Analyzing the Setup
Imagine you are standing in a three-dimensional space with three planes defined by the following linear equations:
2x+3y+6z=8
x+2y+az=5
3x+5y+9z=b
In linear algebra, these equations represent physical surfaces. While these planes typically intersect at a single point, we are investigating the condition for no solution, where the planes never meet at a common point.
The Cramer's Rule Master Key
To unlock this mystery, we utilize the machinery of Cramer's Rule. We define the main determinant of the coefficient matrix, Δ, as:
For a system to have no solution, the first requirement is that the system must be singular, meaning Δ=0. If $\Delta
eq 0$, the system would possess a unique solution.
However, Δ=0 is a necessary but not sufficient condition. We must also ensure the system is not 'dependent' (which would lead to infinite solutions) by verifying that at least one of the numerator determinants—Δx, Δy, or Δz—is non-zero.
The Calculation
Unveiling 'a'
Let us expand Δ along the first row:
Δ=2(18−5a)−3(9−3a)+6(5−6)
Performing the algebraic expansion:
Combining the terms, we find:
Setting this to zero, we immediately see that a=3. This is the value that renders our system singular.
The Twist
The Condition for 'b'
To ensure the system is truly inconsistent, we construct Δz by replacing the third column of the coefficient matrix with the constants from the right side of our equations:
Expanding this along the first row:
Δz=2(2b−25)−3(b−15)+8(5−6)
Simplifying this expression:
This collapses into:
For the system to have no solution, we must satisfy the condition $\Delta_z
eq 0$. Therefore, $b
eq 13$.
The Final Synthesis
By setting Δ=0, we forced the system to be singular, yielding a=3. By ensuring $\Delta_z
eq 0$, we guaranteed that the system is inconsistent rather than dependent, yielding $b
eq 13$.
The values for which the system has no solution are a=3 and $b
eq 13$. This result describes a configuration where the planes exist in space while forever avoiding a common intersection.