Answer
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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
$\begin{align}
& \det (A)=1\left( 1-0 \right)-a\left( 0-{{a}^{2}} \right) \\
& =1+{{a}^{3}} \\
\end{align}$
For infinitely many solutions we have det(A) = 0
$\Rightarrow 1+{{a}^{3}}=0$
Subtracting 1 from both sides, we get
$\begin{align}
& {{a}^{3}}=-1 \\
& \Rightarrow a=-1 \\
\end{align}$
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
$\begin{align}
& x+ay=0 \\
& \Rightarrow x=-ay\text{ (A)} \\
\end{align}$
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
$\begin{align}
& a\left( {{a}^{2}}z \right)+z=0 \\
& \Rightarrow z\left( {{a}^{3}}+1 \right)=0 \\
\end{align}$
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.
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
$\begin{align}
& \det (A)=1\left( 1-0 \right)-a\left( 0-{{a}^{2}} \right) \\
& =1+{{a}^{3}} \\
\end{align}$
For infinitely many solutions we have det(A) = 0
$\Rightarrow 1+{{a}^{3}}=0$
Subtracting 1 from both sides, we get
$\begin{align}
& {{a}^{3}}=-1 \\
& \Rightarrow a=-1 \\
\end{align}$
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
$\begin{align}
& x+ay=0 \\
& \Rightarrow x=-ay\text{ (A)} \\
\end{align}$
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
$\begin{align}
& a\left( {{a}^{2}}z \right)+z=0 \\
& \Rightarrow z\left( {{a}^{3}}+1 \right)=0 \\
\end{align}$
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.
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