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

Let $A$ be a square matrix of order $3$. Choose the correct option regarding the following statements:
$I$. There exists a matrix $B$ of order $3$ such that $AB = I_3$
$II$. There exists a matrix $C$ of order $3$ such that $CA = I_3$
$III$. $A$ is invertible

Find the inverse of the matrix,if it exists: $\left[\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right]$

The element in the third row and the second column in the inverse matrix of a matrix $A = \begin{bmatrix} 1 & 3 & 3 \\ 3 & 1 & 3 \\ 3 & 3 & 4 \end{bmatrix}$ is

$\begin{aligned} & A(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^\beta\end{array}\right] \\ & \Rightarrow[A(\alpha, \beta)]^{-1}=\end{aligned}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module