Let A = begin{bmatrix} 2 & 1 & 1 0 & 1 & 0 | vedclass

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

Question

(B) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$. $|A - \lambda I| = \begin{vmatrix} 2-\lambda & 1 & 1 \ 0 & 1-\lambda

Vedclass · Accepted Answer

$\frac{-1}{3}$

Vedclass · Answer

(B) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$. $|A - \lambda I| = \begin{vmatrix} 2-\lambda & 1 & 1 \ 0 & 1-\lambda & 0 \ 1 & 1 & 2-\lambda \end{vmatrix} = (1-\lambda) \begin{vmatrix} 2-\lambda & 1 \ 1 & 2-\lambda \end{vmatrix} = (1-\lambda)((2-\lambda)^2 - 1) = (1-\lambda)(\lambda^2 - 4\lambda + 3) = (1-\lambda)(\lambda-1)(\lambda-3) = -(\lambda-1)^2(\lambda-3) = -(\lambda^2 - 2\lambda + 1)(\lambda-3) = -(\lambda^3 - 3\lambda^2 - 2\lambda^2 + 6\lambda + \lambda - 3) = -(\lambda^3 - 5\lambda^2 + 7\lambda - 3) = -\lambda^3 + 5\lambda^2 - 7\lambda + 3 = 0$. By Cayley-Hamilton theorem, $A^3 - 5A^2 + 7A - 3I = 0$. Multiplying by $A^{-1}$, we get $A^2 - 5A + 7I - 3A^{-1} = 0$, which implies $3A^{-1} = A^2 - 5A + 7I$, or $A^{-1} = \frac{1}{3}A^2 - \frac{5}{3}A + \frac{7}{3}I$. Comparing this with $A^{-1} = \alpha A^2 + \beta A + \gamma I$, we get $\alpha = \frac{1}{3}$, $\beta = -\frac{5}{3}$, and $\gamma = \frac{7}{3}$. Thus, $17\alpha + 5\beta + \gamma = 17(\frac{1}{3}) + 5(-\frac{5}{3}) + \frac{7}{3} = \frac{17 - 25 + 7}{3} = \frac{-1}{3}$.

Let $A = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2 \end{bmatrix}$. If $A^{-1} = \alpha A^2 + \beta A + \gamma I$, where $\alpha, \beta, \gamma$ are real numbers and $I$ is a $3 \times 3$ identity matrix, then $17 \alpha + 5 \beta + \gamma =$

Similar Questions

If $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$,then the inverse of the matrix $A^3$ is

Let $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$ and $A^{-1} = xA + yI_2$,(where $I_2$ is the unit matrix of order $2$),then

Let $P=[p_{ij}]$ and $Q=[q_{ij}]$ be two square matrices of order $3$ such that $q_{ij}=2^{(i+j-1)}p_{ij}$ and $\det(Q)=2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:

Given $A$ and $C$ are involutory matrices and $B$ is a non-singular matrix,then $(AB^{-1}C)^{-1}$ is equal to -

If $A$ is a square matrix of order $3$ and $|A|=5$,then $|A \text{ adj. } A|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module