Question

Consider the matrices: \[ A = \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}, \quad B = \begin{bmatrix} 20 \\ m \end{bmatrix}, \quad \text{and} \quad X = \begin{bmatrix} x \\ y \end{bmatrix}. \] Let the set of all \( m \), for which the system of equations \( AX = B \) has a negative solution (i.e., \( x < 0 \) and \( y <0 \)), be the interval \( (a, b) \). Then \[ 8 \int_a^b |\det(A)| \, dm \] is equal to _____ .

Answer 1

Correct Answer: 450

To solve the problem, we first interpret the system of linear equations given by \(AX = B\):
\[ A = \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}, \quad B = \begin{bmatrix} 20 \\ m \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix} \].
The matrix equation \(AX = B\) translates to:
\[ \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 20 \\ m \end{bmatrix} \].
This corresponds to the system:
\[ 2x - 5y = 20 \quad \text{(1)} \]
\[ 3x + my = m \quad \text{(2)} \]
To find a negative solution \(x < 0\) and \(y < 0\), we solve for \(x\) and \(y\). From equation (1), solve for \(x\) in terms of \(y\):
\[ 2x = 20 + 5y \quad \Rightarrow \quad x = 10 + \frac{5}{2}y \].
Substitute this into equation (2):
\[ 3(10 + \frac{5}{2}y) + my = m \].
Simplifying:
\[ 30 + \frac{15}{2}y + my = m \],
\[ \frac{15}{2}y + my = m - 30 \],
\[ y(\frac{15}{2} + m) = m - 30 \],
\[ y = \frac{m - 30}{\frac{15}{2} + m} \].
For \(y < 0\), \( \frac{m - 30}{\frac{15}{2} + m} < 0 \). Analyze the signs:
1. If \(m - 30 > 0\), then \(\frac{15}{2} + m > 0\) must be false.
2. If \(m - 30 < 0\), then \(\frac{15}{2} + m > 0\) yields \(m > -\frac{15}{2}\) which simplifies the range as \( -\frac{15}{2} < m < 30\).
Now, solve for \(x < 0\) using \(x = 10 + \frac{5}{2}y\):
Substituting \(y\):
\[ x = 10 + \frac{5}{2} \cdot \frac{m - 30}{\frac{15}{2} + m} < 0 \quad \Rightarrow \quad 150 + 5(m-30) < 0 \],
\[ 150 + 5m - 150 < 0 \quad \Rightarrow \quad 5m < 0 \],
This results in \(m < 0\), further restricting \(m\) to \(-\frac{15}{2} < m < 0\). Verify this by evaluating the determinant of \(A\):
\[ \det(A) = 2m + 15 \].
So, \(|\det(A)| = |2m + 15|\). Integrate this over \(m \in \left(-\frac{15}{2}, 0\right)\):
\[ 8 \int_{-\frac{15}{2}}^0 |2m + 15| \, dm \].
Since \(2m + 15\) is positive over the range,
\[ 8 \int_{-\frac{15}{2}}^0 (2m + 15) \, dm \].
Calculate:
\[ \int (2m + 15) \, dm = m^2 + 15m \quad \text{from} \quad -\frac{15}{2} \quad \text{to} \quad 0\],
\[ = \left[ 0^2 + 15(0) \right] - \left[ \left(-\frac{15}{2}\right)^2 + 15\left(-\frac{15}{2}\right) \right] \],
\[ = 0 - \left[\frac{225}{4} - \frac{225}{2} \right] \],
\[ = -\left[\frac{225}{4} - \frac{450}{4} \right] = -\left[-\frac{225}{4} \right] = \frac{225}{4} \],
\[ 8 \times \frac{225}{4} = 450 \].
Therefore, the solution is 450, which confirms the solution fits within the given range of 450,450.

Answer 2

Given:

\[A = \begin{pmatrix} 2 & -5 \\ 3 & m \end{pmatrix}, \quad B = \begin{pmatrix} 20 \\ m \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}\]

From the equations:

\(2x - 5y = 20\)       (1)

\(3x + my = m\)      (2)

We get:

\[y = \frac{2m - 60}{2m + 15}\]

For \(y < 0\), \(m \in \left(-\frac{15}{2}, 30\right)\).

Similarly:

\[x = \frac{25m}{2m + 15}\]

For \(x < 0\), \(m \in \left(-\frac{15}{2}, 0\right)\).

Thus, combining conditions:

\[m \in \left(-\frac{15}{2}, 0\right)\]

The determinant of matrix \(A\) is:

\[|A| = 2m + 15\]

Now:

\[8 \int_{-\frac{15}{2}}^{0} (2m + 15) \, dm = 8 \left[ m^2 + 15m \right]_{-\frac{15}{2}}^{0}\]

\[= 8 \left\{ \frac{225}{4} - \frac{225}{2} \right\}\]

\[= 8 \times \frac{225}{4} = 450\]

Final Answer: 450