(i) A=[ 1 2 − 3 2 3 15 ] \begin{bmatrix}
\frac{1}{2} & -3 \\
2 & \frac{3}{15}
\end{bmatrix} [ 2 1 2 − 3 15 3 ]
Let I=[ 1 0 0 1 ] \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} [ 1 0 0 1 ]
Applying IA=A
[ 1 0 0 1 ] \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} [ 1 0 0 1 ] [ 1 2 − 3 2 3 15 ] \begin{bmatrix}
\frac{1}{2} & -3 \\
2 & \frac{3}{15}
\end{bmatrix} [ 2 1 2 − 3 15 3 ] [ 1 2 − 3 2 3 15 ] \begin{bmatrix}
\frac{1}{2} & -3 \\
2 & \frac{3}{15}
\end{bmatrix} [ 2 1 2 − 3 15 3 ]
Applying R 1 → 2 R 1 R_1\to2R_1 R 1 → 2 R 1
[ 2 0 0 1 ] [ 1 2 − 3 2 3 15 ] = [ 1 − 6 2 3 15 ] \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} [ 2 0 0 1 ] [ 2 1 2 − 3 15 3 ] = [ 1 2 − 6 15 3 ]
Applying R 2 → R 2 − 2 R 1 R_2\to R_2-2R_1 R 2 → R 2 − 2 R 1
[ 2 0 − 2 1 ] [ 1 2 − 3 2 3 15 ] = [ 1 − 6 0 183 15 ] \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} [ 2 − 2 0 1 ] [ 2 1 2 − 3 15 3 ] = [ 1 0 − 6 15 183 ]
Applying R 1 → R 1 + 90 183 R 2 R_1\to R_1+\frac{90}{183}R_2 R 1 → R 1 + 183 90 R 2
[ 186 183 90 183 − 2 1 ] [ 1 2 − 3 2 3 15 ] = [ 1 0 0 183 15 ] \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} [ 183 186 − 2 183 90 1 ] [ 2 1 2 − 3 15 3 ] = [ 1 0 0 15 183 ]
applying R 2 → 15 183 R 2 R_2\to \frac{15}{183}R_2 R 2 → 183 15 R 2
[ 186 183 90 183 − 30 183 15 183 ] [ 1 2 − 3 2 3 15 ] = [ 1 0 0 1 ] \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} [ 183 186 183 − 30 183 90 183 15 ] [ 2 1 2 − 3 15 3 ] = [ 1 0 0 1 ]
Sequence of elementary matrix are
[ 1 0 0 1 ] \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} [ 1 0 0 1 ] [ 2 0 0 1 ] \begin{bmatrix}
2 & 0 \\
0 & 1
\end{bmatrix} [ 2 0 0 1 ] [ 2 0 − 2 1 ] [ 186 183 90 183 − 2 1 ] [ 186 183 90 183 − 30 183 15 183 ] \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} [ 2 − 2 0 1 ] [ 183 186 − 2 183 90 1 ] [ 183 186 183 − 30 183 90 183 15 ]
(ii) Given equation are
x 1 + x 2 + x 3 = 3 x_1+x_2+x_3=3 x 1 + x 2 + x 3 = 3
x 1 − x 2 + 4 x 3 = 4 x_1-x_2+4x_3=4 x 1 − x 2 + 4 x 3 = 4
2 x 1 + 3 x 2 − 5 x 3 = 0 2x_1+3x_2-5x_3=0 2 x 1 + 3 x 2 − 5 x 3 = 0
A=[ 1 1 1 1 − 1 4 2 3 − 5 ] \begin{bmatrix}
1 & 1 &1\\
1 & -1&4\\
2 & 3 &-5
\end{bmatrix} ⎣ ⎡ 1 1 2 1 − 1 3 1 4 − 5 ⎦ ⎤ [ 3 4 0 ] \begin{bmatrix}
3 \\
4\\
0
\end{bmatrix} ⎣ ⎡ 3 4 0 ⎦ ⎤ [ x 1 x 2 x 3 ] \begin{bmatrix}
x_1 \\
x_2\\
x_3
\end{bmatrix} ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤
Transforming matrix A to upper triangular matrix(U) using elementary row transformation and Lower triangular matrix(L) simultaneously
Applying R 2 → R 2 − R 1 R_2\to R_2-R_1 R 2 → R 2 − R 1
U = [ 1 1 1 0 − 2 3 2 3 − 5 ] U=\begin{bmatrix}
1 & 1 &1\\
0 & -2&3\\
2 & 3 &-5
\end{bmatrix} U = ⎣ ⎡ 1 0 2 1 − 2 3 1 3 − 5 ⎦ ⎤ [ 1 0 0 1 1 0 0 0 1 ] \begin{bmatrix}
1 & 0 &0\\
1 & 1&0\\
0 & 0 &1
\end{bmatrix} ⎣ ⎡ 1 1 0 0 1 0 0 0 1 ⎦ ⎤
Appling R 3 → R 3 − 2 R 1 R_3\to R_3-2R_1 R 3 → R 3 − 2 R 1
U = [ 1 1 1 0 − 2 3 0 1 − 7 ] U=\begin{bmatrix}
1 & 1 &1\\
0 & -2&3\\
0 & 1 &-7
\end{bmatrix} U = ⎣ ⎡ 1 0 0 1 − 2 1 1 3 − 7 ⎦ ⎤ [ 1 0 0 1 1 0 2 0 1 ] \begin{bmatrix}
1 & 0 &0\\
1 & 1&0\\
2 & 0 &1
\end{bmatrix} ⎣ ⎡ 1 1 2 0 1 0 0 0 1 ⎦ ⎤
Applying R 3 → R 3 + 1 2 R 2 R_3\to R_3+\frac{1}{2}R_2 R 3 → R 3 + 2 1 R 2
U = [ 1 1 1 0 − 2 3 0 0 − 11 2 ] U=\begin{bmatrix}
1 & 1 &1\\
0 & -2&3\\
0 & 0 &\frac{-11}{2}
\end{bmatrix} U = ⎣ ⎡ 1 0 0 1 − 2 0 1 3 2 − 11 ⎦ ⎤ [ 1 0 0 1 1 0 2 − 1 3 1 ] \begin{bmatrix}
1 & 0 &0\\
1 & 1&0\\
2 & \frac{-1}{3}&1
\end{bmatrix} ⎣ ⎡ 1 1 2 0 1 3 − 1 0 0 1 ⎦ ⎤
So LY=B
[ 1 0 0 1 1 0 2 0 1 ] [ y 1 y 2 y 3 ] = [ 3 4 0 ] \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} ⎣ ⎡ 1 1 2 0 1 0 0 0 1 ⎦ ⎤ ⎣ ⎡ y 1 y 2 y 3 ⎦ ⎤ = ⎣ ⎡ 3 4 0 ⎦ ⎤
[ y 1 y 1 + y 2 2 y 1 + y 3 ] \begin{bmatrix}
y_1\\
y_1+y_2\\
2y_1+y_3
\end{bmatrix} ⎣ ⎡ y 1 y 1 + y 2 2 y 1 + y 3 ⎦ ⎤ [ 3 4 0 ] \begin{bmatrix}
3\\
4\\
0
\end{bmatrix} ⎣ ⎡ 3 4 0 ⎦ ⎤
y 1 = 2 , y 1 + y 2 = 4 , y 2 = 2 y_1=2,y_1+y_2=4,y_2=2 y 1 = 2 , y 1 + y 2 = 4 , y 2 = 2
2 y 1 + y 3 = 0 , y 3 = − 2 2y_1+y_3=0, y_3=-2 2 y 1 + y 3 = 0 , y 3 = − 2
Now UX=Y
[ 1 1 1 0 − 2 3 0 1 − 7 ] [ x 1 x 2 x 3 ] = [ 2 2 − 2 ] \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} ⎣ ⎡ 1 0 0 1 − 2 1 1 3 − 7 ⎦ ⎤ ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ 2 2 − 2 ⎦ ⎤
x 1 + x 2 + x 3 = 2 , − 2 x 2 + 3 x 3 = 2 x_1+x_2+x_3=2, -2x_2+3x_3=2 x 1 + x 2 + x 3 = 2 , − 2 x 2 + 3 x 3 = 2
x 2 − 7 x 3 = − 2 x_2-7x_3=-2 x 2 − 7 x 3 = − 2
Solving these equation we get,
x 1 = 28 11 , x 2 = − 8 11 , x 3 = 2 11 x_1=\frac{28}{11},x_2=\frac{-8}{11},x_3=\frac{2}{11} x 1 = 11 28 , x 2 = 11 − 8 , x 3 = 11 2
#340153 on Dec 2023
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