Suppose, the system of linear equations -2x + y + z = l x - 2y + z = m x + y - 2z = n is such that l + m + n = 0, then the system has:
Infinitely many solutions.
Let's analyze the given system of linear equations and determine the nature of its solutions based on the condition provided.
System of Linear Equations Analysis
The system of linear equations is given as:
- $-2x + y + z = l$
- $x - 2y + z = m$
- $x + y - 2z = n$
This system can be represented in the matrix form $AX = B$, where:
$$ A = \begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} l \\ m \\ n \end{pmatrix} $$Determining Solution Characteristics
To understand the nature of the solutions, we first examine the determinant of the coefficient matrix $A$.
Calculating the Determinant of A
The determinant of matrix $A$, denoted as $det(A)$, is calculated as follows:
$$ det(A) = -2 \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} + 1 \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} $$Now, we compute the values of the determinants of the 2x2 matrices:
$$ det(A) = -2((-2)(-2) - (1)(1)) - 1((1)(-2) - (1)(1)) + 1((1)(1) - (-2)(1)) $$ $$ det(A) = -2(4 - 1) - 1(-2 - 1) + 1(1 + 2) $$ $$ det(A) = -2(3) - 1(-3) + 1(3) $$ $$ det(A) = -6 + 3 + 3 $$ $$ det(A) = 0 $$Implication of a Zero Determinant
When the determinant of the coefficient matrix $A$ is zero ($det(A) = 0$), the system of linear equations does not have a unique solution. This means the system either has infinitely many solutions or no solution at all.
Verifying System Consistency with the Given Condition
The problem states a condition $l + m + n = 0$. Let's see how this condition affects the system.
We can add the three given equations:
$$ (-2x + y + z) + (x - 2y + z) + (x + y - 2z) = l + m + n $$Combine the terms involving $x$, $y$, and $z$ on the left side:
$$ (-2 + 1 + 1)x + (1 - 2 + 1)y + (1 + 1 - 2)z = l + m + n $$ $$ 0x + 0y + 0z = l + m + n $$ $$ 0 = l + m + n $$The problem provides the condition that $l + m + n = 0$. Substituting this into the equation we derived:
$$ 0 = 0 $$This result ($0 = 0$) indicates that the system is consistent. The condition $l + m + n = 0$ ensures that the equations are not contradictory.
Conclusion on the Nature of Solutions
We have established two key points:
- The determinant of the coefficient matrix $A$ is $0$, meaning there is no unique solution.
- The condition $l + m + n = 0$ makes the system consistent.
When a system of linear equations has a determinant of the coefficient matrix equal to zero and is consistent, it implies that there are infinitely many solutions.
Comparing this conclusion with the given options, the system has Infinitely many solutions.
Important Questions from Application of Determinants
The equations 3x - 4y = 5 and 12x - 16y = 20 have:
The system of equations
2x + y - 3z = 5
3x - 2y + 2z = 5 and
5x - 3y - z = 16If (a, b), (x1, y1) and (x2, y2) are the vertices of a triangle such that the x-coordinates a, x1, x2 are in geometric progression with common ratio r and the y-coordinates b, y1, y2 are also in geometric progression with common ratio s, then the area of the triangle is:
If A is a 2 × 2 matrix and |A| = 5, what is |5A| ? (| | denotes determinant)
If the system of equation x + 2y - 3z = 2, (k + 3) z = 3, (2k + 1) y + z = 2 is consistent, then K is
