# Solving a system of equations with a parameter.

We consider for $a \in \mathbb{C}$ the system of equations:

$$(S_a) : \begin{cases} x + y + z = \alpha \\ x + a^2y + z = \beta \\ x + y + a^3z = \gamma \end{cases}$$

1. What the matrix $A_a$ of the system $(S_a)$?

2. What is the rank of $A_a$?

3. For what values of $a$ does the system admit a unique solution? Determine this solution.

4. Solve the system of equations for $a = -1$.

1. The matrix of the system is $$A_a = \begin{pmatrix} 1 & 1 & 1 \\ 1 & a^2 & 1 \\ 1 & 1 & a^3 \\ \end{pmatrix}$$

2. Rank of $A_a$. Reducing $A_a$ to the echelon form gives: $$A_a = \begin{pmatrix} 1 & 1 & 1 \\ 0 & a^2 - 1 & 0 \\ 0 & 0 & a^3 - 1 \\ \end{pmatrix}$$ hence $\text{rank}(A_a) = 3$.

3. Values for which the system admits a unique solution. We have: $$\det(A_a) = (a^2 - 1)(a^{3} - 1)$$

The system admits a unique solution if and only if $a \in \mathbb{C} \setminus \{ -1, 1 \}$. The solution is: $(x,y,z)$ such that:

$$\begin{split} x =& \frac{\det(A_x)}{\det(A_a)}\\ y =& \frac{\det(A_y)}{\det(A_a)}\\ z =& \frac{\det(A_z)}{\det(A_a)}. \end{split}$$ with:

$\det(A_x) = \begin{vmatrix} \alpha & 1 & 1 \\ \beta & a^2 & 1 \\ \gamma & 1 & a^3 \\ \end{vmatrix} = \begin{vmatrix} \alpha & 1 & 1 \\ \beta - \alpha & a^2 - 1 & 0 \\ \gamma - \alpha & 0 & a^3 - 1 \\ \end{vmatrix} = \gamma(a^2 - 1) - (\beta - \alpha)$.

$\det(A_y) = (\beta - 1)(a^3 - 1)$.

$\det(A_z) = (a^2 - 1)(\gamma - 1)$.

My questions:

1. Is my calculation of $\det(A_x)$ coorect?
2. Is there a simpler/ quicker way to determine the solution?

3. Solving $(S_a)$ for $a = -1$:

We have $\det(A_{-1}) = 0$, which means the matrix $A_a$ is not invertible and I can't solve it with the matrix method.

$$(S_{-1}): \begin{cases} x + y + z = \alpha \\ x + y + z = \beta \\ x + y - z = \gamma \\ \end{cases} \iff \begin{cases} x + y + z = \alpha \\ x + y - z = \gamma \\ z \in \mathbb{C} \end{cases}\\ \iff \begin{cases} x + y = \alpha - z \\ x + y = \gamma + z \\ z \in \mathbb{C} \end{cases} \iff \begin{cases} x + y = \alpha - z \\ y \in \mathbb{C} \\ z \in \mathbb{C} \\ \end{cases}$$

The set of solutions of $S_{-1}$ is:

$S = \{ (\alpha - y - z, y, z)\;|\; y,z \in \mathbb{C} \}$

Are my answers correct?

• Is there a $z$ missing in your last equation of the given System? – Dr. Sonnhard Graubner Jun 30 '18 at 19:22
• @Dr.SonnhardGraubner Yes, thank you. – Zouhair El Yaagoubi Jun 30 '18 at 19:24
• The rank of $A$ depends on the value of $a$. There are values of $a$ for which the matrix does not have full rank, as you ought to have have noticed when solving part 3. – amd Jun 30 '18 at 19:28
• @amd The rank of $A$ is 1 if $a \in \{-1,1\}$ and 3 if it's not. Correct? – Zouhair El Yaagoubi Jun 30 '18 at 19:31
• $a\in\mathbb C$, so there are other roots of the polynomial besides these. – amd Jun 30 '18 at 19:31

## 1 Answer

1. $\operatorname{rank}A_a=3$ if $a^2\ne 1$ and $a^3\ne -1$, i.e. if $a\ne 1,-1, j, j^2$ (the non-real cube roots of $1$.

If $a=1$, the matrix$A_a$ has rank $1$.

If $a=-1, j, j^2$, it has rank $2$.

The determinants are false: you should find, applying Sarrus' rule: \begin{align}\det A_x&=\alpha(a^2-1)(a^3-1)-(\gamma-\alpha)(a^2-1)-(\beta-\alpha)(a^3-1)\\ &=\alpha(a^5-1) \end{align} Also, $$\det A_y=(\beta-\alpha)(a^3-1), \qquad\det A_z=(\gamma-\alpha)(a^2-1).$$

Edit :

For the solutions of $S_{-1}$, you have a linear system of rank $2$, so the set of solutions is either empty, if the compatibility conditions are not satisfied, or a affine subspace of dimension $1$. You forgot the compatibility conditions and obtain a set of solutions of dimension $2$.

The simplest way to solve is to obtain the reduced row echelon form of the augmented matrix: \begin{align} &\begin{vmatrix} \begin{array}{ccc|c}1&1&1&\alpha\\1&1&1&\beta\\1&1&-1&\gamma \end{array} \end{vmatrix}\rightsquigarrow \begin{vmatrix} \begin{array}{ccc|c}1&1&1&\alpha\\0&0&0&\beta-\alpha\\0&0&-2&\gamma-\alpha \end{array} \end{vmatrix}\rightsquigarrow \begin{vmatrix} \begin{array}{ccc|c}1&1&1&\alpha\\0&0&0&\beta-\alpha\\0&0&1&\frac12(\alpha-\gamma) \end{array} \end{vmatrix}\\ \rightsquigarrow{} &\begin{vmatrix} \begin{array}{ccc|c}1&1&0&\frac12(\alpha+\gamma)\\0&0&0&\beta-\alpha\\0&0&1&\frac12(\alpha-\gamma) \end{array} \end{vmatrix}\rightsquigarrow \begin{vmatrix} \begin{array}{ccc|c}1&1&0&\frac12(\alpha+\gamma)\\0&0&1&\frac12(\alpha-\gamma)\\0&0&0&\beta-\alpha \end{array} \end{vmatrix} \end{align}

so the compatibility condition is $\alpha=\beta$, and the solutions are given by $$z=\frac12(\alpha-\gamma), \quad y=\frac12(\alpha+\gamma)-x.$$

• Can you check my answer for question 4. Solving the system for $a = -1$? Thank you for your answer. – Zouhair El Yaagoubi Jun 30 '18 at 20:10
• The final part is wrong. The system has rank $2$, so the space of solutions is an affine subpace of dimension $3-2=1$, and you find a solution which depends on $y$ and $z$. Furthermore, the sole parameter $\alpha$ appears,where is only one compatibility condition: $\alpha=\beta$. – Bernard Jun 30 '18 at 20:24
• Can you edit your answer by elaborating more on this please? – Zouhair El Yaagoubi Jun 30 '18 at 20:28
• $\begin{cases} x + y = \alpha - z \\ x + y = \gamma + z \\ z \in \mathbb{C} \end{cases} \iff \begin{cases} x + y = \frac{\alpha + \gamma}{2} \\ y \in \mathbb{C} \\ z = \frac{\alpha - \gamma}{2} \end{cases} \iff \begin{cases} x = \frac{\alpha + \gamma}{2} - y \\ y \in \mathbb{C} \\ z = \frac{\alpha + \gamma}{2} \end{cases}$. Can I take $y = \frac{\alpha + \gamma}{2}$ so the set of solutions can be one dimensional? – Zouhair El Yaagoubi Jun 30 '18 at 21:00
• Yes it is? There's a small error for the final $z$. Please see the way to solve using only the augmented matrix. – Bernard Jun 30 '18 at 21:10