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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

If $\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = I_2$,then $A =$

Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = \begin{bmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{bmatrix}$,$|A| = -1$. Let $B$ be the inverse of the matrix $\operatorname{adj}(A \operatorname{adj}(A^2))$. Then $|\lambda B + I|$ is equal to . . . . . . .

If $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$ and $A + A^{-1} = I$,then $\alpha =$

If $A=\begin{bmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix}$ and $A^{-1}=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, then $\sum_{1 \leq i, j \leq 3} a_{ij} =$

If $A = \begin{bmatrix} i & 0 \\ 0 & i/2 \end{bmatrix}$ where $i = \sqrt{-1}$,then $A^{-1} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module