Question

find an \\(lu\\) decomposition of \\(\

$$\begin{bmatrix} 3 & 1 \\\\ -6 & -4 \\end{bmatrix}$$

\\).

Explanation:

🆕 New Concept Discovered: LU Decomposition
Splitting a matrix into lower and upper triangular parts.

Step 1: Set up the LU structure

An \( LU \) decomposition factors a matrix \( A \) into a lower triangular matrix \( L \) (with \( 1 \)s on its diagonal) and an upper triangular matrix \( U \).

For a \( 2 \times 2 \) matrix:

$$ A =
LATEXBLOCK0
$$

We define:

$$ L =
LATEXBLOCK1
, \quad U =
LATEXBLOCK2
$$

We want to find the values such that \( L \times U = A \):

$$
LATEXBLOCK3

LATEXBLOCK4
=
LATEXBLOCK5
$$

Step 2: Determine the first row of U

Multiplying the first row of \( L \) by the columns of \( U \) gives the first row of \( A \):

$$ 1 \cdot u_{11} + 0 \cdot 0 = 3 \implies u_{11} = 3 $$

$$ 1 \cdot u_{12} + 0 \cdot u_{22} = 1 \implies u_{12} = 1 $$

So, the first row of \( U \) is identical to the first row of \( A \):

$$ U =
LATEXBLOCK6
$$

Step 3: Determine the multiplier in L

Next, we look at the first element of the second row of \( A \), which is \( -6 \). This is obtained by multiplying the second row of \( L \) by the first column of \( U \):

$$ l_{21} \cdot u_{11} + 1 \cdot 0 = -6 $$

$$ l_{21} \cdot 3 = -6 \implies l_{21} = -2 $$

This multiplier \( l_{21} = -2 \) represents the factor we multiply the first row by to eliminate the entry below the pivot during Gaussian elimination.

Step 4: Determine the remaining entry of U

Finally, we find \( u_{22} \) by multiplying the second row of \( L \) by the second column of \( U \):

$$ l_{21} \cdot u_{12} + 1 \cdot u_{22} = -4 $$

Substitute the known values \( l_{21} = -2 \) and \( u_{12} = 1 \):

$$ (-2)(1) + u_{22} = -4 $$

$$ -2 + u_{22} = -4 \implies u_{22} = -2 $$

Answer:

An \( LU \) decomposition of the matrix is:

$$ L =
LATEXBLOCK0
, \quad U =
LATEXBLOCK1
$$