CN107092581A - The transport equation response matrix Block Diagonalization method discussed based on symmetric group - Google Patents

Abstract

The transport equation response matrix Block Diagonalization method discussed based on symmetric group, is comprised the following steps：1st, divided for whole reactor core domain according to the geometry of rule, obtain the locking nub of rule, based on variation Nodal method, functional is set up to neutron-transport equation, the weak form of transport equation is constructed；Discrete to the progress of neutron angular flax density using spheric harmonic function and orthogonal polynomial, spheric harmonic function together constitutes the basic function of flux with orthogonal space multinomial；2nd, according to the geometry of locking nub, the symmetric group that all symmetry changes of locking nub are constituted is determined；Basic function is projected using the projection operator of symmetric group, so as to construct the basic function of irreducible symmetrization；3rd, by the spheric harmonic function and orthogonal space multinomial in the basic function replacement step 1 of irreducible symmetrization, the matrix that discrete, new response matrix is diagonal piecemeal is carried out to variable using new basic function；The floating-point operation amount of matrix is substantially reduced using the inventive method, the calculating internal memory needed for memory response matrix is greatly reduced.

Description

Block Diagonalization Method of Response Matrix of Transport Equation Based on Symmetric Group Theory

technical field

The invention relates to the technical field of nuclear reactor core design and safety, in particular to a block diagonalization method of a transport equation response matrix based on symmetric group theory.

Background technique

The calculation of the whole core transport segment is of great significance in the calculation of neutronics. In the solution of the neutron transport equation, if the response relation of the neutron entry and exit jet at the inner boundary of the nodal block can be constructed in advance, then the midpoint of the nodal block can be obtained by scanning and iterating the nodal block until convergence according to the response relation. neutron leakage, so as to solve the neutron flux distribution in each node. This approach is called the response matrix approach. The angle term of the transport equation is usually discretized by using spherical harmonics. If the angle term is expanded to take the first N-order spherical harmonic functions, and the transport equation is projected on the first N-order spherical harmonic functions, (N+1) ² linear algebraic equations can be obtained, that is, the PN approximate equations . Since the total number of unknowns inside a block is positively correlated with the number of spherical harmonics, the total number of unknowns increases quadratically with the order of angle expansion.

The construction of the response matrix usually involves matrix multiplication and inversion operations, and the amount of these floating-point operations is roughly proportional to the cube of the number of unknowns. Therefore, when the number of unknowns is large, the calculation memory and the total amount of floating-point operations will be greatly increased.

In the actual core calculation, for small fast neutron reactors with strong leakage and strong anisotropic scattering effect, if you want to obtain more accurate numerical calculation results, you usually need to use a higher angle order for approximation. In addition, if fuel consumption, transient, etc. are calculated, the cross-sections of each block are different, so that each block needs to calculate its own response matrix separately, which poses a great challenge to computing memory and computing efficiency.

Contents of the invention

In order to overcome the problems in the above-mentioned prior art, the object of the present invention is to provide a method for block diagonalization of the transport equation response matrix based on symmetric group theory, which is based on symmetric group theory and uses irreducible symmetrized basis functions , the solution of the transport equation is simplified into several mutually decoupled sub-problems, and thus the response matrix is decomposed into a series of mutually decoupled sub-blocks located on the diagonal positions of the original matrix. In the matrix multiplication and inversion operations, the multiplication and inversion operations can be performed on the sub-block matrix independently, thereby greatly reducing the amount of floating-point operations of the matrix. In addition, due to the block-diagonalization of the matrix, the new matrix has significantly fewer non-zero elements compared to the original matrix, and the computational memory required to store the response matrix is also greatly reduced.

In order to achieve the above object, the present invention has taken the following technical solutions to implement:

A block diagonalization method of the response matrix of the transport equation based on symmetric group theory, the method includes the following steps:

Step 1: Divide the entire core solution area according to regular geometry to obtain regular nodal blocks. Based on the variational nodal method, a functional is established for the neutron transport equation to construct a weak solution form of the transport equation; Spherical harmonic functions and orthogonal polynomials discretize the neutron angular flux density, and spherical harmonic functions and spatial orthogonal polynomials together constitute the basis function of the flux, as shown in formula (8)

where _ψ (r,Ω) is the even flux density; J(r,Ω) is the Lagrangian multiplier term at the block boundary; g(Ω)=[Y _0,0 ,Y _2,-2 , Y _2,-1 ,Y _2,0 ,Y _2,1 ,Y _2,2 ...] ^T is the even order normalized spherical harmonic function; f(r)=[f ₁ ,f ₂ ,f ₃ ...] ^T is an orthogonal polynomial defined inside the node v; h(r)=[h ₁ , h ₂ ...] ^T is an orthogonal polynomial defined at the boundary of the node v; Indicates the boundary basis function located at the No. 1 boundary surface, Indicates the boundary basis function located at the No. 2 boundary surface; And ξ is the expansion moment of the even neutron angular flux density and the expansion moment of the neutron current density respectively; T is the transpose symbol;

Bring the flux in discrete form and the Lagrangian multiplier term into the functional to sort out the response matrix equation:

j ⁺ ＝Bs+Rj ^-

(16)

in j ⁺ is the expanded moment of the outgoing partial neutron flux density; j ^- is the expanded moment of the incident partial neutron flux density; _s is the expanded moment of the source term; R, H, C, B are Four response matrices for the variational block method;

Step 2: According to the geometric shape of the nodal block in step 1, determine the symmetry group formed by all the symmetry transformations of the nodal block; use the projection operator of the symmetry group to project the basis functions in step 1, thereby constructing an irreducible symmetry The basis function of the transformation;

projection operator composed of symmetric operators

In the formula

_n - the order of the group, that is, the number of group elements in the finite group;

s _α ——the dimension of the αth irreducible representation matrix;

R - group element;

P _R ——the symmetry operator corresponding to the R group element;

G - the name of the group;

——the complex conjugate of the α-th irreducible representation matrix element R corresponding to the j-th row and j-th column element;

Use the projection operator to project the basis functions to obtain the irreducible components of the basis functions in different symmetry spaces; if the basis functions of the boundary surface form the basis function vector use projection operator Projecting the function vector gives:

Projection operators consist of symmetric operators that operate on vectors of basis functions It is equivalent to multiplying a certain matrix U _R by a function vector, namely:

Thus available:

where matrix

function vector The functions of are linearly related, and the matrix needs to be Perform Gaussian elimination to obtain a new full-rank matrix its size is for the matrix rank;

get the matrix Then the basis function of the jth column of the αth irreducible representation is obtained:

Repeat the above method and use other projection operators to finally obtain all irreducible symmetrization basis functions;

Step 3, replace the spherical harmonic functions and spatial orthogonal polynomials in step 1 with irreducibly symmetrized basis functions,

Among them, K(r, Ω) and H(r, Ω) are the basis functions of irreducible symmetry; using the new basis function to discretize the variables, a new response relationship can be obtained:

new response matrix is a diagonally block matrix.

Compared with the prior art, the present invention has the following outstanding advantages:

1. Using the symmetry of the nodal block, the complex transport problem can be decomposed into a series of decoupled sub-problems, thus simplifying the solution of the problem.

2. After symmetrizing the basis function, the response matrix of diagonal blocks can be obtained, thereby greatly reducing the calculation time of the response matrix.

3. The number of non-zero elements of the response matrix is greatly reduced, thereby greatly reducing the computational memory required by the response matrix.

Description of drawings

Figure 1 Schematic diagram of the symmetry of a hexagonal segment.

Figure 2 Response Matrix Schematic diagram of the location of non-zero elements.

Figure 3 The response matrix after using irreducibly symmetric basis functions Schematic diagram of the location of non-zero elements.

detailed description

Below in conjunction with accompanying drawing and specific embodiment the present invention is described in further detail:

The method of the present invention will be based on the theory of symmetry groups, and adopt irreducible symmetrization basis functions to simplify the solution of the transport equation into several mutually decoupled sub-problems, thereby decomposing the response matrix into a series of mutually decoupled sub-problems located in the original matrix The subblocks at the diagonal positions of . In the matrix multiplication and inversion operations, the multiplication and inversion operations can be performed on the sub-block matrix independently, thereby greatly reducing the amount of floating-point operations of the matrix. In addition, due to the block-diagonalization of the matrix, the new matrix has significantly fewer non-zero elements compared to the original matrix, and the computational memory required to store the response matrix is also greatly reduced.

The whole calculation specifically includes the following steps:

Step 1. Based on the variational nodal method, a functional is established for the neutron transport equation, and the weak solution form of the transport equation is constructed. The angle and space terms of the odd and even fluxes are discretized by spherical harmonics and orthogonal polynomials, respectively.

The neutron transport equation in the unienergetic second-order even symmetric form can be written as follows:

r - space position variable;

Ω——angle variable;

ψ _g — neutron angular flux density of energy group g;

Φ _g — neutron standard flux density;

g—energy group identification;

Σ _t,g ——the total cross-section of the gth energy group;

——the self-scattering cross section of the gth energy group;

S _g ——the source term of the neutron transport equation of the gth energy group, including the intergroup scattering source term and the fission source term,

Here it is assumed that the scattering is isotropic, and the odd and even fluxes of the gth group are defined as follows

ψ _g (r,Ω)——the even flux density of the gth group;

χ _g (r,Ω)——the odd flux density of the gth group;

make

J _γ (r,Ω)＝n _γ ·Ωχ _γ (r,Ω) (3)

n _γ —— is the outer normal direction on the boundary of the node γ;

χ _γ —— is the odd flux on the boundary of the node γ;

A global functional is established for the area to be solved, and the global functional is the sum of the local functionals of each node.

The local functional is

where I[ψ,J] is the boundary integral term, for the internal boundary

For the vacuum boundary

In the above formula, Γ _i is the internal boundary, and Γ _j is the vacuum boundary.

Therefore, the problem of solving the neutron transport equation is transformed into the variational problem of solving the global functional minimum, and the solution of the variational problem is the weak form solution of the second-order odd-even equation.

In the formula (5), J(r, Ω) is the Lagrange multiplier term, its function is to eliminate the constraints brought by the boundary conditions, through the Lagrange multiplier term, the boundary conditions are added to the functional, so as to realize coupling between nodes.

Expand the even flux and the angle term of the Lagrange multiplier term into the form of spherical harmonic functions, and the space term into the form of orthogonal basis functions:

Where g(Ω)=[Y _0,0 ,Y _2,-2 ,Y _2,-1 ,Y _2,0 ,Y _2,1 ,Y _2,2 ...] ^T is the normalization of even order spherical harmonic function of ; f(r)=[f ₁ ,f ₂ ,f ₃ ...] ^T is an orthogonal polynomial defined inside the nodal block v, h(r)=[h ₁ ,h ₂ ... ] ^T is an orthogonal polynomial defined at the boundary of the node block v; Indicates the boundary basis function located at the No. 1 boundary surface, Indicates the boundary basis function located at the No. 2 boundary surface; and ξ are the expansion moments of the even neutron angular flux density and the expansion moments of the neutron current density respectively; T is the transpose symbol.

Bring (8) into the functional (5) to get the response matrix of the traditional variational block method.

in

M＝[M ₁ M ₂ ... M _γ ...] (10)

The derivative of the functional with respect to the flux expansion moment is zero, and the relationship between the even flux expansion moment and the Lagrangian multiplier term can be obtained:

Define the split flow expansion moment of the γ surface:

in

In the strategy of solving the expansion moments, the four-color checkerboard scanning and the calculation method of the response matrix are adopted. The formula of the response matrix is as follows:

j ⁺ ＝Bs+Rj ^- (16)

in

B＝[G+I] ^-1 C ^T (18)

R＝[G+I] ^-1 [GI] (19)

H＝A ^-1 (22)

Step 2: Determine the symmetry group formed by all the symmetry transformations of the nodal block according to the geometric shape of the nodal block. There are 12 total symmetry transformations of a two-dimensional hexagon, as shown in Figure 1. These symmetry transformations include 6 rotation transformation operators They represent counterclockwise rotation of 0 degrees, 60 degrees, 120 degrees, 180 degrees, 240 degrees, and 300 degrees, respectively. in It can also be recorded as the unit operator E. The six reflection transformation operators are {σ _v1 ,σ _v2 ,σ _v3 ,σ _d1 ,σ _d2 ,σ _d3 }, which are reflection transformations about their respective symmetry axes. All symmetric operators form the C _6v group. For the hexagonal prism block, it is also necessary to consider the symmetry about the xoy plane, so there are 24 symmetry operators in total, Constitute group D _6h .

The basis functions are projected by using the projection operator of the symmetry group, so as to construct the irreducible symmetrization basis functions.

It may be assumed that all symmetric transformations form a group G, and the group G has several projection operators, among which is the projection operator for the jth column of the αth irreducible representation matrix.

projection operator Consists of symmetry operators

In the formula

n——the order of the group, that is, the number of group elements in the finite group;

s _α ——the dimension of the αth irreducible representation matrix;

R - group element;

P _R ——the symmetry operator corresponding to the R group element;

G - the name of the group;

——the complex conjugate of the α-th irreducible representation matrix element R corresponding to the j-th row and j-th column element;

Taking the two-dimensional hexagonal block as an example, according to the group theory, there are 8 projection operators in the group C _6v , and they are respectively

in

——the projection operator of the irreducible representation matrix A ₁ ;

——the projection operator of the irreducible representation matrix A ₂ ;

——the projection operator _of the irreducible representation matrix B1;

——the projection operator of the irreducible representation matrix B2 _;

- the projection operator of the _first column of the irreducible representation matrix E1;

- the projection operator _of the second column of the irreducible representation matrix E1;

- the projection operator of the first column of the irreducible representation matrix _E2 ;

- the projection operator of the _second column of the irreducible representation matrix E2;

The operator at the right end of the formula is the symmetry transformation operator in step 2.

The irreducible components of the basis functions in different symmetry spaces can be obtained by using the projection operator to project the basis functions. If the basis functions of the boundary surface form the basis function vector May wish to set use Projecting the function vector gives:

Projection operators consist of symmetric operators, and a symmetric operator acts on the basis function vector A new set of basis functions can be obtained. If the basis function space is closed to the symmetric operators, then the new basis functions obtained still belong to the original function space. So the symmetric operator acts on the basis function vector It is equivalent to multiplying a certain matrix U _R by a function vector, namely:

where the matrix U _R can be obtained by

According to formula (32) and formula (34), we can get

where matrix

function vector The length of the original basis function vector are equal in length. and The function in is only of the form The component of the jth column of the αth irreducible representation of , so it can be seen that functions are linearly related. At this time, the matrix needs to be Perform Gaussian elimination to obtain a new full-rank matrix its size is for the matrix rank. If yes Summing over all irreducible representation spaces and distinct columns gives:

get the matrix Then the basis function of the jth column of the αth irreducible representation can be obtained:

Repeat the above method and use other projection operators to finally obtain all irreducible basis functions.

For two-dimensional hexagonal blocks, the irreducible basis functions of the C _6v group can be constructed according to the projection operators in formula (24) to formula (31). And for the three-dimensional hexagonal prism block, due to Therefore, the irreducible basis functions of C _6v can be generated first, and then the projection operator of C _1h can be used to further project the irreducible basis functions of C _6v , and finally the irreducible symmetrization basis functions of group D _6h can be obtained.

Step 3, replace the spherical harmonics and orthogonal polynomials in formula (8) with irreducible symmetrized basis functions:

The relationship between the irreducible basis function and the original basis function is as follows:

If the basis functions of irreducible symmetry are adopted, a new response relation can be obtained:

Due to the use of irreducibly symmetric basis functions and the use of the symmetry of nodal blocks, the new response matrix is a diagonal block matrix, as shown in Figure 3.

Taking the response matrix G (formula (20)) of the three-dimensional hexagonal block as an example, if the original basis function is used, the matrix is shown in Fig. 2 . The calculation adopts the angle approximation of P3, the space polynomial out of the boundary adopts the 2nd order expansion, and the polynomial inside the volume adopts the 6th order expansion. The matrix G in Figure 2 has a total of 11504 non-zero elements. However, if irreducible basis functions are used, a new matrix can be obtained The relationship between these two matrices is:

matrix As shown in Figure 3. It can be seen from the figure that the matrix There are 16 sub-blocks located diagonally. When matrix inversion and multiplication are performed, sub-blocks can be independently inverted and multiplied, which will greatly save matrix calculation time. At the same time, the non-zero elements of the matrix are also greatly reduced, from 11504 to 1454, so the memory required for calculation is greatly reduced. Table 1 shows the comparison of the total time for calculating the TAKETA problem using different angle expansion orders. For the angle unwrapping of P7, the calculation speedup can reach 15 times.

Table 1 Comparison of the total calculation time of the expansion order problem from different angles

Claims (1)

1. The block diagonalization method of the transport equation response matrix based on symmetric group theory, is characterized in that: comprise the following steps: Step 1: Divide the entire core solution area according to regular geometry to obtain regular nodal blocks. Based on the variational nodal method, a functional is established for the neutron transport equation to construct a weak solution form of the transport equation; Spherical harmonic functions and orthogonal polynomials discretize the neutron angular flux density, and spherical harmonic functions and spatial orthogonal polynomials together constitute the basis function of the flux, as shown in formula (8) where ψ(r,Ω) is the even flux density; J(r,Ω) is the Lagrangian multiplier term at the block boundary; g(Ω)=[Y _0,0 ,Y _2,-2 , Y _2,-1 ,Y _2,0 ,Y _2,1 ,Y _2,2 ...] ^T is the even order normalized spherical harmonic function; f(r)=[f ₁ ,f ₂ ,f ₃ ...] ^T is an orthogonal polynomial defined inside the node v; h(r)=[h ₁ , h ₂ ...] ^T is an orthogonal polynomial defined at the boundary of the node v; Indicates the boundary basis function located at the No. 1 boundary surface, Indicates the boundary basis function located at the No. 2 boundary surface; And ξ is the expansion moment of the even neutron angular flux density and the expansion moment of the neutron current density respectively; T is the transpose symbol; Bring the flux in discrete form and the Lagrangian multiplier term into the functional to sort out the response matrix equation: j ⁺ ＝Bs+Rj ^- (16) in j ⁺ is the expanded moment of the outgoing partial neutron flux density; j ^- is the expanded moment of the incident partial neutron flux density; s is the expanded moment of the source term; R, H, C, B are Four response matrices for the variational block method; Step 2: According to the geometric shape of the nodal block in step 1, determine the symmetry group formed by all the symmetry transformations of the nodal block; use the projection operator of the symmetry group to project the basis functions in step 1, thereby constructing an irreducible symmetry The basis function of the transformation; projection operator composed of symmetric operators <mrow> <msubsup> <mi>P</mi> <mrow> <mi>j</mi> <mi>j</mi> </mrow> <mi>&amp;alpha;</mi> </msubsup> <mo>=</mo> <mfrac> <msub> <mi>s</mi> <mi>&amp;alpha;</mi> </msub> <mi>n</mi> </mfrac> <munder> <mo>&amp;Sigma;</mo> <mrow> <mi>R</mi> <mo>&amp;Element;</mo> <mi>G</mi> </mrow> </munder> <msubsup> <mi>&amp;Gamma;</mi> <mrow> <mi>j</mi> <mo>,</mo> <mi>j</mi> </mrow> <mrow> <mo>(</mo> <mi>&amp;alpha;</mi> <mo>)</mo> </mrow> </msubsup> <msup> <mrow> <mo>(</mo> <mi>R</mi> <mo>)</mo> </mrow> <mo>*</mo> </msup> <msub> <mi>P</mi> <mi>R</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>23</mn> <mo>)</mo> </mrow> </mrow> In the formula _n - the order of the group, that is, the number of group elements in the finite group; s _α ——the dimension of the αth irreducible representation matrix; R - group element; P _R ——the symmetry operator corresponding to the R group element; G - the name of the group; ——the complex conjugate of the α-th irreducible representation matrix element R corresponding to the j-th row and j-th column element; Use the projection operator to project the basis functions to obtain the irreducible components of the basis functions in different symmetry spaces; if the basis functions of the boundary surface form the basis function vector use projection operator Projecting the function vector gives: <mrow> <msup> <mover> <mi>H</mi> <mo>&amp;RightArrow;</mo> </mover> <mo>&amp;prime;</mo> </msup> <mrow> <mo>(</mo> <mi>r</mi> <mo>,</mo> <mi>&amp;Omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mi>P</mi> <mrow> <mi>j</mi> <mi>j</mi> </mrow> <mi>&amp;alpha;</mi> </msubsup> <mover> <mi>H</mi> <mo>&amp;RightArrow;</mo> </mover> <mrow> <mo>(</mo> <mi>r</mi> <mo>,</mo> <mi>&amp;Omega;</mi> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>32</mn> <mo>)</mo> </mrow> </mrow> Projection operators consist of symmetric operators that operate on vectors of basis functions It is equivalent to multiplying a certain matrix U _R by a function vector, namely: <mrow> <msub> <mi>P</mi> <mi>R</mi> </msub> <mover> <mi>H</mi> <mo>&amp;RightArrow;</mo> </mover> <mrow> <mo>(</mo> <mi>r</mi> <mo>,</mo> <mi>&amp;Omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>U</mi> <mi>R</mi> </msub> <mover> <mi>H</mi> <mo>&amp;RightArrow;</mo> </mover> <mrow> <mo>(</mo> <mi>r</mi> <mo>,</mo> <mi>&amp;Omega;</mi> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>33</mn> <mo>)</mo> </mrow> </mrow> Thus available: <mrow> <msup> <mover> <mi>H</mi> <mo>&amp;RightArrow;</mo> </mover> <mo>&amp;prime;</mo> </msup> <mrow> <mo>(</mo> <mi>r</mi> <mo>,</mo> <mi>&amp;Omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mi>Q</mi> <mi>j</mi> <mi>&amp;alpha;</mi> </msubsup> <mover> <mi>H</mi> <mo>&amp;RightArrow;</mo> </mover> <mrow> <mo>(</mo> <mi>r</mi> <mo>,</mo> <mi>&amp;Omega;</mi> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>35</mn> <mo>)</mo> </mrow> </mrow> where matrix function vector The functions of are linearly related, and the matrix needs to be Perform Gaussian elimination to obtain a new full-rank matrix its size is for the matrix rank; get the matrix Then the basis function of the jth column of the αth irreducible representation is obtained: <mrow> <msubsup> <mover> <mi>H</mi> <mo>&amp;RightArrow;</mo> </mover> <mi>j</mi> <mi>&amp;alpha;</mi> </msubsup> <mrow> <mo>(</mo> <mi>r</mi> <mo>,</mo> <mi>&amp;Omega;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mover> <mi>Q</mi> <mo>~</mo> </mover> <mi>j</mi> <mi>&amp;alpha;</mi> </msubsup> <mover> <mi>H</mi> <mo>&amp;RightArrow;</mo> </mover> <mrow> <mo>(</mo> <mi>r</mi> <mo>,</mo> <mi>&amp;Omega;</mi> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>37</mn> <mo>)</mo> </mrow> </mrow> Repeat the above method and use other projection operators to finally obtain all irreducible symmetrization basis functions; Step 3, replace the spherical harmonic functions and spatial orthogonal polynomials in step 1 with irreducibly symmetrized basis functions, Among them, K(r, Ω) and H(r, Ω) are the basis functions of irreducible symmetry; using the new basis function to discretize the variables, a new response relationship can be obtained: <mrow> <msup> <mover> <mi>j</mi> <mo>~</mo> </mover> <mo>+</mo> </msup> <mo>=</mo> <mover> <mi>B</mi> <mo>~</mo> </mover> <mover> <mi>s</mi> <mo>~</mo> </mover> <mo>+</mo> <mover> <mi>R</mi> <mo>~</mo> </mover> <msup> <mover> <mi>j</mi> <mo>~</mo> </mover> <mo>-</mo> </msup> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>40</mn> <mo>)</mo> </mrow> </mrow> new response matrix is a diagonally block matrix.