Consider the matrices : A= [ … ] and X= [ … ]. Let the set of all m, for which the system of equations A X=B has a negative solution (i.e., x < 0 and…

$\begin{aligned} & A=\left(\begin{array}{cc}2 & -5 \\ 3 & m\end{array}\right), B=\left(\begin{array}{c}20 \\ m\end{array}\right) \\ & X=\left(\begin{array}{l}x \\ y\end{array}\right) \\ & 2 x-5 y=20 \ldots(1)\\ & 3 x+m y=m \ldots(2)\\ & \Rightarrow y=\frac{2 m-60}{2 m+15} \\ & y < 0 \Rightarrow m \in\left(\frac{-15}{2}, 30\right)\end{aligned}$​A=(23​−5m​),B=(20m​)X=(xy​)2x−5y=20…(1)3x+my=m…(2)⇒y=2m+152m−60​y<0⇒m∈(2−15​,30)​ $\begin{aligned}
& \mathrm{x}=\frac{25 \mathrm{m}}{2 \mathrm{m}+15} \
& \mathrm{x} < 0 \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 0\right) \
& \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 0\right) \
& |\mathrm{A}|=2 \mathrm{~m}+15
\end{aligned}$

Now, $\begin{aligned}
& 8 \int_{\frac{-15}{2}}^0(2 \mathrm{m}+15) \mathrm{dm}=8\left{\mathrm{m}^2+15 \mathrm{~m}\right}_{\frac{-15}{2}}^0 \
& \Rightarrow 8\left{-\left(\frac{225}{4}-\frac{225}{2}\right)\right} \
& =8 \times \frac{225}{4}=450
\end{aligned}$