# 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

Last updated date: 19th Mar 2023

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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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