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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices3 Marks
Question
Find the matrix $X$ so that $X\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$
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Answer

Let $X = \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]$
$\therefore \;\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ 4 \end{array}\;\;\begin{array}{*{20}{c}} 2 \\ 5 \end{array}\;\;\begin{array}{*{20}{c}} 3 \\ 6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 7} \\ 2 \end{array}\;\;\begin{array}{*{20}{c}} { - 8} \\ 4 \end{array}\;\;\begin{array}{*{20}{c}} { - 9} \\ 6 \end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}} {a + 4b} \\ {c + 4d} \end{array}\;\;\begin{array}{*{20}{c}} {2a + 5b} \\ {2c + 5d} \end{array}\;\;\begin{array}{*{20}{c}} {3a + 6b} \\ {3c + 6d} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 7} \\ 2 \end{array}\;\;\begin{array}{*{20}{c}} { - 8} \\ 4 \end{array}\;\;\begin{array}{*{20}{c}} { - 9} \\ 6 \end{array}} \right]$
On solving a + 4b = -7 and  2a + 5b = -8 & c + 4d = 2 and 2c + 5d = 4
we get a = 1, b = -2, c = 2, d = 0
$X = \left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 2&0 \end{array}} \right]$

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