Matrices — Class 12 Maths Solution

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$X\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&3\\4&5&6\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]$

We can say that X is a 2 × 2 matrix.

Let $X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&b\\c&d\end{array}} \right]$

$\therefore$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&b\\c&d\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&3\\4&5&6\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{a + 4b}&{2a + 5b}&{3a + 6b}\\{c + 4d}&{2c + 5d}&{3c + 6d}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 7}&{ - 8}&{ - 9}\\2&4&6\end{array}} \right]$

$\Rightarrow$ $a + 4b = - 7$ ....

(i) and $c + 4d = 2$ ...(ii)

$2a + 5b = - 8$ .... (iii)

and $2c + 5d = 4$ ...(iv)

Solving(i) and (iii), we get a = 1, b =$-$2
Solving (ii) and (iv), we get c = 2,d =0

Hence, $X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}\\2&0\end{array}} \right]$