Consider the matrices : A=[\beginarraycc2 & -5 3 & m\endarray], B=[\beginarrayl20 m\endarray] and X=[\beginarraylx y\endarray] . Let the set of all m , for which the system of equations A X=B has a negative solution (i.e., x\lt 0 and y\lt 0 ), be the interval (a, b) . Then 8 \int\limits_a^b |A| \, dm is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2024 · 9 Apr · Shift 2 · Q53 with Solution

View written solutionFree

Correct answer: 450

Write the system explicitly

We have $A=\begin{bmatrix}2 & -5\\ 3 & m\end{bmatrix},\quad X=\begin{bmatrix}x\\ y\end{bmatrix},\quad B=\begin{bmatrix}20\\ m\end{bmatrix}.$

So $AX=B$ gives

2x-5y=20 \\ 3x+my=m \end{cases}$$ We need those values of $m$ for which the solution satisfies $$x<0,\quad y<0.$$ --- 2. **Find $x,y$ in terms of $m$** First, determinant of $A$ is $$|A|=2m-(-15)=2m+15.$$ For a unique solution, we need $$2m+15\ne 0 \implies m\ne -\frac{15}{2}.$$ Using Cramer's rule: $$x=\frac{\begin{vmatrix}20 & -5\\ m & m\end{vmatrix}}{2m+15} =\frac{20m+5m}{2m+15} =\frac{25m}{2m+15}.$$ $$y=\frac{\begin{vmatrix}2 & 20\\ 3 & m\end{vmatrix}}{2m+15} =\frac{2m-60}{2m+15} =\frac{2(m-30)}{2m+15}.$$ So $$x=\frac{25m}{2m+15},\qquad y=\frac{2(m-30)}{2m+15}.$$ --- 3. **Impose negativity conditions** We need both $$\frac{25m}{2m+15}<0$$ and $$\frac{2(m-30)}{2m+15}<0.$$ Since $25>0$ and $2>0$, this becomes $$\frac{m}{2m+15}<0 \quad\text{and}\quad \frac{m-30}{2m+15}<0.$$ Now solve each. ### (i) For $x<0$: Critical points are $m=0$ and $m=-\frac{15}{2}$. Sign analysis of $$\frac{m}{2m+15}<0$$ gives $$-\frac{15}{2}<m<0.$$ ### (ii) For $y<0$: Critical points are $m=30$ and $m=-\frac{15}{2}$. Sign analysis of $$\frac{m-30}{2m+15}<0$$ gives $$-\frac{15}{2}<m<30.$$ For both $x<0$ and $y<0$, take intersection: $$\left(-\frac{15}{2},0\right).$$ Hence, $$a=-\frac{15}{2},\qquad b=0.$$ --- 4. **Evaluate the integral** Since $$|A|=2m+15,$$ on the interval $\left(-\frac{15}{2},0\right)$ we have $2m+15>0$, so $$|A|=2m+15.$$ Thus $$\int_a^b |A|\,dm=\int_{-15/2}^{0} (2m+15)\,dm.$$ Compute: $$\int (2m+15)\,dm=m^2+15m.$$ So $$\int_{-15/2}^{0} (2m+15)\,dm =\left[m^2+15m\right]_{-15/2}^{0}.$$ At $m=0$: $$0.$$ At $m=-\frac{15}{2}$: $$\left(-\frac{15}{2}\right)^2+15\left(-\frac{15}{2}\right) =\frac{225}{4}-\frac{225}{2} =\frac{225-450}{4} =-\frac{225}{4}.$$ Therefore, $$\int_{-15/2}^{0} (2m+15)\,dm=0-\left(-\frac{225}{4}\right)=\frac{225}{4}.$$ Hence, $$8\int_a^b |A|\,dm=8\cdot \frac{225}{4}=450.$$ --- 5. **Final answer** $$\boxed{450}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: 450

Write the system explicitly

We have $A=\begin{bmatrix}2 & -5\\ 3 & m\end{bmatrix},\quad X=\begin{bmatrix}x\\ y\end{bmatrix},\quad B=\begin{bmatrix}20\\ m\end{bmatrix}.$

So $AX=B$ gives

2x-5y=20 \\ 3x+my=m \end{cases}$$ We need those values of $m$ for which the solution satisfies $$x<0,\quad y<0.$$ --- 2. **Find $x,y$ in terms of $m$** First, determinant of $A$ is $$|A|=2m-(-15)=2m+15.$$ For a unique solution, we need $$2m+15\ne 0 \implies m\ne -\frac{15}{2}.$$ Using Cramer's rule: $$x=\frac{\begin{vmatrix}20 & -5\\ m & m\end{vmatrix}}{2m+15} =\frac{20m+5m}{2m+15} =\frac{25m}{2m+15}.$$ $$y=\frac{\begin{vmatrix}2 & 20\\ 3 & m\end{vmatrix}}{2m+15} =\frac{2m-60}{2m+15} =\frac{2(m-30)}{2m+15}.$$ So $$x=\frac{25m}{2m+15},\qquad y=\frac{2(m-30)}{2m+15}.$$ --- 3. **Impose negativity conditions** We need both $$\frac{25m}{2m+15}<0$$ and $$\frac{2(m-30)}{2m+15}<0.$$ Since $25>0$ and $2>0$, this becomes $$\frac{m}{2m+15}<0 \quad\text{and}\quad \frac{m-30}{2m+15}<0.$$ Now solve each. ### (i) For $x<0$: Critical points are $m=0$ and $m=-\frac{15}{2}$. Sign analysis of $$\frac{m}{2m+15}<0$$ gives $$-\frac{15}{2}<m<0.$$ ### (ii) For $y<0$: Critical points are $m=30$ and $m=-\frac{15}{2}$. Sign analysis of $$\frac{m-30}{2m+15}<0$$ gives $$-\frac{15}{2}<m<30.$$ For both $x<0$ and $y<0$, take intersection: $$\left(-\frac{15}{2},0\right).$$ Hence, $$a=-\frac{15}{2},\qquad b=0.$$ --- 4. **Evaluate the integral** Since $$|A|=2m+15,$$ on the interval $\left(-\frac{15}{2},0\right)$ we have $2m+15>0$, so $$|A|=2m+15.$$ Thus $$\int_a^b |A|\,dm=\int_{-15/2}^{0} (2m+15)\,dm.$$ Compute: $$\int (2m+15)\,dm=m^2+15m.$$ So $$\int_{-15/2}^{0} (2m+15)\,dm =\left[m^2+15m\right]_{-15/2}^{0}.$$ At $m=0$: $$0.$$ At $m=-\frac{15}{2}$: $$\left(-\frac{15}{2}\right)^2+15\left(-\frac{15}{2}\right) =\frac{225}{4}-\frac{225}{2} =\frac{225-450}{4} =-\frac{225}{4}.$$ Therefore, $$\int_{-15/2}^{0} (2m+15)\,dm=0-\left(-\frac{225}{4}\right)=\frac{225}{4}.$$ Hence, $$8\int_a^b |A|\,dm=8\cdot \frac{225}{4}=450.$$ --- 5. **Final answer** $$\boxed{450}$$ This matches the stored correct answer.

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