4.4 Q-17

Question Statement

We are given two matrix equations and are asked to find the system of linear equations corresponding to each. Additionally, we must check if the lines represented by the systems are concurrent.

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Background and Explanation

A system of linear equations can be represented in matrix form. The matrix multiplication gives the equations of the system. For the given matrix equations, we will extract the corresponding linear equations and determine whether the system has any solutions where the lines intersect (concurrent).

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Solution

Part (a)

The matrix equation is:

$$\left[\begin{array}{ccc}1 & 0 & -1 2 & 0 & 1 0 & -1 & 2\end{array}\right] \left[\begin{array}{l}x y z\end{array}\right] = \left[\begin{array}{l}0 0 0\end{array}\right]$$

Multiplying the matrices on the left-hand side, we get the following system of equations:

$$\left[\begin{array}{ccc} x & + 0y & -z 2x & + 0y & +z 0x & -y & +2z \end{array}\right] = \left[\begin{array}{l} 0 0 0 \end{array}\right]$$

This leads to the system:

$$x - z = 0 \quad (1)$$ $$2x + z = 0 \quad (2)$$ $$-y + 2z = 0 \quad (3)$$

Now, solving these equations:

1. From equation (1), we have $x = z$.
2. Substitute $x = z$ into equation (2):

$$2z + z = 0 \quad \Rightarrow \quad 3z = 0 \quad \Rightarrow \quad z = 0.$$

3. Substituting $z = 0$ into $x = z$, we find $x = 0$.
4. From equation (3), substitute $z = 0$:

$$-y + 2(0) = 0 \quad \Rightarrow \quad y = 0.$$

Thus, the solution to this system is $x = 0, y = 0, z = 0$.

Now, let’s check if the lines are concurrent. To check for concurrency, we compute the determinant of the coefficient matrix:

$$A = \left[\begin{array}{ccc}1 & 0 & -1 2 & 0 & 1 0 & -1 & 2\end{array}\right]$$

The determinant of $A$ is:

$$|A| = \left|\begin{array}{ccc}1 & 0 & -1 2 & 0 & 1 0 & -1 & 2\end{array}\right| = \left|\begin{array}{cc}1 & -1 2 & 1\end{array}\right| = (1)(1) - (2)(-1) = 1 + 2 = 3 \neq 0.$$

Since the determinant is not zero, the system of equations does not represent concurrent lines.

Part (b)

The second matrix equation is:

$$\left[\begin{array}{ccc}1 & 1 & 2 2 & 4 & -3 3 & 6 & -5\end{array}\right] \left[\begin{array}{l}x y 1\end{array}\right] = \left[\begin{array}{l}0 0 0\end{array}\right]$$

Multiplying the matrices on the left-hand side, we get the following system of equations:

$$\begin{aligned} x + y + 2 &= 0 \quad (1) 2x + 4y - 3 &= 0 \quad (2) 3x + 6y - 5 &= 0 \quad (3) \end{aligned}$$

Thus, the system of equations is:

$$x + y + 2 = 0$$ $$2x + 4y - 3 = 0$$ $$3x + 6y - 5 = 0$$

Next, we compute the determinant of the coefficient matrix:

$$A = \left[\begin{array}{ccc}1 & 1 & 2 2 & 4 & -3 3 & 6 & -5\end{array}\right]$$

The determinant of $A$ is:

$$|A| = \left|\begin{array}{ccc}1 & 1 & 2 2 & 4 & -3 3 & 6 & -5\end{array}\right| = 1\left|\begin{array}{cc}4 & -3 6 & -5\end{array}\right| - 1\left|\begin{array}{cc}2 & -3 3 & -5\end{array}\right| + 2\left|\begin{array}{cc}2 & 4 3 & 6\end{array}\right|$$

Calculating the 2x2 determinants:

$$= 1\left[(4)(-5) - (6)(-3)\right] - 1\left[(2)(-5) - (3)(-3)\right] + 2\left[(2)(6) - (3)(4)\right]$$ $$= 1\left[-20 + 18\right] - 1\left[-10 + 9\right] + 2\left[12 - 12\right]$$ $$= (-2) - 1(-1) + 2(0)$$ $$= -2 + 1 = -1 \neq 0.$$

Since the determinant is not zero, the matrix is non-singular, and thus the system of equations does not represent concurrent lines.

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Key Formulas or Methods Used

• Matrix Multiplication: Used to convert the matrix equation into a system of linear equations.
• Determinant of a Matrix: Used to check if the lines represented by the system are concurrent (if the determinant is zero, the system has no unique solution, suggesting concurrency).

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Summary of Steps

1. Part (a):
   • Multiply the matrices to form the system of equations.
   • Solve for $x$, $y$, and $z$.
   • Compute the determinant of the coefficient matrix.
   • Since the determinant is non-zero, the lines are not concurrent.
2. Part (b):
   • Multiply the matrices to form the system of equations.
   • Compute the determinant of the coefficient matrix.
   • Since the determinant is non-zero, the lines are not concurrent.