Question

For what value of a does the following system of equations
x + ay = 0, y + az = 0, z + ax = 0, has infinitely many solutions
[a] a = 1
[b] a = 0
[c] a = -1
[d] None of these

Answer 1

Hint: Write the system of equations in matrix form AX=B. Use the fact that a homogeneous system of equations has infinitely many solutions if det(A) = 0. Hence find det(A) in terms of a and put det(A) = 0 to get the value of a.

Complete step-by-step answer:

The given system of equations can be written as $\left[ \begin{matrix} 1 & a & 0 \\\ 0 & 1 & a \\\ a & 0 & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\\ 0 \\\ 0 \\\ \end{matrix} \right]$.
So, we have $A=\left[ \begin{matrix} 1 & a & 0 \\\ 0 & 1 & a \\\ a & 0 & 1 \\\ \end{matrix} \right],B=\left[ \begin{matrix} 0 \\\ 0 \\\ 0 \\\ \end{matrix} \right]$ and $X=\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]$
Finding det(A):
We have
$\begin{aligned} & \det (A)=1\left( 1-0 \right)-a\left( 0-{{a}^{2}} \right) \\\ & =1+{{a}^{3}} \\\ \end{aligned}$
For infinitely many solutions we have det(A) = 0
$\Rightarrow 1+{{a}^{3}}=0$
Subtracting 1 from both sides, we get
$\begin{aligned} & {{a}^{3}}=-1 \\\ & \Rightarrow a=-1 \\\ \end{aligned}$
Hence when a = - 1 the given system of equations has infinitely many solutions.

Note: Alternatively we can convert the given system of equations to an equation in variable and find the value of “a” so that the formed equation goes identically to 0.
Step I: From the first equation express x in terms of y
We have
$\begin{aligned} & x+ay=0 \\\ & \Rightarrow x=-ay\text{ (A)} \\\ \end{aligned}$
Step II: From the second equation express y in terms of z
We have
y+az = 0
y = -az (B)
Hence we have $x=-a\left( -az \right)={{a}^{2}}z$
Step III: Substitute the value of x and y in the third equation
We have
$\begin{aligned} & a\left( {{a}^{2}}z \right)+z=0 \\\ & \Rightarrow z\left( {{a}^{3}}+1 \right)=0 \\\ \end{aligned}$
To make the equation go identically to 0, we must have $1+{{a}^{3}}=0$
Hence a = -1.
Hence the given system of equations has infinitely many solutions when a = -1.