If A=left[begin{array}{ccc}1 & 1 & 3 5 & 2 & | vedclass

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

Question

(B) Given $A=\left[\begin{array}{ccc}1 & 1 & 3 \ 5 & 2 & 6 \ -2 & -1 & -3\end{array}ight]$.The characteristic equation is given by $|A-\lambda I|=

Vedclass · Accepted Answer

$\left[\begin{array}{ccc}4 & 1 & 3 \ 5 & 5 & 6 \ -2 & -1 & 0\end{array}ight]$

Vedclass · Answer

(B) Given $A=\left[\begin{array}{ccc}1 & 1 & 3 \ 5 & 2 & 6 \ -2 & -1 & -3\end{array}ight]$.The characteristic equation is given by $|A-\lambda I|=0$.$\left|\begin{array}{ccc}1-\lambda & 1 & 3 \ 5 & 2-\lambda & 6 \ -2 & -1 & -3-\lambda\end{array}ight|=0$Expanding the determinant:$(1-\lambda)[(2-\lambda)(-3-\lambda) - (-6)] - 1[5(-3-\lambda) - (-12)] + 3[5(-1) - (-2)(2-\lambda)] = 0$$(1-\lambda)[\lambda^2+\lambda-6+6] - 1[-15-5\lambda+12] + 3[-5+4-2\lambda] = 0$$(1-\lambda)(\lambda^2+\lambda) - 1(-5\lambda-3) + 3(-2\lambda-1) = 0$$\lambda^2+\lambda-\lambda^3-\lambda^2 + 5\lambda+3 - 6\lambda-3 = 0$$-\lambda^3 = 0 \Rightarrow \lambda^3 = 0$.By the Cayley-Hamilton theorem,every matrix satisfies its characteristic equation,so $A^3 = 0$.Since $A^3 = 0$,it follows that $A^4 = A^3 \cdot A = 0$ and $A^5 = A^3 \cdot A^2 = 0$.Therefore,$A+A^3+A^4+A^5+3I = A + 0 + 0 + 0 + 3I = A + 3I$.$A+3I = \left[\begin{array}{ccc}1 & 1 & 3 \ 5 & 2 & 6 \ -2 & -1 & -3\end{array}ight] + \left[\begin{array}{ccc}3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3\end{array}ight] = \left[\begin{array}{ccc}4 & 1 & 3 \ 5 & 5 & 6 \ -2 & -1 & 0\end{array}ight]$.

If $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$,then $A+A^3+A^4+A^5+3 I=$

Similar Questions

The value of $\sum\limits_{n = 1}^N {{U_n}} $ if ${U_n} = \left| {\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}} \right|$ is

If $D_1$ and $D_2$ are two $3 \times 3$ diagonal matrices,then

If $A$ is a square matrix,such that $A^2=A$,then $(I+A)^3$ is equal to

The equation $\left| \begin{array}{ccc} (1+x)^2 & (1-x)^2 & -(2+x^2) \\ 2x+1 & 3x & 1-5x \\ x+1 & 2x & 2-3x \end{array} \right| + \left| \begin{array}{ccc} (1+x)^2 & 2x+1 & x+1 \\ (1-x)^2 & 3x & 2x \\ 1-2x & 3x-2 & 2x-3 \end{array} \right| = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module