whaat is $LU$ decomposition of a matrix

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Algebra 1

Simplify exponential expressions using exponent rules

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Q. whaat is

LU

decomposition of a matrix

Pick Matrix: First, we gotta pick a matrix to decompose. Let's say we have a $2 \times 2$ matrix $A$ : $A = \left[\begin{array}{cc}a & b \ c & d\end{array}\right]$
Decompose Matrix: Now, we're gonna write $A$ as the product of a lower triangular matrix $L$ and an upper triangular matrix $U$ . So, $A = LU$
Lower Triangular Matrix: $L$ is gonna look like this: $\newline$ $L = \left[\begin{array}{cc} 1 & 0 \ x & 1 \end{array}\right]$ $\newline$ 'Cause the diagonal of $L$ is always $1$ s.
Upper Triangular Matrix: And $U$ is like: $U = \left[\begin{array}{cc} y & z \ 0 & w \end{array}\right]$ Upper triangular means zeros below the diagonal.
Multiply L and U: We multiply L and U to get A back: $\begin{bmatrix} 1 & 0 \ x & 1 \end{bmatrix} * \begin{bmatrix} y & z \ 0 & w \end{bmatrix} = \begin{bmatrix} a & b \ c & d \end{bmatrix}$
Solve Equations: Do the multiplication: $\newline$ \left[\begin{array}{cc}$\newline1y + 0\cdot 0 & 1z + 0w (\newline$xy + 1\cdot 0 & xz + 1w $\newline$ \end{array}\right] = \left[\begin{array}{cc} $\newline$ a & b (\newline\)c & d $\newline$ \end{array}\right]\)
Plug in Values: This gives us the equations: $\newline$ $y = a$ $\newline$ $z = b$ $\newline$ $x*y = c$ $\newline$ $x*z + w = d$
Final L and U Matrices: Solve for $y$ , $z$ , $x$ , and $w$ : $y = a$ $z = b$ $x = \frac{c}{y}$ $w = d - x \cdot z$
Final L and U Matrices: Solve for $y$ , $z$ , $x$ , and $w$ :
$y = a$
$z = b$
$x = \frac{c}{y}$
$w = d - x \cdot z$ Plug in the values:
$y = a$
$z = b$
$z$ $0$
$z$ $1$
Final L and U Matrices: Solve for $y$ , $z$ , $x$ , and $w$ : $y = a$ $z = b$ $x = \frac{c}{y}$ $w = d - x \cdot z$ Plug in the values: $y = a$ $z = b$ $x = \frac{c}{a}$ $w = d - \left(\frac{c}{a}\right) \cdot b$ So our L and U matrices are: $L = \begin{bmatrix}1 & 0\ \frac{c}{a} & 1\end{bmatrix}$ $U = \begin{bmatrix}a & b\ 0 & d - \left(\frac{c}{a}\right)\cdot b\end{bmatrix}$