MDETERM


Returns the determinant of a matrix.

Syntax:

MDETERM(array)


returns the determinant of the square matrix array, which may be either an inline array or a range, containing all numbers.

MDETERM returns a single value. It need not be entered as an array formula.

Example:

MDETERM({4,1;2,5})

returns 18, that is 4*5 - 1*2.

MDETERM(A1:B2)

where A1, B1, A2, B2 contain 4, 2, 2, 3 respectively, returns 8, that is 4*3 - 2*2.


Application:

Structural Engineering - Analyzing a Truss System


Let's consider a simple truss system in structural engineering. A truss is a framework of interconnected straight members (like beams) that is typically used in bridges and roofs. Engineers need to analyze these systems to ensure they are stable and can support the required loads.


One method for analyzing a truss is using the stiffness matrix method. This method involves setting up a system of linear equations that describe the relationship between the forces applied to the truss and the resulting displacements (movement) of its joints.


The determinant of the stiffness matrix is a critical value. Here’s why:


  • If the determinant is non-zero (det ≠ 0): This means the stiffness matrix is invertible. The system of equations has a unique solution, and the truss is statically determinate and stable. The structure will not collapse under the applied load, and an engineer can calculate the forces in each member.
  • If the determinant is zero (det = 0): This indicates that the stiffness matrix is singular (non-invertible). The system of equations does not have a unique solution. In a physical sense, this means the truss is unstable or a mechanism. A small force can cause an infinitely large displacement, leading to a collapse.

Scenario

Imagine a simple 2D truss with three members and two joints (excluding supports). We want to check the stability of this truss using its stiffness matrix. We've simplified the problem to a 2x2 matrix for this example.


The stiffness matrix K represents the relationship between applied forces and joint displacements.



Where:


  • k11​ and k22​ relate the forces to the displacements in a single direction.
  • k12​ and k21​ relate the forces in one direction to the displacements in another.


We will use the MDETERM function to find the determinant of this matrix to check the stability of our simplified truss.

Spreadsheet Table

Here is how you would set this up in a spreadsheet:


Table 1: Stable Truss System




A
B
C
1

Stiffness Matrix (K)

2

k11​

k12

3

k21

1000

200

4

k22

200

500

5
6

Determinant (det(K))

MDETERM(B3:C4)

7

Result

460000

8

Conclusion

Stable (det ≠ 0)


Explanation:


  • The matrix values are entered in cells B3 to C4.
  • The formula =MDETERM(B3:C4) is entered into cell B6.
  • The function calculates the determinant: (1000×500)−(200×200)=500000−40000=460000.
  • Since the determinant is 460,000 (non-zero), an engineer can conclude that the truss is stable under these conditions.


Table 2: Unstable Truss System


Now, let's change one of the stiffness values to represent an instability, perhaps due to a poorly designed member or a faulty connection.




A
B
C
1

Stiffness Matrix (K)

2

k11

k12

3

k21

1000

500

4

k22

200

100

5
6

Determinant (det(K))

MDETERM(B3:C4)

7

Result

0

8

Conclusion

Unstable (det = 0)


Explanation:


  • The matrix values are entered in cells B3 to C4.
  • The formula MDETERM(B3:C4) is entered into cell B6.
  • The function calculates the determinant: (1000×100)−(500×200)=100000−100000=0.
  • Since the determinant is 0, an engineer knows immediately that this configuration is unstable and would fail under load. This is a critical check that prevents catastrophic structural failures.


In this application, the MDETERM function is not just a mathematical tool; it is a fundamental part of the safety analysis for physical structures.






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