Question:
The positive value of \(a\) for which the system of linear homogeneous equations
\[
x+ay+z=0,
\qquad
ax+2y-z=0,
\qquad
2x+3y+z=0
\]
has non-trivial solution is
The positive value of \(a\) for which the system of linear homogeneous equations
\[
x+ay+z=0,
\qquad
ax+2y-z=0,
\qquad
2x+3y+z=0
\]
has non-trivial solution is
Show Hint
For a homogeneous system:
\[
AX=0
\]
non-trivial solutions exist only when:
\[
|A|=0
\]
Always form the coefficient matrix first and then evaluate its determinant carefully.
Updated On: Jun 3, 2026
- \(0\)
- \(1\)
- \(\dfrac{1+\sqrt{5}}{2}\)
- \(\dfrac{\sqrt{5}-1}{2}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
A homogeneous system of linear equations has a non-trivial solution if and only if the determinant of the coefficient matrix is zero.
That is,
\[
|A|=0
\]
For the system:
\[
AX=0
\]
non-trivial solutions exist when the coefficient matrix is singular.
Step 1: Write the coefficient matrix. The given equations are: \[ x+ay+z=0 \] \[ ax+2y-z=0 \] \[ 2x+3y+z=0 \] Therefore, the coefficient matrix is: \[ A= \begin{bmatrix} 1 & a & 1 a & 2 & -1 2 & 3 & 1 \end{bmatrix} \] For non-trivial solution: \[ |A|=0 \]
Step 2: Evaluate the determinant. \[ \begin{vmatrix} 1 & a & 1 a & 2 & -1 2 & 3 & 1 \end{vmatrix}=0 \] Expand along the first row: \[ 1 \begin{vmatrix} 2 & -1 3 & 1 \end{vmatrix} -a \begin{vmatrix} a & -1 2 & 1 \end{vmatrix} +1 \begin{vmatrix} a & 2 2 & 3 \end{vmatrix} =0 \] Now compute each minor. First minor: \[ \begin{vmatrix} 2 & -1 3 & 1 \end{vmatrix} =(2)(1)-(-1)(3) \] \[ =2+3=5 \] Second minor: \[ \begin{vmatrix} a & -1 2 & 1 \end{vmatrix} =a(1)-(-1)(2) \] \[ =a+2 \] Third minor: \[ \begin{vmatrix} a & 2 2 & 3 \end{vmatrix} =3a-4 \] Substituting: \[ 5-a(a+2)+(3a-4)=0 \]
Step 3: Simplify the equation. \[ 5-a^2-2a+3a-4=0 \] \[ 1-a^2+a=0 \] Multiply by \(-1\): \[ a^2-a-1=0 \]
Step 4: Solve the quadratic equation. Using quadratic formula: \[ a=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)} \] \[ a=\frac{1\pm\sqrt{1+4}}{2} \] \[ a=\frac{1\pm\sqrt{5}}{2} \] Since positive value is required: \[ \boxed{a=\frac{1+\sqrt{5}}{2}} \] Hence, the correct option is: \[ \boxed{(c)\ \frac{1+\sqrt{5}}{2}} \]
Step 1: Write the coefficient matrix. The given equations are: \[ x+ay+z=0 \] \[ ax+2y-z=0 \] \[ 2x+3y+z=0 \] Therefore, the coefficient matrix is: \[ A= \begin{bmatrix} 1 & a & 1 a & 2 & -1 2 & 3 & 1 \end{bmatrix} \] For non-trivial solution: \[ |A|=0 \]
Step 2: Evaluate the determinant. \[ \begin{vmatrix} 1 & a & 1 a & 2 & -1 2 & 3 & 1 \end{vmatrix}=0 \] Expand along the first row: \[ 1 \begin{vmatrix} 2 & -1 3 & 1 \end{vmatrix} -a \begin{vmatrix} a & -1 2 & 1 \end{vmatrix} +1 \begin{vmatrix} a & 2 2 & 3 \end{vmatrix} =0 \] Now compute each minor. First minor: \[ \begin{vmatrix} 2 & -1 3 & 1 \end{vmatrix} =(2)(1)-(-1)(3) \] \[ =2+3=5 \] Second minor: \[ \begin{vmatrix} a & -1 2 & 1 \end{vmatrix} =a(1)-(-1)(2) \] \[ =a+2 \] Third minor: \[ \begin{vmatrix} a & 2 2 & 3 \end{vmatrix} =3a-4 \] Substituting: \[ 5-a(a+2)+(3a-4)=0 \]
Step 3: Simplify the equation. \[ 5-a^2-2a+3a-4=0 \] \[ 1-a^2+a=0 \] Multiply by \(-1\): \[ a^2-a-1=0 \]
Step 4: Solve the quadratic equation. Using quadratic formula: \[ a=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)} \] \[ a=\frac{1\pm\sqrt{1+4}}{2} \] \[ a=\frac{1\pm\sqrt{5}}{2} \] Since positive value is required: \[ \boxed{a=\frac{1+\sqrt{5}}{2}} \] Hence, the correct option is: \[ \boxed{(c)\ \frac{1+\sqrt{5}}{2}} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Matrices and Determinants Questions
- If the cofactors of the elements 3, 7, and 6 of the matrix \( \begin{bmatrix} 1 & 2 & 3 \\ 4 & -1 & 7 \\ 2 & 4 & 6 \end{bmatrix} \) are \( a, b, c \) respectively, then evaluate: \[ \left[ a \ b \ c \right] \begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix} + \left[ a \ b \ c \right] \begin{bmatrix} 3 \\ 7 \\ 6 \end{bmatrix} \]
- Let \( B = \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & -1 & -1 \end{bmatrix} \), \( C = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix} \), and if a matrix \( A \) is such that \( BAC = I \), then \( A^{-1} = \)
- If \( \det(AB) = \det(A)\det(B) \) and \( A \) is a non-singular matrix of order \( 3 \times 3 \), then \( \det(\text{adj}(A)) \) is:
- If \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), where \( a, b, c, d \in \{-1, 1\} \), then the number of singular matrices in \( A \) is:
The value of the determinant \[ \begin{vmatrix} 2 & 3 & 5 \\ 1 & 0 & 4 \\ 7 & 2 & 1 \end{vmatrix} \] is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions