Question #139244
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
1
Expert's answer
2020-10-22T17:57:01-0400

(i) A=[1232315]\begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix}


Let I=[1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Applying IA=A

[1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} [1232315]\begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix} =[1232315]\begin{bmatrix} \frac{1}{2} & -3 \\ 2 & \frac{3}{15} \end{bmatrix}


Applying R12R1R_1\to2R_1

[2001][1232315]=[162315]\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 R2R22R1R_2\to R_2-2R_1

[2021][1232315]=[16018315]\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 R1R1+90183R2R_1\to R_1+\frac{90}{183}R_2


[1861839018321][1232315]=[10018315]\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 R215183R2R_2\to \frac{15}{183}R_2

[186183901833018315183][1232315]=[1001]\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

[1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} [2001]\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} [2021][1861839018321][186183901833018315183]\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

x1+x2+x3=3x_1+x_2+x_3=3

x1x2+4x3=4x_1-x_2+4x_3=4

2x1+3x25x3=02x_1+3x_2-5x_3=0


A=[111114235]\begin{bmatrix} 1 & 1 &1\\ 1 & -1&4\\ 2 & 3 &-5 \end{bmatrix} ,B=[340]\begin{bmatrix} 3 \\ 4\\ 0 \end{bmatrix} ,X=[x1x2x3]\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 R2R2R1R_2\to R_2-R_1


U=[111023235]U=\begin{bmatrix} 1 & 1 &1\\ 0 & -2&3\\ 2 & 3 &-5 \end{bmatrix} ,L=[100110001]\begin{bmatrix} 1 & 0 &0\\ 1 & 1&0\\ 0 & 0 &1 \end{bmatrix}


Appling R3R32R1R_3\to R_3-2R_1

U=[111023017]U=\begin{bmatrix} 1 & 1 &1\\ 0 & -2&3\\ 0 & 1 &-7 \end{bmatrix} ,L=[100110201]\begin{bmatrix} 1 & 0 &0\\ 1 & 1&0\\ 2 & 0 &1 \end{bmatrix}



Applying R3R3+12R2R_3\to R_3+\frac{1}{2}R_2


U=[11102300112]U=\begin{bmatrix} 1 & 1 &1\\ 0 & -2&3\\ 0 & 0 &\frac{-11}{2} \end{bmatrix} ,L=[1001102131]\begin{bmatrix} 1 & 0 &0\\ 1 & 1&0\\ 2 & \frac{-1}{3}&1 \end{bmatrix}


So LY=B

[100110201][y1y2y3]=[340]\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}


[y1y1+y22y1+y3]\begin{bmatrix} y_1\\ y_1+y_2\\ 2y_1+y_3 \end{bmatrix} =[340]\begin{bmatrix} 3\\ 4\\ 0 \end{bmatrix}


y1=2,y1+y2=4,y2=2y_1=2,y_1+y_2=4,y_2=2


2y1+y3=0,y3=22y_1+y_3=0, y_3=-2


Now UX=Y


[111023017][x1x2x3]=[222]\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}


x1+x2+x3=2,2x2+3x3=2x_1+x_2+x_3=2, -2x_2+3x_3=2

x27x3=2x_2-7x_3=-2


Solving these equation we get,

x1=2811,x2=811,x3=211x_1=\frac{28}{11},x_2=\frac{-8}{11},x_3=\frac{2}{11}




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