Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 121 Question 121 of 472 Medium NUMERICAL
Consider the matrices : A = [ 2 − 5 3 m ] , B = [ 20 m ] A=[\begin{array}{|ccc|}2 & -5 \\ 3 & m\end{array}],B=[\begin{gathered} 20 \\ m \end{gathered}] A = [ 2 3 − 5 m ] , B = [ 20 m ] and X = [ x y ] X=[\begin{gathered} x \\ y \end{gathered}] X = [ x y ] . Let the set of all m m m , for which the system of equations A X = B AX=B A X = B has a negative solution (i.e., x < 0 x<0 x < 0 and y < 0 y<0 y < 0 ), be the interval ( a , b ) (a,b) ( a , b ) . Then 8 ∫ a b ∣ A ∣ d m 8\int _{a}^{b}|A|dm 8 ∫ a b ∣ A ∣ d m is equal to \_\_\_\_\_\_\_\_\_.
Answer: 450
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A X = B 2 x − 5 y = 20 3 x + m y = m ⇒ 3 ( 20 + 5 y 2 ) + m y = m \begin{gathered} AX=B \\ 2x-5y=20 \\ 3x+my=m \\ \Rightarrow 3(\frac{20+5y}{2})+my=m \end{gathered} A X = B 2 x − 5 y = 20 3 x + m y = m ⇒ 3 ( 2 20 + 5 y ) + m y = m
\[ & ⇒ 30 + 15 2 y + m y = m & ⇒ y ( 15 2 + m ) = m − 30 & ⇒ y = m − 30 15 2 + m S i m i l a r l y : \[\& \Rightarrow 30+\frac{15}{2}y+my=m \ \& \Rightarrow y(\frac{15}{2}+m)=m-30 \ \& \Rightarrow y=\frac{m-30}{\frac{15}{2}+m}Similarly : \[ & ⇒ 30 + 2 15 y + m y = m & ⇒ y ( 2 15 + m ) = m − 30 & ⇒ y = 2 15 + m m − 30 S imi l a r l y : 3x+m(\frac{2x-20}{5})=m \]
\[\begin{gathered} \Rightarrow 3x+\frac{2mx}{5}-\frac{20m}{5}=m \\ \Rightarrow \frac{15x+2mx}{5}=5m\Rightarrow x=\frac{25m}{15+2m} \\ x<0\Rightarrow \frac{25m}{15+2m}<0\Rightarrow m\in (-\frac{15}{2},0) \\ \therefore \, m\in (-\frac{15}{2},0) \\ a=-\frac{15}{2},b=0 \\ 8\int _{-\frac{15}{2}}^{0}(2m+15)dm=450 \end{gathered}\] Answer the question to unlock