On using elementary row operation R 1 - 3 R 2 in the following matrix equation = , we have

On using elementary row operation R 1 - 3 R 2 in the followi — Class 12 Maths Solution

Matrices — Class 12 Maths Solution

exemplar objective MCQ NCERT,Exemp,Q.no.67,Page 62

Question

On using elementary row operation ${R_1} \to {R_1} - 3{R_2}$ in the following matrix equation $\left[ {\begin{array}{llllllllllllllllllll}4&2\\3&3\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&2\\0&3\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}2&0\\1&1\end{array}} \right]$, we have

(a) $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&{ - 7}\\3&3\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&{ - 7}\\0&3\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}2&0\\1&1\end{array}} \right]$ ✓ Correct
(b) $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&{ - 7}\\3&3\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&2\\0&3\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&{ - 3}\\1&1\end{array}} \right]$
(c) $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&{ - 7}\\3&3\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\1&{ - 7}\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}2&0\\1&1\end{array}} \right]$
(d) $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2\\{ - 5}&{ - 7}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\{ - 3}&{ - 3}\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}2&0\\1&1\end{array}} \right]$

Step-by-step Solution

We have, $\left[ {\begin{array}{llllllllllllllllllll}4&2\\3&3\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&2\\0&3\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}2&0\\1&1\end{array}} \right]$

Using elementary row operation ${R_1} \to {R_1} - 3{R_2}$,

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

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.

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Matrices. Curated by Sachin Sharma. Free for all students.