written 5.0 years ago by | • modified 5.0 years ago |
Solution:
Given, P(x) = $7+8x+9x^2$
$P_1(x)=2+x+4x^2,$
$P_2(x)=1-x+3x^2,$
$P_3(x)=2+x+5x^2$
Let $P=K_1P_1+k_2P_2+k_3P_3$
$(7+8x+9x^2)= k_1(2+x+4x^2)+k_2(1-x+3x^2)+k_3(2+x+5x^2)$-------(1)
$(7+8x+9x^2)= (2k_1+k_2+2k_3)+(K_1-k_2+K_3)x+(4k_1+3K_2+5k_3)x^2$
equating both side,
$2k_1+k_2+2k_3=7$-----(2)
$k_1-k_2+k_3=8$------(3)
$4k_1+3k_2+5k_3=9$-------(4)
$\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 4 & 3 & 5 \end{bmatrix}\begin{bmatrix} k_1\\ k_2 \\ k_3 \end{bmatrix}=\begin{bmatrix} 8\\ 7 \\ 9 \end{bmatrix}$
$R_2-2R_1,\\ R_3-2R_2 $ $\begin{bmatrix} 1 & -1 & 1 \\ 0 & 3 & 0 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} k_1\\ k_2 \\ k_3 \end{bmatrix}=\begin{bmatrix} 8\\ -9 \\ -5 \end{bmatrix}$
$R_{23}$ $\begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & 1 \\ 0 & 3 & 0 \end{bmatrix}\begin{bmatrix} k_1\\ k_2 \\ k_3 \end{bmatrix}=\begin{bmatrix} 8\\ -5 \\ -9 \end{bmatrix}$
$k_1-k_2+k_3=8$-----(5)
$k_2+k_3=-5$----(6)
$3k_2=-9$
$k_2=-3$
Put $k_2=-3$ in eqn(6)
$-3+k_3=-5$
$k_3=-5+3$
$k_3=-2$
put $k_2 \ and \ k_3$ values in (5)
$k_1-(-3)-2=8 $
$k_1+3-2=8$
$k_1=8-1$
$k_1 =7 $
put $k_1 ,\ k_2 \ and \ k_3$ values in eqn(1)
$7+8x+9x^2=7(2+x+4x^2)-3(1-x+3x^2)-2(2+x+5x^2)$