Analytical solution of the spacecraft magnetometer calibration problem using the method of least squares

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Introduction

Magnetometers are part of the orientation and stabilization system of low-orbit small-sized spacecraft. They are widely used due to the fact that they are lightweight, inexpensive and reliable. However, due to the physical properties of the sensitive element, modern magnetometers require mathematical calibration. At present, various methods for calibrating magnetometers have been proposed, and a considerable number of scientific papers are devoted to these methods [1–12], in particular, article [9], which provides an overview of various methods for performing such operations.

In the above-mentioned works, the problem of calibrating the magnetometer of a spacecraft was solved using numerical methods. In this article, an analytical method for solving this problem for the model considered in [1] is proposed.

1. Model of errors in measurements of magnetic induction vector

We denote by h the value of the measured magnetic induction vector at a certain spatial position of the magnetometer unit (MU). We use the simplified measurement model considered in  [1]:

$\text{h}=\left(\begin{array}{c}{h}_1\\ h_2\\ h_3\end{array}\right)=QP\text{B}+\text{b}+\text{n}.$                                                                                       (1)

The following notations [1] are used in(1):

B = (B₁, B₂, B₃)^T is a true magnetic induction vector;

b = (b₁, b₂, b₃)^T is a constant vector corresponding to zero offsets for each of the measuring axes of the MU;

n is a random vector corresponding to uncorrelated noise for each of the measurement axes;

Р is a matrix, the rows of which are the unit vectors of the measuring axes of the MU, written in the “base” MU system;

Q is a diagonal matrix containing on the main diagonal the scaling factors k₁, k₂, k₃  for measuring axes MU i.e

.$Q=\left(\begin{array}{ccc}{k}_1& 0& 0\\ 0& k_2& 0\\ 0& 0& k_3\end{array}\right).$

In other words, the matrix P describes the non-orthogonality of the MU measurement axes, and the matrix Q corresponds to scaling along these axes.

The task of calibrating the MU measuring axes comes down to finding the elements of the matrices P and Q, as well as the zero offset vector b.

2. Development of an algorithm for determining the calibration parameters of the MU

 When solving the problem of MU determining the calibration parameters, we will use the fact that for measurements with any  MU spatial orientation, the value of the measured magnetic induction vector B is preserved and is a known model value.

We will assume that in flight, as a result of magnetometer measurements at discrete moments in time, a set of vectors is obtained h^(l) = (h₁^(l), h₂^(l), h₃^(l))^T, l = 1, …, N. Provided there is no measurement noise, from formula (1) we obtain:

${\text{h}}^{\left(l\right)}=QP{\text{B}}^{\left(l\right)}+\text{b},l=\mathrm{1,}\dots ,N,$                                                                          (2)

where B^(l) = (B₁^(l), B₂^(l), B₃^(l))^T is the true vector of magnetic induction at the same point in space as the measured vector h^(l), l = 1, …, N. We express vectors   (l = 1, …, N) from equalities  (2):

${\text{B}}^{\left(l\right)}=S\left({\text{h}}^{\left(l\right)}-\text{b}\right),$                                                                                               (3)

l = 1, …, N, где

$S={\left(QP\right)}^{-1}=P^{-1}{Q}^{-1}.$                                                                                    (4)

As noted in [1], without loss of generality, the non-orthogonality matrix P can be represented with a minimum number of unknown elements as follows:

$P=\left(\begin{array}{ccc}1& 0& 0\\ {\text{sin\hspace{0.17em}ε}}_1& {\text{cos\hspace{0.17em}ε}}_1& 0\\ {\text{sin\hspace{0.17em}ε}}_2& {\text{cos\hspace{0.17em}ε}}_2{\text{sin\hspace{0.17em}ε}}_3& {\text{cos\hspace{0.17em}ε}}_2{\text{cos\hspace{0.17em}ε}}_3\end{array}\right),$

where ε₁, ε₂, ε₃ are small angles. Then the inverse matrix P ^{– 1} will take the form

$P^{-1}=\left(\begin{array}{ccc}1& 0& 0\\ -{\text{α}}_1& {\text{β}}_1& 0\\ {\text{α}}_1{\text{α}}_3-{\text{α}}_2{\text{β}}_3& -{\text{β}}_1{\text{α}}_3& {\text{β}}_2{\text{β}}_3\end{array}\right),$

where α_i = tg ε_i, β_i = sec ε_i, i = 1, 2, 3, and since

$Q^{-1}=\left(\begin{array}{ccc}{k_1}^{-1}& 0& 0\\ 0& {k_2}^{-1}& 0\\ 0& 0& {k_3}^{-1}\end{array}\right),$

then in accordance with (4) we obtain:

$S=\left(\begin{array}{ccc}{\text{γ}}_1& 0& 0\\ -{\text{α}}_1{\text{γ}}_1& {\text{β}}_1{\text{γ}}_2& 0\\ \left({\text{α}}_1{\text{α}}_3-{\text{α}}_2{\text{β}}_3\right){\text{γ}}_1& -{\text{β}}_1{\text{α}}_3{\text{γ}}_2& {\text{β}}_2{\text{β}}_3{\text{γ}}_3\end{array}\right),$                                        (5)

where γ_i = k_i ^{– 1}, i = 1, 2, 3.

We rewrite equality (3) taking into account (5):

$\left(\begin{array}{c}{B}_1^{\left(l\right)}\\ B_2^{\left(l\right)}\\ B_3^{\left(l\right)}\end{array}\right)=\left(\begin{array}{ccc}{\text{γ}}_1& 0& 0\\ -{\text{α}}_1{\text{γ}}_1& {\text{β}}_1{\text{γ}}_2& 0\\ \left({\text{α}}_1{\text{α}}_3-{\text{α}}_2{\text{β}}_3\right){\text{γ}}_1& -{\text{β}}_1{\text{α}}_3{\text{γ}}_2& {\text{β}}_2{\text{β}}_3{\text{γ}}_3\end{array}\right)\left(\begin{array}{c}{h}_1^{\left(l\right)}-b_1\\ h_2^{\left(l\right)}-b_2\\ h_3^{\left(l\right)}-b_3\end{array}\right),$               (6)

l = 1, …, N. We write each of the N vector equalities (6) as a system of three scalar equalities:

$\left\{\begin{array}{l}{\text{γ}}_1\left(h_1^{\left(l\right)}-b_1\right)=B_1^{\left(l\right)},\\ -{\text{α}}_1{\text{γ}}_1\left(h_1^{\left(l\right)}-b_1\right)+{\text{β}}_1{\text{γ}}_2\left(h_2^{\left(l\right)}-b_2\right)=B_2^{\left(l\right)},\\ \left({\text{α}}_1{\text{α}}_3-{\text{α}}_2{\text{β}}_3\right){\text{γ}}_1\left(h_1^{\left(l\right)}-b_1\right)-{\text{β}}_1{\text{α}}_3{\text{γ}}_2\left(h_2^{\left(l\right)}-b_2\right)+{\text{β}}_2{\text{β}}_3{\text{γ}}_3\left(h_3^{\left(l\right)}-b_3\right)=B_3^{\left(l\right)},\end{array}\right.$                 (7)

l = 1, …, N. Due to the first equality of system (7), we rewrite the second equality of this system in the form

$-{\text{α}}_1{B}_1^{\left(l\right)}+{\text{β}}_1{\text{γ}}_2\left(h_2^{\left(l\right)}-b_2\right)=B_2^{\left(l\right)},$

${\text{β}}_1{\text{γ}}_2\left(h_2^{\left(l\right)}-b_2\right)={\text{α}}_1{B}_1^{\left(l\right)}+B_2^{\left(l\right)},$

as a result of which the third equality of system (7) is written as follows:

$-{\text{α}}_2{\text{β}}_3{B}_1^{\left(l\right)}-{\text{α}}_3{B}_2^{\left(l\right)}+{\text{β}}_2{\text{β}}_3{\text{γ}}_3\left(h_3^{\left(l\right)}-b_3\right)=B_3^{\left(l\right)},$

l = 1, …, N.

We introduce into consideration a function of twelve variables α_i, β_i, γ_i, b_i (i = 1, 2, 3)

$\Phi =\sum _{l=1}^N$$\left\{\begin{array}{l}{\left[B_1^{\left(l\right)}\text{}-{\text{γ}}_1\left(h_1^{\left(l\right)}-b_1\right)\right]}^2\text{}+{\left[B_2^{\left(l\right)}\text{}+{\text{α}}_1{B}_1^{\left(l\right)}\text{}-{\text{β}}_1{\text{γ}}_2\left(h_2^{\left(l\right)}-b_2\right)\right]}^2\text{}+\\ +{\left[B_3^{\left(l\right)}\text{}+{\text{α}}_2{\text{β}}_3{B}_1^{\left(l\right)}\text{}+{\text{α}}_3{B}_2^{\left(l\right)}\text{}-{\text{β}}_2{\text{β}}_3{\text{γ}}_3\left(h_3^{\left(l\right)}\text{}-b_3\right)\right]}^2\end{array}\right\},$

where the variables  α_i и β_i are related by the trigonometric identity

${\text{tg}}^2\text{φ}\equiv {\sec }^2\text{φ}-1$

by the equalities

${{\text{α}}_i}^2-{{\text{β}}_i}^2+1=\mathrm{0,}$                                                                         (8)

i = 1, 2, 3.

The algorithm for solving the problem of calibrating the MU measuring axes is reduced to searching by the least squares method [13] taking into account (8) such values of variables α_i, β_i, γ_i, b_i (i = 1, 2, 3), which, for a given set of measurement vectors,   {h^(l)} (l = 1, …, N) provide a minimum of the function F. For this purpose, it is necessary to investigate the function F for a conditional extremum [14] in the presence of three constraint equations (8).

We compose the Lagrange function

$F=\Phi +\sum _{i=1}^3{\lambda }_i\left({{\text{α}}_i}^2-{{\text{β}}_i}^2+1\right)$,                                                                                  (9)

fifteen variables  α_i, β_i, γ_i, b_i, λ_i (i = 1, 2, 3) and we write down the necessary condition for the local extremum of this function:

It is required to find the stationary points of the function F, i.e. the solution of the system of equations (10). Let us introduce the following notations:

$C_i=\sum _{l=1}^N{B}_i^{\left(l\right)},H_i=\sum _{l=1}^N{h}_i^{\left(l\right)},F_i=\sum _{l=1}^N{\left(h_i^{\left(l\right)}\right)}^2,D_{i\text{}j}=\sum _{l=1}^N{B}_i^{\left(l\right)}{h}_j^{\left(l\right)},G_{i\text{}j}=G_{j\text{\hspace{0.05em}}i}=\sum _{l=1}^N{B}_i^{\left(l\right)}{B}_j^{\left(l\right)},i,j=\mathrm{1,2,3.}$

Taking into account these notations, for convenience we will divide the system of equations (10) into three nonlinear systems – a system of equations for two unknowns b₁, γ₁:

$\left\{\begin{array}{l}{D}_{11}-C_1{b}_1+\left(-F_1+2H_1{b}_1-N{b_1}^2\right){\text{γ}}_1=\mathrm{0,}\\ \left[C_1+\left(N{b}_1-H_1\right){\text{γ}}_1\right]{\text{γ}}_1=\mathrm{0,}\end{array}\right.$                                                       (11)

a system of equations for five unknowns  b₂, α₁, β₁, γ₂, λ₁:

$C_i=\sum _{l=1}^N{B}_i^{\left(l\right)},H_i=\sum _{l=1}^N{h}_i^{\left(l\right)},F_i=\sum _{l=1}^N{\left(h_i^{\left(l\right)}\right)}^2,D_{i\text{}j}=\sum _{l=1}^N{B}_i^{\left(l\right)}{h}_j^{\left(l\right)},G_{i\text{}j}=G_{j\text{\hspace{0.05em}}i}=\sum _{l=1}^N{B}_i^{\left(l\right)}{B}_j^{\left(l\right)},i,j=\mathrm{1,2,3.}$

$\left\{\begin{array}{l}\left(C_1{b}_2-D_{12}\right){\text{β}}_1{\text{γ}}_2+\left(G_{11}+{\lambda }_1\right){\text{α}}_1+G_{12}=\mathrm{0,}\\ \left[D_{22}-C_2{b}_2+\left(D_{12}-C_1{b}_2\right){\text{α}}_1+\left(-F_2+2H_2{b}_2-N{b_2}^2\right){\text{β}}_1{\text{γ}}_2\right]{\text{γ}}_2+{\lambda }_1{\text{β}}_1=\mathrm{0,}\\ \left[D_{22}-C_2{b}_2+\left(D_{12}-C_1{b}_2\right){\text{α}}_1+\left(-F_2+2H_2{b}_2-N{b_2}^2\right){\text{β}}_1{\text{γ}}_2\right]{\text{β}}_1=\mathrm{0,}\\ \left[C_2+C_1{\text{α}}_1+\left(N{b}_2-H_2\right){\text{β}}_1{\text{γ}}_2\right]{\text{β}}_1{\text{γ}}_2=\mathrm{0,}\\ {\text{β}}_1=\sqrt{{{\text{α}}_1}^2+1}\end{array}\right.$       (12)

and a system of equations for eight unknowns.  b₃, α₂, α₃, β₂, β₃, γ₃, λ₂, λ₃:

$\left\{\begin{array}{l}\left[G_{13}+G_{11}{\text{α}}_2{\text{β}}_3+G_{12}{\text{α}}_3+\left(C_1{b}_3-D_{13}\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3\right]{\text{β}}_3+{\lambda }_2{\text{α}}_2=\mathrm{0,}\\ \left[D_{33}-C_3{b}_3+\left(D_{13}-C_1{b}_3\right){\text{α}}_2{\text{β}}_3+\left(D_{23}-C_2{b}_3\right){\text{α}}_3+\left(-F_3+2H_3{b}_3-N{b_3}^2\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3\right]{\text{β}}_3{\text{γ}}_3+{\lambda }_2{\text{β}}_2=\mathrm{0,}\\ \left[D_{33}-C_3{b}_3+\left(D_{13}-C_1{b}_3\right){\text{α}}_2{\text{β}}_3+\left(D_{23}-C_2{b}_3\right){\text{α}}_3+\left(-F_3+2H_3{b}_3-N{b_3}^2\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3\right]{\text{β}}_2{\text{β}}_3=\mathrm{0,}\\ \left[C_3+C_1{\text{α}}_2{\text{β}}_3+C_2{\text{α}}_3+\left(N{b}_3-H_3\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3\right]{\text{β}}_2{\text{β}}_3{\text{γ}}_3\text{=0,}\\ G_{23}+G_{12}{\text{α}}_2{\text{β}}_3+G_{22}{\text{α}}_3+\left(C_2{b}_3-D_{23}\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3+{\lambda }_3{\text{α}}_3=\mathrm{0,}\\ \left[G_{13}+G_{11}{\text{α}}_2{\text{β}}_3+G_{12}{\text{α}}_3+\left(C_1{b}_3-D_{13}\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3\right]{\text{α}}_2-{\lambda }_3{\text{β}}_3-\\ -\left[D_{33}-C_3{b}_3+\left(D_{13}-C_1{b}_3\right){\text{α}}_2{\text{β}}_3+\left(D_{23}-C_2{b}_3\right){\text{α}}_3+\left(-F_3+2H_3{b}_3-N{b_3}^2\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3\right]{\text{β}}_2{\text{γ}}_3=\mathrm{0,}\\ {\text{β}}_i=\sqrt{{{\text{α}}_i}^2+1},i=\mathrm{2,3.}\end{array}\right.$(13)

неравенства Коши

(13)

In writing the expressions β_i through α_i in the last equations of systems (12) and (13), we used the fact that β_i = sec ε_i > 0 due to the obvious inequalities–π/2 < ε_i < π/2, i = 1, 2, 3. We solve each of the systems of equations (11), (12), (13).

First, we note that the inequality is valid

$-F_i+2H_i{b}_i-N{b_i}^2\ne \mathrm{0,}$                                                                                    (14)

i = 1, 2, 3. Indeed, the discriminant of the quadratic equation is – F_i + 2H_ib_i – Nb_i² = 0 with respect to the unknown b_i, being equal to

${\delta }_i=4{H_i}^2-4N{F}_i=4\left[{\left(\sum _{l=1}^N{h}_i^{\left(l\right)}\right)}^2-N\sum _{l=1}^N{\left(h_i^{\left(l\right)}\right)}^2\right],$

is negative due to the obvious consequence

${\left(\sum _{l=1}^N{h}_i^{\left(l\right)}\right)}^2<N\sum _{l=1}^N{\left(h_i^{\left(l\right)}\right)}^2$

of the Cauchy–Bunyakovsky inequality [15]  (here the sign “<” is used instead of “≤”, since at least one of the terms h_i⁽¹⁾, …, h_i^(N) is obviously not equal to zero), therefore the quadratic equation under consideration has no real roots, i = 1, 2, 3.

We solve the system of equations (11). Taking into account inequality (14), we express γ₁ through b₁ from the first equation of this system:

${\text{γ}}_1=\frac{D_{11}-C_1{b}_1}{F_1-2H_1{b}_1+N{b_1}^2}.$

Substituting the obtained expression for γ₁ into the second equation of system (11) and taking into account the inequality γ₁ = k₁^{– 1} ≠ 0, after simple transformations we arrive at an equality,

$b_1=\frac{C_1{F}_1-D_{11}{H}_1}{C_1{H}_1-N{D}_{11}},$

taking into account which the equality written above for γ1 is reduced to the form

$b_1=\frac{C_1{F}_1-D_{11}{H}_1}{C_1{H}_1-N{D}_{11}},$

Thus, the solution to the system of equations (11) is found.

We consider the system of equations (12). Since  γ₂ = k₂ ^{– 1} ≠ 0, the second equation of this system can be written as

$D_{22}-C_2{b}_2+\left(D_{12}-C_1{b}_2\right){\text{α}}_1+\left(-F_2+2H_2{b}_2-N{b_2}^2\right){\text{β}}_1{\text{γ}}_2+\frac{{\lambda }_1{\text{β}}_1}{{\text{γ}}_2}=\mathrm{0,}$                 (15)

since β₁ = sec ε₁ ≠ 0, the third equation of system (12) is equivalent to the following equation:

$D_{22}-C_2{b}_2+\left(D_{12}-C_1{b}_2\right){\text{α}}_1+\left(-F_2+2H_2{b}_2-N{b_2}^2\right){\text{β}}_1{\text{γ}}_2=0.$                                (16)

From (15), taking into account (16), we obtain that λ₁β₁γ₂^{– 1} = 0 and then, by virtue of the inequality β₁ ≠ 0, we have

λ₁ = 0                                                                         (17)

The inequality D₁₂ – C₁b₂ ≠ 0 holds, since a direct check shows that the value b₂ = D₁₂C₁^{– 1}, which is the root of the equation D₁₂ – C₁b₂ = 0, does not satisfy the equations of system (12)

First, we consider the case H₂ – Nb₂ ≠ 0, i.e. b₂ ≠ H₂N^{– 1}. We express β₁γ₂ from (16), as well as from the first (taking into account equality (17)) and fourth (taking into account the inequality β₁γ₂ ≠ 0) equations of system (12):

${\text{β}}_1{\text{γ}}_2=\frac{\left(D_{12}-C_1{b}_2\right){\text{α}}_1+D_{22}-C_2{b}_2}{F_2-2H_2{b}_2+N{b_2}^2}, {\text{β}}_1{\text{γ}}_2=\frac{G_{11}{\text{α}}_1+G_{12}}{D_{12}-C_1{b}_2}, {\text{β}}_1{\text{γ}}_2=\frac{C_1{\text{α}}_1+C_2}{H_2-N{b}_2}.$        (18)

As proven, the denominator of the fraction in the second of the equalities (18) is not equal to zero, the denominator of the fraction in the first of the equalities (18) is different from zero by virtue of (14). From the first and third equalities (18) follows the equation

$\frac{C_1{\text{α}}_1+C_2}{H_2-N{b}_2}=\frac{\left(D_{12}-C_1{b}_2\right){\text{α}}_1+D_{22}-C_2{b}_2}{F_2-2H_2{b}_2+N{b_2}^2},$

we express b₂ through α₁:

$b_2=\frac{\left(D_{12}{H}_2-C_1{F}_2\right){\text{α}}_1+D_{22}{H}_2-C_2{F}_2}{\left(N{D}_{12}-C_1{H}_2\right){\text{α}}_1+N{D}_{22}-C_2{H}_2}.$                                                                          (19)

Similarly, from the second and third equalities (18) follows the equation

$\frac{C_1{\text{α}}_1+C_2}{H_2-N{b}_2}=\frac{G_{11}{\text{α}}_1+G_{12}}{D_{12}-C_1{b}_2},$

we express b₂ through α₁:

$b_2=\frac{\left(G_{11}{H}_2-C_1{D}_{12}\right){\text{α}}_1+G_{12}{H}_2-C_2{D}_{12}}{\left(N{G}_{11}-{C_1}^2\right){\text{α}}_1+N{G}_{12}-C_1{C}_2}.$                                                                        (20)

Equating the right-hand sides of equalities (19) and (20), after simple transformations we obtain a quadratic equation with respect to α₁:

$\begin{array}{l}\left[{C_1}^2\left(C_1{F}_2-D_{12}{H}_2\right)+C_1{G}_{11}\left({H_2}^2-N{F}_2\right)+C_1{D}_{12}\left(N{D}_{12}-C_1{H}_2\right)\right]{{\text{α}}_1}^2+\\ +\left[2C_1{C}_2\left(C_1{F}_2-D_{12}{H}_2\right)+\left(N{D}_{12}-C_1{H}_2\right)\left(C_1{D_2}_2+C_2{D}_{12}\right)+\left({H_2}^2-N{F}_2\right)\left(C_1{G_1}_2+C_2{G}_{11}\right)\right]{\text{α}}_1+\\ +{C_2}^2\left(C_1{F}_2-D_{12}{H}_2\right)+C_2{G}_{12}\left({H_2}^2-N{F}_2\right)+C_2{D}_{22}\left(N{D}_{12}-C_1{H}_2\right)=0.\end{array}$  (21)

The discriminant of equation (21) is equal to

and the roots of this equation are calculated using the formulae

$\left[\begin{array}{l}{\text{α}}_1=-\frac{C_2}{C_1},\\ {\text{α}}_1=-\frac{C_2\left(C_1{F}_2-D_{12}{H}_2\right)+G_{12}\left({H_2}^2-N{F}_2\right)+D_{22}\left(N{D}_{12}-C_1{H}_2\right)}{C_1\left(C_1{F}_2-D_{12}{H}_2\right)+G_{11}\left({H_2}^2-N{F}_2\right)+D_{12}\left(N{D}_{12}-C_1{H}_2\right)}.\end{array}\right.$                              (22)

 We note that the values (22) of the unknown α₁ are not the roots of the denominators of the fractions in equalities (19) and (20). Substituting the first of the values (22) of the unknown α1 into the fourth equation of system (12), taking into account the inequality β₁γ₂ ≠ 0, we arrive at the equality H₂ – Nb₂ = 0, which contradicts the case under consideration.

 Now H₂ – Nb₂ = 0, i.e. b₂ = H₂N^{– 1}. Taking into account the inequality β₁γ₂ ≠ 0, from the fourth equation of system (12) we obtain that α₁ = – C₂C₁^{– 1}, i.e. the found value of α1 coincides with the first of the roots (22) of equation (21). It is easy to show that equalities (19) and (20), proven under the assumption b₂ ≠ H₂N^{– 1}, are also valid forb₂ = H₂N^{– 1}. In the case b₂ = H₂N^{– 1}, the first two equalities from (18) also hold, since no restrictions on the value of the unknown b₂ were taken into account when deriving them.

From the above reasoning it follows that the system of equations (12) has two solutions. The components of both solutions of the specified system are obtained in the following order:

• – the values of the unknown α₁ are found by formulae (22);
• – the values of β₁ corresponding to the found values of α₁ are calculated by the formula written as the last equation of the system (12);
• – the values of b₂ – by any of the formulae (19), (20);
•  – the values of γ₂ – by any of the two formulae

${\text{γ}}_2=\frac{\left(D_{12}-C_1{b}_2\right){\text{α}}_1+D_{22}-C_2{b}_2}{{\text{β}}_1\left(F_2-2H_2{b}_2+N{b_2}^2\right)}, {\text{γ}}_2=\frac{G_{11}{\text{α}}_1+G_{12}}{{\text{β}}_1\left(D_{12}-C_1{b}_2\right)},$   

which follow respectively from the first and second equalities (18);

– the value of the unknown λ₁ in both solutions of the system of equations (12) due to (17) is taken to be equal to zero.

We move on to finding solutions to the system of equations (13). Due to the inequalities γ₃ = k₃^– 1 ≠ 0, β₂ = sec ε₂ ≠ 0 and β₃ = sec ε₃ ≠ 0 the second and third equations of this system can be written as

$D_{33}-C_3{b}_3+\left(D_{13}-C_1{b}_3\right){\text{α}}_2{\text{β}}_3+\left(D_{23}-C_2{b}_3\right){\text{α}}_3+\left(-F_3+2H_3{b}_3-N{b_3}^2\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3+\frac{{\lambda }_2{\text{β}}_2}{{\text{β}}_3{\text{γ}}_3}=\mathrm{0,}$    (23)

$D_{33}-C_3{b}_3+\left(D_{13}-C_1{b}_3\right){\text{α}}_2{\text{β}}_3+\left(D_{23}-C_2{b}_3\right){\text{α}}_3+\left(-F_3+2H_3{b}_3-N{b_3}^2\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3=0$                  (24)

respectively. From (23), taking into account (24), we obtain: λ₂β₂(β₃γ₃)^–1 = 0. Consequently, the equality is valid

${\lambda }_2=0.$                                                                                                         (25)

 Then the first equation of system (13), taking into account the inequalityβ₃ ≠ 0, can be written in the form

$G_{13}+G_{11}{\text{α}}_2{\text{β}}_3+G_{12}{\text{α}}_3+\left(C_1{b}_3-D_{13}\right){\text{β}}_2{\text{β}}_3{\text{γ}}_3=0.$                                      (26)

 From the sixth equation of system (13), taking into account (24) and (26), we find – λ₃β₃ = 0, from which the equality follows

${\lambda }_3=0.$                                                                                                         (27)

The inequalities H₃ – Nb₃ ≠ 0, D_i3 – C_ib₃ ≠ 0 take place, since a direct check shows that neither the root b₃ = H₃N^{– 1} of the equation H₃ – Nb₃ = 0, nor the roots b₃ = D_i3C_i^{– 1} of the equations D_i3 – C_ib₃ = 0 satisfy the equations of system (13), i = 1, 2.

Taking into account equalities (25), (27) and inequalities β₂ ≠ 0, β₃ ≠ 0, γ₃ ≠ 0, we express β₂β₃γ₃ from the first, third, fourth and fifth equations of system (13):

$\begin{array}{l}{\text{β}}_2{\text{β}}_3{\text{γ}}_3=\frac{G_{13}+G_{11}{\text{α}}_2{\text{β}}_3+G_{12}{\text{α}}_3}{D_{13}-C_1{b}_3},\\ {\text{\hspace{1em}β}}_2{\text{β}}_3{\text{γ}}_3=\frac{D_{33}-C_3{b}_3+\left(D_{13}-C_1{b}_3\right){\text{α}}_2{\text{β}}_3+\left(D_{23}-C_2{b}_3\right){\text{α}}_3}{F_3-2H_3{b}_3+N{b_3}^2},\\ {\text{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}β}}_2{\text{β}}_3{\text{γ}}_3=\frac{C_3+C_1{\text{α}}_2{\text{β}}_3+C_2{\text{α}}_3}{H_3-N{b}_3},\\ {\text{\hspace{1em}β}}_2{\text{β}}_3{\text{γ}}_3=\frac{G_{23}+G_{12}{\text{α}}_2{\text{β}}_3+G_{22}{\text{α}}_3}{D_{23}-C_2{b}_3}.\end{array}$                         (28)

As proven, the denominators of the fractions in the first, third and fourth equalities (28) are not equal to zero, the denominator of the fraction in the second equalities (28) is different from zero by virtue of (14). From the first and third equalities (28) follows the equation

$\frac{G_{13}+G_{11}{\text{α}}_2{\text{β}}_3+G_{12}{\text{α}}_3}{D_{13}-C_1{b}_3}=\frac{C_3+C_1{\text{α}}_2{\text{β}}_3+C_2{\text{α}}_3}{H_3-N{b}_3},$

from the second and third equalities (28) follows the equation

$\frac{D_{33}-C_3{b}_3+\left(D_{13}-C_1{b}_3\right){\text{α}}_2{\text{β}}_3+\left(D_{23}-C_2{b}_3\right){\text{α}}_3}{F_3-2H_3{b}_3+N{b_3}^2}=\frac{C_3+C_1{\text{α}}_2{\text{β}}_3+C_2{\text{α}}_3}{H_3-N{b}_3},$

from the third and thorthequalities (28) follows the equation

$\frac{C_3+C_1{\text{α}}_2{\text{β}}_3+C_2{\text{α}}_3}{H_3-N{b}_3}=\frac{G_{23}+G_{12}{\text{α}}_2{\text{β}}_3+G_{22}{\text{α}}_3}{D_{23}-C_2{b}_3}.$

From the last three equations we express α₂β₃ through α₃ and b₃:

${\text{α}}_2{\text{β}}_3=$$\frac{\left[G_{12}\left(H_3-N{b}_3\right)-C_2\left(D_{13}-C_1{b}_3\right)\right]{\text{α}}_3+G_{13}\left(H_3-N{b}_3\right)-C_3\left(D_{13}-C_1{b}_3\right)}{C_1\left(D_{13}-C_1{b}_3\right)-G_{11}\left(H_3-N{b}_3\right)},$           (29)

${\text{α}}_2{\text{β}}_3\text{}=$$\frac{\left[\text{}\left(H_3\text{}-\text{}N{b}_3\right)\left(D_{23}\text{}-C_2{b}_3\right)-C_2\text{}\left(\text{}N{b_3}^2\text{}-2H_3{b}_3\text{}+\text{}{F}_3\right)\text{}\right]{\text{α}}_3\text{}+\left(H_3\text{}-N{b}_3\right)\left(D_{33}\text{}-C_3{b}_3\right)-C_3\text{}\left(\text{}N{b_3}^2\text{}-2H_3{b}_3\text{}+\text{}{F}_3\right)}{C_1\left(N{b_3}^2-2H_3{b}_3+F_3\right)-\left(H_3-N{b}_3\right)\left(D_{13}-C_1{b}_3\right)},$ (30)

${\text{α}}_2{\text{β}}_3=$$\frac{\left[C_2\left(D_{23}-C_2{b}_3\right)-G_{22}\left(H_3-N{b}_3\right)\right]{\text{α}}_3+C_3\left(D_{23}-C_2{b}_3\right)-G_{23}\left(H_3-N{b}_3\right)}{G_{12}\left(H_3-N{b}_3\right)-C_1\left(D_{23}-C_2{b}_3\right)}.$           (31)

By equating the right-hand side of equality (29) to the right-hand sides of equalities (30) and (31), after simple transformations we obtain the equations

$\begin{array}{l}\left\{\left[C_2\text{}{\left(C_1{b}_3-D_{13}\right)}^2\text{}+G_{11}\left(\text{}N{b}_3-H_3\right)\left(C_2{b}_3-D_{23}\text{}\right)-G_{12}\text{}\left(\text{}N{b}_3-H_3\right)\left(C_1{b}_3-D_{13}\right)-C_1\text{}\left(C_1{b}_3-D_{13}\right)\left(C_2{b}_3-D_{23}\text{}\right)+\right.\right.\\ \left.+\left(C_1{G}_{12}\text{}-C_2{G}_{11}\right)\left(N{b_3}^2\text{}-2H_3{b}_3\text{} +F_3\right)\right]{\text{α}}_3\text{}-\left(N{b_3}^2-2H_3{b}_3+F_3\right)\left(C_3{G}_{11}\text{}-C_1{G}_{13}\right)-C_1\left(C_1{b}_3-D_{13}\right)\left(C_3{b}_3-D_{33}\right)-\\ \left.-G_{13}\left(\text{}N{b}_3-H_3\right)\left(C_1{b}_3-D_{13}\right)+G_{11}\left(\text{}N{b}_3-H_3\right)\left(C_3{b}_3-D_{33}\right)+C_3{\left(C_1{b}_3-D_{13}\right)}^2\right\}\left(\text{}{H}_3-N{b}_3\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}\left\{\left[\left(C_2{G}_{12}-C_1{G}_{22}\right)\left(C_1{b}_3-D_{13}\text{}\right)+\left(G_{11}{G}_{22}-{G_{12}}^2\right)\left(\text{}N{b}_3-H_3\right)+\left(C_1{G}_{12}-C_2{G}_{11}\right)\left(C_2{b}_3-D_{23}\right)\right]\right.{\text{α}}_3+\\ \left.+\left(C_3{G}_{12}-C_1{G}_{23}\right)\left(C_1{b}_3-D_{13}\text{}\right)+\left(G_{11}{G}_{23}-G_{12}{G}_{13}\right)\left(\text{}N{b}_3-H_3\right)+\left(C_1{G}_{13}-C_3{G}_{11}\right)\left(C_2{b}_3-D_{23}\right)\right\}\left(\text{}{H}_3-N{b}_3\right)=\mathrm{0,}\end{array}$

from which we express α3 through b3, having previously divided both parts of each of them by (H₃ – Nb₃):

${\text{α}}_3=$$\frac{\left(H_3{b}_3-F_3\right)\left(C_1{G}_{13}-C_3{G}_{11}\right)+\left(N{b}_3-H_3\right)\left(D_{33}{G}_{11}-D_{13}{G}_{13}\right)+\left(C_1{b}_3-D_{13}\right)\left(C_3{D}_{13}-C_1{D}_{33}\right)}{\left(H_3{b}_3-F_3\right)\left(C_2{G}_{11}-C_1{G}_{12}\right)+\left(N{b}_3-H_3\right)\left(D_{13}{G}_{12}-D_{23}{G}_{11}\right)+\left(C_1{b}_3-D_{13}\right)\left(C_1{D}_{23}-C_2{D}_{13}\right)},$  (32)

${\text{α}}_3=$$\frac{\left(D_{23}-C_2{b}_3\right)\left(C_1{G}_{13}-C_3{G}_{11}\right)+\left(N{b}_3-H_3\right)\left(G_{12}{G}_{13}-G_{11}{G}_{23}\right)+\left(C_1{b}_3-D_{13}\right)\left(C_1{G}_{23}-C_3{G}_{12}\right)}{\left(D_{23}-C_2{b}_3\right)\left(C_2{G}_{11}-C_1{G}_{12}\right)+\left(\text{}N{b}_3-H_3\right)\left(G_{11}{G}_{22}-{G_{12}}^2\right)+\left(C_1{b}_3-D_{13}\right)\left(C_2{G}_{12}-C_1{G}_{22}\right)}.$ (33)

Having equated the right-hand sides of equalities (32) and (33), after the simplest transformations we arrive at a quadratic equation with respect to b₃

$\Gamma {b_3}^2+\Lambda b_3+\Omega =\mathrm{0,}$                                                           (34)

the coefficients of which are determined as follows:

$\begin{array}{l}\mathrm{\Gamma }=\left[G_{13}\left(C_1{H}_3\text{}-N{D}_{13}\right)+C_3\left(C_1{D}_{13}\text{}-G_{11}{H}_3\right)+D_{33}\left(N{G}_{11}\text{}-{C_1}^2\right)\right]\left[G_{22}\left(N{G}_{11}-{C_1}^2\right)+G_{12}\left(C_1{C}_2-N{G}_{12}\right)+\right.\\ \left.+C_2\left(C_1{G}_{12}-C_2{G}_{11}\right)\right]+\left[G_{23}\left(N{G}_{11}-{C_1}^2\right)+C_3\left(C_1{G}_{12}-C_2{G}_{11}\right)+G_{13}\left(C_1{C}_2-N{G}_{12}\right)\right]\left[D_{23}\left({C_1}^2-N{G}_{11}\right)+\right.\\ \left.+C_2\left(G_{11}{H}_3-C_1{D}_{13}\right)+G_{12}\left(N{D}_{13}-C_1{H}_3\right)\right],\end{array}$    

$\begin{array}{l}\Lambda =\left[G_{13}\left(C_1{H}_3\text{}-N{D}_{13}\right)+C_3\left(C_1{D}_{13}\text{}-G_{11}{H}_3\right)+D_{33}\left(N{G}_{11}\text{}-{C_1}^2\right)\right]\left[G_{22}\left(C_1{D}_{13}\text{}-G_{11}{H}_3\right)+G_{12}\left(G_{12}{H}_3\text{}-C_2{D}_{13}\right)+\right.\\ \left.+D_{23}\left(C_2{G}_{11}\text{}-C_1{G}_{12}\right)\right]+\left[F_3\left(C_3{G}_{11}\text{}-C_1{G}_{13}\right)+D_{33}\left(C_1{D}_{13}\text{}-G_{11}{H}_3\right)+D_{13}\left(G_{13}{H}_3\text{}-C_3{D}_{13}\right)\right]\left[G_{12}\left(C_1{C}_2\text{}-N{G}_{12}\right)+\right.\\ \left.+G_{22}\left(N{G}_{11}-{C_1}^2\right)+C_2\left(C_1{G}_{12}-C_2{G}_{11}\right)\right]+\left[G_{23}\left(N{G}_{11}-{C_1}^2\right)+C_3\left(C_1{G}_{12}-C_2{G}_{11}\right)+G_{13}\left(C_1{C}_2-N{G}_{12}\right)\right]\times \\ \times \left[F_3\text{}\left(C_1{G}_{12}\text{}-C_2{G}_{11}\right)+D_{13}\text{}\left(C_2{D}_{13}\text{}-G_{12}{H}_3\right)+D_{23}\text{}\left(G_{11}{H}_3\text{}-C_1{D}_{13}\right)\right]+\left[G_{12}\text{}\left(G_{13}{H}_3\text{}-C_3{D}_{13}\right)+\right.G_{23}\text{}\left(C_1{D}_{13}\text{}-G_{11}{H}_3\right)+\\ \left.+D_{23}\left(C_3{G}_{11}-C_1{G}_{13}\right)\right]\left[C_2\left(G_{11}{H}_3-C_1{D}_{13}\right)+G_{12}\left(N{D}_{13}-C_1{H}_3\right)+D_{23}\left({C_1}^2-N{G}_{11}\right)\right],\end{array}$

$\begin{array}{l}\Omega =\left[F_3\text{}\left(C_3{G}_{11}\text{}-C_1{G}_{13}\right)+D_{33}\text{}\left(C_1{D}_{13}\text{}-G_{11}{H}_3\text{}\right)+D_{13}\text{}\left(G_{13}{H}_3\text{}-C_3{D}_{13}\text{}\right)\right]\left[G_{22}\text{}\left(C_1{D}_{13}\text{}-G_{11}{H}_3\text{}\right)+G_{12}\text{}\left(G_{12}{H}_3\text{}-C_2{D}_{13}\right)+\right.\\ \left.+D_{23}\left(C_2{G}_{11}-C_1{G}_{12}\right)\right]-\left[G_{12}\left(C_3{D}_{13}\text{}-G_{13}{H}_3\right)+G_{23}\left(G_{11}{H}_3\text{}-C_1{D}_{13}\text{}\right)+D_{23}\left(C_1{G}_{13}\text{}-C_3{G}_{11}\right)\right]\left[F_3\text{}\left(C_1{G}_{12}\text{}-C_2{G}_{11}\right)\right.+\\ \left.+D_{13}\left(C_2{D}_{13}\text{}-G_{12}{H}_3\right)+D_{23}\left(G_{11}{H}_3-C_1{D}_{13}\right)\right].\end{array}$

The discriminant of equation (34) is equal to

$\begin{array}{l}\tilde{\Delta }=\left\{\left[{C_2}^2\left(D_{13}{G}_{13}-D_{33}{G}_{11}\right)+G_{12}{H}_3\left(C_3{G}_{12}-C_1{G}_{23}\right)+C_2{G}_{11}{G}_{23}{H}_3\right]\right.\left(G_{11}{H}_3-C_1{D}_{13}\right)+\left[F_3{G}_{11}\times \right.\\ \times \left(C_2{G}_{23}-C_3{G}_{22}\right)+F_3{G}_{12}\left(C_3{G}_{12}-C_1{G}_{23}\right)+F_3{G}_{13}\left(C_1{G}_{22}-C_2{G}_{12}\right)+C_1{D}_{23}\left(D_{33}{G}_{12}-D_{23}{G}_{13}\right)+C_2{D}_{23}\times \\ \left.\times \left(D_{13}{G}_{13}-D_{33}{G}_{11}\right)+C_3{D}_{23}\left(D_{23}{G}_{11}-D_{13}{G}_{12}\right)\right]\left({C_1}^2-N{G}_{11}\right)+\left[C_2{H}_3\text{}\left(D_{23}{G}_{11}\text{}-D_{13}{G}_{12}\right)+G_{22}{H}_3\left(C_1{D}_{13}-G_{11}{H}_3\right)-\right.\\ \left.-C_1{C}_2{D}_{13}{D}_{23}\right]\left(C_3{G}_{11}\text{}-C_1{G}_{13}\right)+\left[C_2{G}_{11}\left(D_{13}{G}_{23}-D_{33}{G}_{12}\right)+C_3{G}_{11}\left(D_{23}{G}_{12}-D_{13}{G}_{22}\right)+C_1{G}_{13}\left(D_{13}{G}_{22}-D_{23}{G}_{12}\right)+\right.\\ {\left.\left.+C_1{G}_{12}\left(D_{33}{G}_{12}-D_{13}{G}_{23}\right)\right]\left(N{D}_{13}-C_1{H}_3\right)+C_2{G}_{12}\left[C_1{D}_{13}\left(C_3{D}_{13}-C_1{D}_{33}\right)+G_{11}{H}_3\left(C_1{D}_{33}-G_{13}{H}_3\right)\right]\right\}}^2,\end{array}$

and the roots of this equation are calculated using the formulae

$\left[\begin{array}{l}{b}_3=\frac{C_1{D}_{13}-G_{11}{H}_3}{{C_1}^2-N{G}_{11}},\\ b_3=\left[C_3{G}_{11}\left({D_{23}}^2-F_3{G}_{22}\right)+D_{33}{G}_{11}\left(G_{22}{H}_3-C_2{D}_{23}\right)+G_{11}{G}_{23}\left(C_2{F}_3-D_{23}{H}_3\right)+D_{13}{G}_{12}\left(G_{23}{H}_3-C_3{D}_{23}\right)+\right.\\ +D_{13}{D}_{23}\left(C_1{G}_{23}-C_3{G}_{12}\right)+G_{13}{H}_3\left(D_{23}{G}_{12}-D_{13}{G}_{22}\right)+D_{13}{D}_{33}\left(C_2{G}_{12}-C_1{G}_{22}\right)+F_3{G}_{12}\left(C_3{G}_{12}-C_1{G}_{23}\right)+\\ \left.+D_{33}{G}_{12}\left(C_1{D}_{23}-G_{12}{H}_3\right)+D_{23}{G}_{13}\left(C_2{D}_{13}-C_1{D}_{23}\right)+{D_{13}}^2\left(C_3{G}_{22}-C_2{G}_{23}\right)+F_3{G}_{13}\left(C_1{G}_{22}-C_2{G}_{12}\right)\right]\times \\ \times \left[N{D}_{23}\left(G_{12}{G}_{13}-G_{11}{G}_{23}\right)+N{D}_{33}\left(G_{11}{G}_{22}-{G_{12}}^2\right)+N{D}_{13}\left(G_{12}{G}_{23}-G_{13}{G}_{22}\right)+C_3{G}_{11}\left(C_2{D}_{23}-G_{22}{H}_3\right)+\right.\\ +C_2{G}_{11}\left(G_{23}{H}_3-C_2{D}_{33}\right)+C_3{D}_{13}\left(C_1{G}_{22}-C_2{G}_{12}\right)+C_3{G}_{12}\left(G_{12}{H}_3-C_1{D}_{23}\right)+C_2{G}_{13}\left(C_2{D}_{13}-G_{12}{H}_3\right)+\\ {\left.+C_1{G}_{23}\left(C_1{D}_{23}-G_{12}{H}_3\right)+C_1{G}_{13}\left(G_{22}{H}_3-C_2{D}_{23}\right)+C_1{D}_{33}\left(C_2{G}_{12}-C_1{G}_{22}\right)+C_1{C}_2\left(D_{33}{G}_{12}-D_{13}{G}_{23}\right)\right]}^{-1}.\end{array}\right.$(35)

Since equation (34) has two solutions, the system of equations (13) also has two solutions. The components of both solutions of this system are obtained in the following order:

• – the values of the unknown b₃ are found using formulae (35);
• – the values of the unknown α₃ corresponding to the found values of b₃ are calculated using any of the formulae (33), (34);
• – the values of β₃ – using the formula written as the last equation of system (13);
• – the values of α₂ – using any of the three formulae

${\text{α}}_2=\frac{\left[G_{12}\left(H_3-N{b}_3\right)-C_2\left(D_{13}-C_1{b}_3\right)\right]{\text{α}}_3+G_{13}\left(H_3-N{b}_3\right)-C_3\left(D_{13}-C_1{b}_3\right)}{{\text{β}}_3\left[C_1\left(D_{13}-C_1{b}_3\right)-G_{11}\left(H_3-N{b}_3\right)\right]},$