For what value of k do the following...

MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Advanced 1979Moderate

For what value of $k$ do the following system of equations possess a non trivial (i.e., not all zero) solution over the set of rationals $Q$ ? $x + k y + 3 z = 0, 3 x + k y - 2 z = 0, 2 x + 3 y - 4 z = 0$ . For that value of $k$ , find all the solutions for the system.

Answer:

k = 33/2

; solutions are

x = b, y = - 2 b /15, z = 2 b /5

for

b \in Q

Visualized Solution (English)

System of Homogeneous Equations

Given system of homogeneous equations:
$x + k y + 3 z = 0$
$3 x + k y - 2 z = 0$
$2 x + 3 y - 4 z = 0$
A non-trivial solution exists if the planes intersect along a line or a plane, rather than just at the origin $(0, 0, 0)$ .

Condition for Non-Trivial Solution

For a non-trivial solution in a homogeneous system $A X = 0$ , we must have:
$Δ = ∣ A ∣ = 0$
If $Δ \neq = 0$ , the system has only the trivial solution $(0, 0, 0)$ .

Setting up the Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

Constructing the determinant from coefficients:
$Δ = 132 k k 3 3 - 2 - 4 = 0$

Expanding the Determinant

Expanding along the first row:
$1 [k (- 4) - 3 (- 2)] - k [3 (- 4) - 2 (- 2)] + 3 [3 (3) - 2 (k)] = 0$
Simplifying the inner terms:
$1 (- 4 k + 6) - k (- 12 + 4) + 3 (9 - 2 k) = 0$

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Combining terms:
$- 4 k + 6 + 8 k + 27 - 6 k = 0$
$- 2 k + 33 = 0$
Transposing and solving:
$2 k = 33 ⟹ k = \frac{33}{2}$

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k to find Solutions

Substitute $k = \frac{33}{2}$ back into the system:
1) $x + \frac{33}{2} y + 3 z = 0$
2) $3 x + \frac{33}{2} y - 2 z = 0$
3) $2 x + 3 y - 4 z = 0$

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x, y, and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Subtracting eq (1) from eq (2):
$(3 x - x) + (\frac{33}{2} y - \frac{33}{2} y) + (- 2 z - 3 z) = 0$
$2 x - 5 z = 0 ⟹ z = \frac{2 x}{5}$
Substitute $z$ into eq (3):
$2 x + 3 y - 4 (\frac{2 x}{5}) = 0$
$2 x + 3 y - \frac{8 x}{5} = 0 ⟹ 3 y + \frac{2 x}{5} = 0 ⟹ y = - \frac{2 x}{15}$

Final Parametric Solution

Let $x = b$ , where $b \in Q$ :
Then $y = - \frac{2 b}{15}$ and $z = \frac{2 b}{5}$
Final Answer: $k = \frac{33}{2}$ ; solutions are $(b, - \frac{2 b}{15}, \frac{2 b}{5})$ for $b \in Q$ .
Geometrically, this is the line of intersection of the three planes.

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