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प्रश्न
Find x, y, a and b if
`[[2x-3y,a-b,3],[1,x+4y,3a+4b]]`=`[[1,-2,3],[1,6,29]]`
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उत्तर
\[\begin{bmatrix}2x - 3y & a - b & 3 \\ 1 & x + 4y & 3a + 4b\end{bmatrix} = \begin{bmatrix}1 & - 2 & 3 \\ 1 & 6 & 29\end{bmatrix}\]\[\Rightarrow 2x - 3y = 1 . . . \left( 1 \right) \]
\[x + 4y = 6 \]
\[ \Rightarrow x = 6 - 4y . . . \left( 2 \right)\]
Putting the value of x in eq. (1), we get
\[2\left( 6 - 4y \right) - 3y = 1 \]
\[ \Rightarrow 12 - 8y - 3y = 1 \]
\[ \Rightarrow 12 - 11y = 1 \]
\[ \Rightarrow - 11y = - 11 \]
\[ \Rightarrow y = \frac{- 11}{- 11} = 1\]
\[\]
Putting the value of y in eq. (2), we get
\[x = 6 - 4\left( 1 \right)\]
\[ \Rightarrow x = 6 - 4 \]
\[ \Rightarrow x = 2 \]
\[Now, \]
\[a - b = - 2 \]
\[ \Rightarrow a = - 2 + b . . . \left( 3 \right) \]
\[3a + 4b = 29 . . . \left( 4 \right) \]
Putting the value of a in eq. (4), we get\[3\left( - 2 + b \right) + 4b = 29 \]
\[ \Rightarrow - 6 + 3b + 4b = 29 \]
\[ \Rightarrow - 6 + 7b = 29 \]
\[ \Rightarrow 7b = 29 + 6\]
\[ \Rightarrow 7b = 35 \]
\[ \Rightarrow b = \frac{35}{7} = 5 \]
Putting the value of b in eq. (1), we get
\[a = - 2 + 5 \]
\[ \Rightarrow a = 3 \]
`∴a=3,b=5,x=2` and `y=1`
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