The characteristic equation of a matrix A is | vedclass

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

Question

(D) The characteristic equation of a $3 	imes 3$ matrix $A$ is given by $|A - \lambda I| = 0$,which results in $-\lambda^3 + 	ext{tr}(A)\lambda^2 -

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The characteristic equation of a $3 	imes 3$ matrix $A$ is given by $|A - \lambda I| = 0$,which results in $-\lambda^3 + 	ext{tr}(A)\lambda^2 - (\dots)\lambda + |A| = 0$. Given the characteristic equation $\lambda^{3}-5 \lambda^{2}-3 \lambda+2=0$,we compare this with the standard form $\lambda^3 - 	ext{tr}(A)\lambda^2 + (\dots)\lambda - |A| = 0$. Comparing the constant terms,we get $-|A| = 2$,which implies $|A| = -2$. We know that for a square matrix $A$ of order $n$,$|	ext{adj}(A)| = |A|^{n-1}$. Here,$n = 3$,so $|	ext{adj}(A)| = |A|^{3-1} = |A|^2$. Substituting the value of $|A|$,we get $|	ext{adj}(A)| = (-2)^2 = 4$.

The characteristic equation of a matrix $A$ is $\lambda^{3}-5 \lambda^{2}-3 \lambda+2=0$. Then $|\text{adj}(A)|$ is equal to:

Similar Questions

If $A = \begin{bmatrix} 1 & 2 & i \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix}$,then $[\operatorname{adj}(\operatorname{adj} A)]^{-1} = $

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

If $A = \begin{bmatrix} 2 & 1 & 3 \\ -1 & 2 & 0 \\ 4 & 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 2 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix}$,then $\operatorname{det}(2 B^{-1} A^{-1})$ is equal to

Using elementary transformations,find the inverse of the following matrix,if it exists: $A = \begin{bmatrix} -1 & 2 \\ -3 & 5 \end{bmatrix}$

Find the inverse of the matrix $A = \left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]$,if it exists.

Vedclass Test Series

Exam Paper Generator

Online Exam Module