If frac{x^2+5x+1}{(x+1)(x+2)(x+3)}=frac{a}{x+ | vedclass

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(C) Given the equation: $\frac{x^2+5x+1}{(x+1)(x+2)(x+3)} = \frac{a}{x+1} + \frac{b}{(x+1)(x+2)} + \frac{c}{(x+1)(x+2)(x+3)}$.Multiply both sides by $

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$\left[\begin{array}{ll}1 & 0 \ 5 & 1\end{array}ight]$

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(C) Given the equation: $\frac{x^2+5x+1}{(x+1)(x+2)(x+3)} = \frac{a}{x+1} + \frac{b}{(x+1)(x+2)} + \frac{c}{(x+1)(x+2)(x+3)}$.Multiply both sides by $(x+1)(x+2)(x+3)$:$x^2+5x+1 = a(x+2)(x+3) + b(x+3) + c$.For $x = -1$: $(-1)^2 + 5(-1) + 1 = a(1)(2) + b(2) + c \implies 1-5+1 = 2a+2b+c \implies -3 = 2a+2b+c$.For $x = -2$: $(-2)^2 + 5(-2) + 1 = a(0) + b(1) + c \implies 4-10+1 = b+c \implies -5 = b+c$.For $x = -3$: $(-3)^2 + 5(-3) + 1 = a(0) + b(0) + c \implies 9-15+1 = c \implies c = -5$.Using $b+c = -5$,we get $b-5 = -5 \implies b = 0$.Using $2a+2b+c = -3$,we get $2a+0-5 = -3 \implies 2a = 2 \implies a = 1$.The matrix is $M = \left[\begin{array}{ll}a & b \ c & 1\end{array}ight] = \left[\begin{array}{ll}1 & 0 \ -5 & 1\end{array}ight]$.The inverse $M^{-1} = \frac{1}{\det(M)} 	ext{adj}(M)$.$\det(M) = (1)(1) - (0)(-5) = 1$.$	ext{adj}(M) = \left[\begin{array}{ll}1 & 0 \ 5 & 1\end{array}ight]$.Thus,$M^{-1} = \left[\begin{array}{ll}1 & 0 \ 5 & 1\end{array}ight]$.

If $\frac{x^2+5x+1}{(x+1)(x+2)(x+3)}=\frac{a}{x+1}+\frac{b}{(x+1)(x+2)}+\frac{c}{(x+1)(x+2)(x+3)}$,then the inverse of the matrix $\left[\begin{array}{ll}a & b \\ c & 1\end{array}\right]$ is

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