The characteristic equation of the matrix A = | vedclass

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

Question

(D) The characteristic equation is given by $|A - xI| = 0$.Given $A = \begin{bmatrix} 2 & 3 & 0 \ 1 & 2 & 5 \ 3 & -1 & 2 \end{bmatrix}$,then $A - xI

Vedclass · Accepted Answer

$x^3 - 6x^2 + 14x - 57 = 0$

Vedclass · Answer

(D) The characteristic equation is given by $|A - xI| = 0$.Given $A = \begin{bmatrix} 2 & 3 & 0 \ 1 & 2 & 5 \ 3 & -1 & 2 \end{bmatrix}$,then $A - xI = \begin{bmatrix} 2-x & 3 & 0 \ 1 & 2-x & 5 \ 3 & -1 & 2-x \end{bmatrix}$.The determinant $|A - xI| = (2-x) \begin{vmatrix} 2-x & 5 \ -1 & 2-x \end{vmatrix} - 3 \begin{vmatrix} 1 & 5 \ 3 & 2-x \end{vmatrix} + 0 \begin{vmatrix} 1 & 2-x \ 3 & -1 \end{vmatrix} = 0$.$(2-x)((2-x)^2 + 5) - 3(2-x - 15) = 0$.$(2-x)(4 - 4x + x^2 + 5) - 3(-x - 13) = 0$.$(2-x)(x^2 - 4x + 9) + 3x + 39 = 0$.$2x^2 - 8x + 18 - x^3 + 4x^2 - 9x + 3x + 39 = 0$.$-x^3 + 6x^2 - 14x + 57 = 0$.Multiplying by $-1$,we get $x^3 - 6x^2 + 14x - 57 = 0$.

The characteristic equation of the matrix $A = \begin{bmatrix} 2 & 3 & 0 \\ 1 & 2 & 5 \\ 3 & -1 & 2 \end{bmatrix}$ is:

Similar Questions

If $f(x) = \left| \begin{array}{ccc} x & x+1 & x+3 \\ x+2 & x+4 & x+7 \\ x+6 & x+9 & x+13 \end{array} \right|$, then $f(5) =$

The value of the determinant $\left|\begin{array}{lll}b^2-a b & b-c & b c-a c \\ a b-a^2 & a-b & b^2-a b \\ b c-a c & c-a & a b-a^2\end{array}\right|$ is

$\left| \begin{array}{ccc} 2 \sin \frac{\pi}{3} & 1 & 0 \\ 1 & 2 \sin \frac{\pi}{3} & 1 \\ 0 & 1 & 2 \cos \frac{\pi}{6} \end{array} \right| = $ . . . . . .

Evaluate $\left|\begin{array}{cc}x & x+1 \\ x-1 & x\end{array}\right|$

Given the system of equations $a(x + y + z) = x$,$b(x + y + z) = y$,$c(x + y + z) = z$ where $a, b, c$ are non-zero real numbers. If the real numbers $x, y, z$ are such that $xyz \neq 0$,then $(a + b + c)$ is equal to-

Vedclass Test Series

Exam Paper Generator

Online Exam Module