Triangular Matrix
2026-02-28 23:45 Diff
https://brightchamps.com/en-us/math/algebra/triangular-matrix

In a linear transformation, changes in one variable lead to proportional changes in others. The determinant of a matrix represents how much a linear transformation scales space. The determinant of a triangular matrix is calculated by multiplying all the entries on its main diagonal. Let’s take an example to understand how to compute the determinant:

Upper triangular Matrix:

\( A = \begin{bmatrix} 2 & 3 & 4 \\ 0 & 5 & 6 \\ 0 & 0 & 7 \end{bmatrix} \)

  • Diagonal elements: 2, 5, 7
     
  • Determinant: det(A) = 2 × 5 × 7 = 70

Lower Triangular Matrix:

\( B = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 3 & 0 \\ 7 & 5 & 6 \end{bmatrix} \)

  • Diagonal elements: 1, 3, 6
     
  • Determinant: det(B)=1 × 3 × 6 = 18
     

In triangular matrices, the determinant remains unchanged by the zeros located outside the main diagonal.

LU Decomposition and Triangular Matrices

LU decomposition (also called factorization) of a square matrix 𝐴 expresses A = LU, where L is lower-triangular with 1s on the diagonal (or unit lower triangular) and U is upper-triangular.

The connection is that solving 𝐴𝑥 = 𝑏 becomes LUx=b, and Ly = b can be solved by forward substitution, then 𝑈𝑥 = 𝑦 by backward substitution. 

For example, we have: \( A = \begin{bmatrix} 4 & 3 \\ 6 & 3 \end{bmatrix} \)

We want to find L and U such that A = LU. 
 

Step 1: Let, \( L = \begin{bmatrix} 1 & 0 \\ l_{21} & 1 \end{bmatrix} \)

                    \( U = \begin{bmatrix} u_{11} & u_{12} \\ 0 & u_{22} \end{bmatrix} \) 
 

Step 2: Multiply 𝐿 and 𝑈. 
 

\( LU = \begin{bmatrix} 1 & 0 \\ l_{21} & 1 \end{bmatrix} \begin{bmatrix} u_{11} & u_{12} \\ 0 & u_{22} \end{bmatrix} = \begin{bmatrix} u_{11} & u_{12} \\ l_{21}u_{11} & l_{21}u_{12} + u_{22} \end{bmatrix} \)
 

Step 3: Compare with A. 

\( \begin{bmatrix} u_{11} & u_{12} \\ l_{21}u_{11} & l_{21}u_{12} + u_{22} \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 6 & 3 \end{bmatrix} \)

So, u11 = 4, u12 = 3, l21u11 = 6, l21u12 + u22 = 3

From l21u11= 6, we get l21 = \(\frac{6}{4}\) = -1.5

Therefor \( L = \begin{bmatrix} 1 & 0 \\ 1.5 & 1 \end{bmatrix} \) 
 

               \( U = \begin{bmatrix} 4 & 3 \\ 0 & -1.5 \end{bmatrix} \) 
 

and \(A = LU\)