# Solving Systems of Linear Equations

By Catalin David

The general form is:

$\begin{cases} a_{1,1}\cdot x_{1} + a_{1,2}\cdot x_{2} + a_{1,3}\cdot x_{3} + \cdots a_{1,n} \cdot x_{n} =b_{1} \\ a_{2,1}\cdot x_{1} + a_{2,2}\cdot x_{2}+ a_{2,3}\cdot x_{3} + \cdots + a_{2,n}\cdot x_{n} = b_{2} \\ a_{3,1}\cdot x_{1} + a_{3,2}\cdot x_{2}+a_{3,3}\cdot x_{3}+ \cdots + a_{3,n}\cdot x_{n}=b_{3} \\ \cdots\\ a_{m,1}\cdot x_{1}+ a_{m,2}\cdot x_{2}+a_{m,3}\cdot x_{3}+\cdots + a_{m,n}\cdot x_{n} =b_{n} \end{cases}$

$A= \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} & . & . & a_{1,n} \\ a_{2,1} & a_{2,2} & a_{2,3} & . & . & a_{2,n} \\ a_{3,1} & a_{3,2} & a_{3,3} & . & . & a_{3,n} \\ \cdots \\ a_{m,1} & a_{m,2} & a_{m,3} & . & . & a_{m,n} \end{pmatrix}$ is the matrix associated to the system and $b_{1}, b_{2},b_{3} \cdots b_{n}$ are the constant terms of the system.

If the constant terms are 0, the system is homogeneous

## The Associated Matrix is Square (m=n)

We calculate the determinant of the associated matrix.

$\Delta = \begin{vmatrix} a_{1,1} & a_{1,2} & a_{1,3} & . & . & a_{1,n} \\ a_{2,1} & a_{2,2} & a_{2,3} & . & . & a_{2,n} \\ a_{3,1} & a_{3,2} & a_{3,3} & . & . & a_{3,n} \\ \cdots \\ a_{n,1} & a_{n,2} & a_{n,3} & . & . & a_{n,n} \end{vmatrix}$

## The Determinant of the Associated Matrix is not 0

The system is called a consistent system with a single solution. To determine the solution of the system we use Cramer's rule.

We calculate $\Delta_{x_{1}}$, the determinant obtained by replacing the column containing the coefficients of the respective variable $x_{1}$ with the column of constant terms.
$\Delta_{x_{1}}= \begin{vmatrix} b_{1} & a_{1,2} & a_{1,3} & . & . & a_{1,n} \\ b_{2} & a_{2,2} & a_{2,3} & . & . & a_{2,n} \\ b_{3} & a_{3,2} & a_{3,3} & . & . & a_{3,n} \\ \cdots \\ b_{n} & a_{n,2} & a_{n,3} & . & . & a_{n,n} \end{vmatrix}$

We get $x_{1} = \dfrac{\Delta_{x_{1}}}{\Delta}$

We calculate $\Delta_{x_{2}}$, the determinant obtained by replacing the column containing the coefficients of the respective variable $x_{2}$ with the column of constant terms.
$\Delta_{x_{2}}= \begin{vmatrix} a_{1,1} & b_{1} & a_{1,3} & . & . & a_{1,n} \\ a_{2,1} & b_{2} & a_{2,3} & . & . & a_{2,n} \\ a_{3,1} & b_{3} & a_{3,3} & . & . & a_{3,n} \\ \cdots \\ a_{n,1} & b_{n} & a_{n,3} & . & . & a_{n,n} \end{vmatrix}$

We get $x_{2} = \dfrac{\Delta_{x_{2}}}{\Delta}$

We calculate $\Delta_{x_{3}}$, the determinant obtained by replacing the column containing the coefficients of the respective variable $x_{3}$ with the column of constant terms.
$\Delta_{x_{3}}= \begin{vmatrix} a_{1,1} & a_{1,2} & b_{1} & . & . & a_{1,n} \\ a_{2,1} & a_{2,2} & b_{2} & . & . & a_{2,n} \\ a_{3,1} & a_{3,2} & b_{3} & . & . & a_{3,n} \\ \cdots \\ a_{n,1} & a_{n,2} & a_{n} & . & . & a_{n,n} \end{vmatrix}$

We get $x_{3} = \dfrac{\Delta_{x_{3}}}{\Delta}$

We keep doing this for the other variables until the last one and then we write down the solution of the system.
$x_{n}=\dfrac{\Delta_{x_{n}}}{\Delta}$

Example 53
$\begin{cases} 2\cdot x + 3\cdot y -5\cdot z = \color{red}{-7}\\ -3 \cdot x + 2\cdot y + z = \color{red}{-9}\\ 4\cdot x - y + 2\cdot z = \color{red}{17} \end{cases}$

The matrix associated to the system is
$\begin{pmatrix} 2 & 3 & -5\\ -3 & 2 & 1\\ 4 & -1 & 2 \end{pmatrix}$

We calculate the determinant of the matrix and we get $\Delta = 8 -15 + 12 +40 +2 + 18 = 65$
We calculate $\Delta_{x}= \begin{vmatrix} \color{red}{-7} & 3 & -5\\ \color{red}{-9} & 2 & 1\\ \color{red}{17} & -1 & 2 \end{vmatrix}= -28 - 45 + 51 + 170 - 7 +54 = 195$

We calculate $\Delta_{y}= \begin{vmatrix} 2 & \color{red}{-7} & -5\\ -3 & \color{red}{-9} & 1\\ 4 & \color{red}{17} & 2 \end{vmatrix}=-36 + 255 -28 -180 -34 -42 = -65$

We calculate $\Delta_{z}= \begin{vmatrix} 2 & 3 &\color{red}{-7}\\ -3 & 2 & \color{red}{-9}\\ 4 & -1 & \color{red}{17} \end{vmatrix}= 68 -21 -108 + 56 -18 + 153 =130$

The solution of the system is:
$x = \dfrac{\Delta_{x}}{\Delta} =\dfrac{195}{65} = 3$
$y = \dfrac{\Delta_{y}}{\Delta} = -\dfrac{65}{65}= -1$
$z = \dfrac{\Delta_{z}}{\Delta} =\dfrac{130}{65}= 2$
$S=\{3;-1;2\}$

Example 54
$\begin{cases} 4\cdot x + 5\cdot y -2\cdot z = \color{red}{3}\\ -2 \cdot x + 3\cdot y - z = \color{red}{-3}\\ -1\cdot x - 2\cdot y + 3\cdot z = \color{red}{-5} \end{cases}$

The matrix associated to the system is $\begin{pmatrix} 4 & 5 & -2\\ -2 & 3 & -1\\ -1 & -2 & 3 \end{pmatrix}$

We calculate the determinant of the matrix and we get $\Delta = 36 -8 + 5 -6 -8 + 30 = 49$

We calculate $\Delta_{x}= \begin{vmatrix} \color{red}{3} & 5 & -2\\ \color{red}{-3} & 3 &-1\\ \color{red}{-5} & -2 & 3 \end{vmatrix}= 27 - 12 + 25 - 30 - 6 + 45 = 49$

We calculate $\Delta_{y}= \begin{vmatrix} 4 & \color{red}{3} & -2\\ -2 & \color{red}{-3} & -1\\ -1 & \color{red}{-5} & 3 \end{vmatrix}=-36 -20+ 3 +6 -20 + 18 = -49$

We calculate $\Delta_{z}= \begin{vmatrix} 4 & 5 & \color{red}{3}\\ -2 & 3 & \color{red}{-3}\\ -1& -2 & \color{red}{-5} \end{vmatrix}= -60 + 12 + 15 + 9 - 24 -50 = - 98$

The solution of the system is:
$x = \dfrac{\Delta_{x}}{\Delta} =\dfrac{49}{49} = 1$
$y = \dfrac{\Delta_{y}}{\Delta} = -\dfrac{-49}{49}= -1$
$z = \dfrac{\Delta_{z}}{\Delta} =\dfrac{-98}{4}= -2$
$S=\{1;-1;-2\}$

If the system is homogeneous, its solution is {0;0;0} because in determinants $\Delta_{x}$,$\Delta_{y}$ and $\Delta_{z}$ there are columns of 0, so they are also equal to 0.

Example 55
$\begin{cases} 2\cdot x + 3\cdot y -5\cdot z = \color{red}{0}\\ -3 \cdot x + 2\cdot y + z = \color{red}{0}\\ 4\cdot x - y + 2\cdot z = \color{red}{0} \end{cases}$

The matrix associated to the system is
$\begin{pmatrix} 2 & 3 & -5\\ -3 & 2 & 1\\ 4 & -1 & 2 \end{pmatrix}$

We calculate the determinant of the matrix and we get $\Delta = 8 -15 + 12 +40 +2 + 18 = 65$

$\Delta_{x}= \begin{vmatrix} \color{red}{0} & 3 & -5\\ \color{red}{0} & 2 & 1\\ \color{red}{0} & -1 & 2 \end{vmatrix}= 0$

$\Delta_{y}= \begin{vmatrix} 2 & \color{red}{0} & -5\\ -3 & \color{red}{0} & 1\\ 4 & \color{red}{0} & 2 \end{vmatrix}= 0$

$\Delta_{z}= \begin{vmatrix} 2 & 3 &\color{red}{0}\\ -3 & 2 & \color{red}{0}\\ 4 & -1 & \color{red}{0} \end{vmatrix}= 0$

The solution of the system is:
$x = \dfrac{\Delta_{x}}{\Delta} =\dfrac{0}{65} = 0$
$y = \dfrac{\Delta_{y}}{\Delta} = -\dfrac{0}{65}= 0$
$z = \dfrac{\Delta_{z}}{\Delta} =\dfrac{0}{65}= 0$
$S = \{0;0;0\}$

## The Determinant of the Associated Matrix is 0.

We calculate the rank of the matrix associated to the system and the rank of the augmented matrix (the initial matrix to which we add the column of constant terms).

We have the following situations:

• If the ranks of the two matrices are different, then the system has no solution. It is an inconsistent system.
• If the ranks are equal, then the system is consistent with an infinitude of solutions.

To solve the system, we follow these steps:
• The minor corresponding to the rank becomes the primary minor.
• The variables having coefficients in the primary minor become the main variables. The other variables become secondary, are noted using other letters and are moved to the other side of the equation.
• The equations containing the primary minor become primary equations.
• We solve the system formed only by the primary equations and we determine the solution of the system depending on the secondary variables.
• We write down the solution.

Exemplu 56
$\begin{cases} 2\cdot x + 3\cdot y +2\cdot z = \color{red}{5}\\ -3 \cdot x + 2\cdot y -3 \cdot z = \color{red}{-1}\\ 4\cdot x - y + 4\cdot z = \color{red}{3} \end{cases}$

The matrix associated to the system is:
$\begin{pmatrix} 2 & 3 & 2\\ -3 & 2 & -3\\ 4 & -1 & 4 \end{pmatrix}$

We determine the rank of the matrix.
$2\neq 0$

$\begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}= 4 + 9 =13 \neq0$

$\begin{vmatrix} 2 & 3 & 2\\ -3 & 2 & -3\\ 4 & -1 & 4 \end{vmatrix}=0$ (because it has two equal columns; consequently, the rank is 2)

The augmented matrix is:
$\begin{pmatrix} 2 & 3 & 2 & \color{red}{5}\\ -3 & 2 & -3 & \color{red}{-1}\\ 4 & -1 & 4 & \color{red}{3} \end{pmatrix}$

We determine the rank of the augmented matrix.
$2\neq 0$

$\begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}= 4 + 9 =13 \neq0$

$\begin{vmatrix} 2 & 3 & 2\\ -3 & 2 & -3\\ 4 & -1 & 4 \end{vmatrix}=0$
$\begin{vmatrix} 2 & 3 & \color{red}{5}\\ -3 & 2 & \color{red}{-1}\\ 4 & -1 & \color{red}{3} \end{vmatrix}=0$ (because it has two equal columns; the rank is also 2)

Since the ranks are equal, the system is consistent with an infinitude of solutions. The minor corresponding to the rank becomes the primary minor.
$\Delta_{p} = \begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}$

The variables x and y having coefficients in the primary minor become primary variables and z becomes a secondary variable. Let $z=\alpha$. The first two equations in which we find the primary minor become primary equations. We solve the system formed by the primary equations.
$\begin{cases} 2\cdot x + 3\cdot y +2\cdot \alpha = 5\\ -3 \cdot x + 2\cdot y -3 \cdot\alpha = -1\\ \end{cases}=$ $\begin{cases} 2\cdot x + 3\cdot y = 5 - 2\cdot \alpha\\ -3 \cdot x + 2\cdot y = -1 + 3\cdot\alpha\\ \end{cases}$

We multiply the first equation by 3 and the second one by 2.
$\begin{cases} 6\cdot x + 9\cdot y = 15 - 6\cdot \alpha\\ -6 \cdot x + 4\cdot y = -2 + 6 \cdot \alpha \\ \end{cases}$

We add up the two equations and we get:
$13\cdot y = 13 \Rightarrow y = \dfrac{13}{13} = 1$

We multiply the first equation by -2 and the second one by 3.
$\begin{cases} -4\cdot x - 6\cdot y = -10 + 4\cdot \alpha\\ -9 \cdot x + 6\cdot y = -3 + 9 \cdot \alpha \\ \end{cases}$

We add up the two equations and we get:
$-13\cdot x = 13 \Rightarrow y = \dfrac{13\cdot\alpha -13}{13} = \alpha -1$
The solution of the system is $\{\alpha-1;1;\alpha \}$

Example 57
$\begin{cases} 2\cdot x + y +5\cdot z = \color{red}{3}\\ 3 \cdot x + 2\cdot y +2 \cdot z = \color{red}{1}\\ 7\cdot x +y + 12\cdot z = \color{red}{2} \end{cases}$

The matrix associated to the system is:
$\begin{pmatrix} 2 & 1 & 5\\ 3 & 2 & 2\\ 7 & 4 & 12 \end{pmatrix}$

We determine the rank of the matrix.
$2\neq 0$
$\begin{vmatrix} 2 & 1\\ 3 & 2 \end{vmatrix}= 4 - 3 =1 \neq0$

$\begin{vmatrix} 2 & 1 & 5\\ 3 & 2 & 2\\ 7 & 4 & 12 \end{vmatrix}= 48 + 60 + 14 - 70 -16 -36 =0$ (consequently, the rank is 2)

The augmented matrix is:
$\begin{pmatrix} 2 & 1 & 5 & \color{red}{3}\\ 3 & 2 & 2 & \color{red}{1}\\ 7 & 4 & 12 & \color{red}{2} \end{pmatrix}$

We determine the rank of the augmented matrix.
$2\neq 0$
$\begin{vmatrix} 2 & 1\\ 3 & 2 \end{vmatrix}= 4 -3 =1 \neq0$

$\begin{vmatrix} 2 & 1 & 5\\ 3 & 2 & 2\\ 7 & 4 & 12 \end{vmatrix}=0$

$\begin{vmatrix} 2 & 1 & \color{red}{3}\\ 3 & 2 & \color{red}{1}\\ 7 & 4 & \color{red}{2} \end{vmatrix}= 8 + 36 + 7 - 42 -8 -6 = -5\neq 0$

The rank of the augmented matrix is 3.

Since the ranks of the two matrices are different, the system has no solutions. It's an inconsistent system. A homogeneous system will always be consistent with an infinitude of solutions, because the augmented matrix containing a column of zeroes will have the same rank as the matrix associated to the system.

Example 58
$\begin{cases} 2\cdot x + 3\cdot y +2\cdot z = \color{red}{0}\\ -3 \cdot x + 2\cdot y -3 \cdot z = \color{red}{0}\\ 4\cdot x - y + 4\cdot z = \color{red}{0} \end{cases}$

The matrix associated to the system is:
$\begin{pmatrix} 2 & 3 & 2\\ -3 & 2 & -3\\ 4 & -1 & 4 \end{pmatrix}$

We determine the rank of the matrix.
$2\neq 0$
$\begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}= 4 + 9 = 13 \neq0$

$\begin{vmatrix} 2 & 3 & 2\\ -3 & 2 & -3\\ 4 & -1 & 4 \end{vmatrix}=0$ (because it has two equal columns. consequently, the rank is 2)

The augmented matrix is:
$\begin{pmatrix} 2 & 3 & 2 & \color{red}{0}\\ -3 & 2 & -3 & \color{red}{0}\\ 4 & -1 & 4 & \color{red}{0} \end{pmatrix}$

We determine the rank of the augmented matrix.
$2\neq 0$
$\begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}= 4 + 9 =13 \neq0$

$\begin{vmatrix} 2 & 3 & 2\\ -3 & 2 & -3\\ 4 & -1 & 4 \end{vmatrix}=0$

$\begin{vmatrix} 2 & 3 & \color{red}{0}\\ -3 & 2 & \color{red}{0}\\ 4 & -1 & \color{red}{0} \end{vmatrix}=0$ (because it has a column of zeroes; consequently, the rank is also 2)

Since the ranks are equal, the system is consistent with an infinitude of solutions. The minor corresponding to the rank becomes the primary minor.
$\Delta_{p} = \begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}$

The variables x and y having coefficients in the primary minor become primary variables and z becomes a secondary variable. Let $z=\alpha$. The first two equations in which we find the primary minor become primary equations. We solve the system formed by the primary equations.
$\begin{cases} 2\cdot x + 3\cdot y +2\cdot \alpha = 0\\ -3 \cdot x + 2\cdot y -3 \cdot\alpha = 0\\ \end{cases}=$ $\begin{cases} 2\cdot x + 3\cdot y = - 2\cdot \alpha\\ -3 \cdot x + 2\cdot y = 3\cdot\alpha\\ \end{cases}$

We multiply the first equation by 3 and the second one by 2.
$\begin{cases} 6\cdot x + 9\cdot y = -6\cdot \alpha\\ -6 \cdot x + 4\cdot y = 6 \cdot \alpha \\ \end{cases}$

We add up the two equations and we get:
$13\cdot y = 0 \Rightarrow y = \dfrac{0}{13} = 0$
We do the same thing to find x. We multiply the first equation by -2 and the second one by 3.
$\begin{cases} -4\cdot x - 6\cdot y = 4\cdot \alpha\\ -9 \cdot x + 6\cdot y =9 \cdot \alpha \\ \end{cases}$

We add up the two equations and we get:
$-13\cdot x = 13 \Rightarrow y = \dfrac{13\cdot\alpha}{-13} = -\alpha$
The solution of the system is$\{-\alpha;0;\alpha \}$

## The Associated Matrix is not Square $(m\neq n)$

We calculate the rank of the matrix associated to the system and the rank of the augmented matrix (the initial matrix to which we add the column of constant terms).

We have the following situations:

• If the ranks of the two matrices are different, then the system has no solution. It is an inconsistent system.
• If the ranks are equal, then the system is consistent with an infinitude of solutions.
To solve the system, we follow these steps:
• The minor corresponding to the rank becomes the primary minor.
• The variables having coefficients in the primary minor become the main variables. The other variables become secondary, are noted using other letters and are moved to the other side of the equation.
• The equations containing the primary minor become primary equations.
• We solve the system formed only by the primary equations and we determine the solution of the system depending on the secondary variables.
• We write down the solution.

Example 59
$\begin{cases} 2\cdot x + 3\cdot y +2\cdot z = \color{red}{5}\\ -3 \cdot x + 2\cdot y -3 \cdot z = \color{red}{-1}\\ \end{cases}$

The matrix associated to the system is:
$\begin{pmatrix} 2 & 3 & 2\\ -3 & 2 & -3\\ \end{pmatrix}$

We determine the rank of the matrix.
$2\neq 0$
$\begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}= 4 + 9 =13 \neq0$ (the rank is 2)

The augmented matrix is:
$\begin{pmatrix} 2 & 3 & 2 & \color{red}{5}\\ -3 & 2 & -3 & \color{red}{-1}\\ \end{pmatrix}$

We determine the rank of the augmented matrix.
$2\neq 0$
$\begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}= 4 + 9 =13 \neq0$ (the rank is also 2)

Since the ranks are equal, the system is consistent with an infinitude of solutions. The minor corresponding to the rank becomes the primary minor.

$\Delta_{p} = \begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}$

The variables x and y having coefficients in the primary minor become primary variables and z becomes a secondary variable. Let $z=\alpha$. The first two equations in which we find the primary minor become primary equations. We solve the system formed by the primary equations.

$\begin{cases} 2\cdot x + 3\cdot y +2\cdot \alpha = 5\\ -3 \cdot x + 2\cdot y -3 \cdot\alpha = -1\\ \end{cases}=$ $\begin{cases} 2\cdot x + 3\cdot y = 5 - 2\cdot \alpha\\ -3 \cdot x + 2\cdot y = -1 + 3\cdot\alpha\\ \end{cases}$

We multiply the first equation by 3 and the second one by 2.
$\begin{cases} 6\cdot x + 9\cdot y = 15 - 6\cdot \alpha\\ -6 \cdot x + 4\cdot y = -2 + 6 \cdot \alpha \\ \end{cases}$

We add up the two equations and we get:
$13\cdot y = 13 \Rightarrow y = \dfrac{13}{13} = 1$
We do the same thing to find x. We multiply the first equation by -2 and the second one by 3.
$\begin{cases} -4\cdot x - 6\cdot y = -10 + 4\cdot \alpha\\ -9 \cdot x + 6\cdot y = -3 + 9 \cdot \alpha \\ \end{cases}$

We add up the two equations and we get:
$-13\cdot x = 13 \Rightarrow y = \dfrac{13\cdot\alpha -13}{13} = \alpha -1$
The solution of the system is $\{\alpha-1;1;\alpha \}$

Example 60
$\begin{cases} 2\cdot x + 3\cdot y = \color{red}{5}\\ -3 \cdot x + 2\cdot y = \color{red}{-1}\\ 4\cdot x - y = \color{red}{3} \end{cases}$

The matrix associated to the system is:
$\begin{pmatrix} 2 & 3 \\ -3 & 2 \\ 4 & -1 \end{pmatrix}$

We determine the rank of the matrix.
$2\neq 0$
$\begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}= 4 + 9 =13 \neq0$ (the rank is 2)

The augmented matrix is:
$\begin{pmatrix} 2 & 3 & \color{red}{5}\\ -3 & 2 & \color{red}{-1}\\ 4 & -1 & \color{red}{3} \end{pmatrix}$

We determine the rank of the augmented matrix.
$2\neq 0$
$\begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}= 4 + 9 =13 \neq0$
$\begin{vmatrix} 2 & 3 & \color{red}{5}\\ -3 & 2 & \color{red}{-1}\\ 4 & -1 & \color{red}{3} \end{vmatrix}=0$ (because it has two equal columns; consequently, the rank is 2)

Since the ranks are equal, the system is consistent with an infinitude of solutions. The minor corresponding to the rank becomes the primary minor.
$\Delta_{p} = \begin{vmatrix} 2 & 3\\ -3 & 2 \end{vmatrix}$

The variables x and y having coefficients in the primary minor become primary variables and z becomes a secondary variable. There are no secondary variables. The first two equations in which we find the primary minor become primary equations. We solve the system formed by the primary equations.
$\begin{cases} 2\cdot x + 3\cdot y = 5\\ -3 \cdot x + 2\cdot y = -1\\ \end{cases}$

We multiply the first equation by 3 and the second one by 2.
$\begin{cases} 6\cdot x + 9\cdot y = 15\\ -6 \cdot x + 4\cdot y = -2 \\ \end{cases}$

We add up the two equations and we get:
$13\cdot y = 13 \Rightarrow y = \dfrac{13}{13} = 1$
We do the same thing to find x. We multiply the first equation by -2 and the second one by 3.
$\begin{cases} -4\cdot x - 6\cdot y = -10\\ -9 \cdot x + 6\cdot y = -3\\ \end{cases}$

We add up the two equations and we get:
$-13\cdot x = -13 \Rightarrow y = \dfrac{-13}{-13} = 1$
We check if the results are valid solutions of the secondary equation.
$4\cdot1 -1\cdot1 = 3$
The solution of the system is $\{1;1 \}$

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