Answer to Question #139244 in Linear Algebra for Aji

Question

1.) Find the sequence of elementary matrices whose product is:
A=[1/2 -3
2 3/15] << 2 x 2 matrix


2.) Solve the given SLE using LU-Factorization:
x(sub 1) + x(sub 2) + x(sub 3) = 3
x(sub1) - x(sub 2) + 4x(sub 3) = 4
2x(sub 1) + 3x(sub 2) - 5x(sub 3) = 0

Answer 1

(i) A=$\begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix}$

Let I=$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Applying IA=A

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix}$ =$\begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix}$

Applying $R_1\to2R_1$

$\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix} =\begin{bmatrix} {1} & -6\\ 2 & \frac{3}{15} \end{bmatrix}$

Applying $R_2\to R_2-2R_1$

$\begin{bmatrix} 2 & 0 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix} =\begin{bmatrix} {1} & -6\\ 0 & \frac{183}{15} \end{bmatrix}$

Applying $R_1\to R_1+\frac{90}{183}R_2$

$\begin{bmatrix} \frac{186}{183}& \frac{90}{183} \\ -2 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix} =\begin{bmatrix} {1} & 0\\ 0 & \frac{183}{15} \end{bmatrix}$

applying $R_2\to \frac{15}{183}R_2$

$\begin{bmatrix} \frac{186}{183}& \frac{90}{183} \\ \frac{-30}{183} & \frac{15}{183} \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix} =\begin{bmatrix} {1} & 0\\ 0 & 1 \end{bmatrix}$

Sequence of elementary matrix are

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix} 2 & 0 \\ -2& 1 \end{bmatrix}\begin{bmatrix} \frac{186}{183} & \frac{90}{183} \\ -2 & 1 \end{bmatrix}\begin{bmatrix} \frac{186}{183} & \frac{90}{183} \\ \frac{-30}{183} & \frac{15}{183} \end{bmatrix}$ =A

(ii) Given equation are

$x_1+x_2+x_3=3$

$x_1-x_2+4x_3=4$

$2x_1+3x_2-5x_3=0$

A=$\begin{bmatrix} 1 & 1 &1\\ 1 & -1&4\\ 2 & 3 &-5 \end{bmatrix}$ ,B=$\begin{bmatrix} 3 \\ 4\\ 0 \end{bmatrix}$ ,X=$\begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix}$

Transforming matrix A to upper triangular matrix(U) using elementary row transformation and Lower triangular matrix(L) simultaneously

Applying $R_2\to R_2-R_1$

$U=\begin{bmatrix} 1 & 1 &1\\ 0 & -2&3\\ 2 & 3 &-5 \end{bmatrix}$ ,L=$\begin{bmatrix} 1 & 0 &0\\ 1 & 1&0\\ 0 & 0 &1 \end{bmatrix}$

Appling $R_3\to R_3-2R_1$

$U=\begin{bmatrix} 1 & 1 &1\\ 0 & -2&3\\ 0 & 1 &-7 \end{bmatrix}$ ,L=$\begin{bmatrix} 1 & 0 &0\\ 1 & 1&0\\ 2 & 0 &1 \end{bmatrix}$

Applying $R_3\to R_3+\frac{1}{2}R_2$

$U=\begin{bmatrix} 1 & 1 &1\\ 0 & -2&3\\ 0 & 0 &\frac{-11}{2} \end{bmatrix}$ ,L=$\begin{bmatrix} 1 & 0 &0\\ 1 & 1&0\\ 2 & \frac{-1}{3}&1 \end{bmatrix}$

So LY=B

$\begin{bmatrix} 1 & 0 &0\\ 1 & 1&0\\ 2 & 0 &1 \end{bmatrix} \begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} =\begin{bmatrix} 3\\ 4\\ 0 \end{bmatrix}$

$\begin{bmatrix} y_1\\ y_1+y_2\\ 2y_1+y_3 \end{bmatrix}$ =$\begin{bmatrix} 3\\ 4\\ 0 \end{bmatrix}$

$y_1=2,y_1+y_2=4,y_2=2$

$2y_1+y_3=0, y_3=-2$

Now UX=Y

$\begin{bmatrix} 1 & 1 &1\\ 0 & -2&3\\ 0 & 1 &-7 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =\begin{bmatrix} 2\\ 2\\ -2 \end{bmatrix}$

$x_1+x_2+x_3=2, -2x_2+3x_3=2$

$x_2-7x_3=-2$

Solving these equation we get,

$x_1=\frac{28}{11},x_2=\frac{-8}{11},x_3=\frac{2}{11}$