and on the first matrix A of the product AB on RHS.

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