For M=left[begin{array}{ll}3 & -4 1 & -1end{ | vedclass

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

Question

(B) Given $M = \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight]$.First,calculate $M^2 = M 	imes M = \left[\begin{array}{ll}3 & -4 \ 1 & -1\

Vedclass · Accepted Answer

$\left[\begin{array}{ll}2 & -4 \ 1 & -2\end{array}ight]$

Vedclass · Answer

(B) Given $M = \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight]$.First,calculate $M^2 = M 	imes M = \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight] \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight] = \left[\begin{array}{ll}9-4 & -12+4 \ 3-1 & -4+1\end{array}ight] = \left[\begin{array}{ll}5 & -8 \ 2 & -3\end{array}ight]$.Now,consider the expression $M^{n+1}-M^n = M^n(M-I)$.For $n=1$,$M^2-M = \left[\begin{array}{ll}5 & -8 \ 2 & -3\end{array}ight] - \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight] = \left[\begin{array}{ll}2 & -4 \ 1 & -2\end{array}ight]$.Note that $M^2 - M = \left[\begin{array}{ll}2 & -4 \ 1 & -2\end{array}ight]$.Since $M^2 - 2M + I = 0$ (characteristic equation $\lambda^2 - 2\lambda + 1 = 0$),we have $M^2 = 2M - I$,which implies $M^2 - M = M - I$.Thus,$M^{n+1}-M^n = M^{n-1}(M^2-M) = M^{n-1}(M-I) = M^n - M^{n-1}$.By induction,$M^{n+1}-M^n = M^2-M = \left[\begin{array}{ll}2 & -4 \ 1 & -2\end{array}ight]$ for all $n \in N$.

For $M=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ and for any $n \in N$,the matrix $M^{n+1}-M^n=$

Similar Questions

Let $A$ be a $3 \times 3$ matrix such that $A^2 - 5A + 7I = 0$.
Statement-$I$: ${A^{-1}} = \frac{1}{7}(5I - A)$.
Statement-$II$: The polynomial $A^3 - 2A^2 - 3A + I$ can be reduced to $5(A - 4I)$.

If $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$,prove that $A^{3} - 6A^{2} + 7A + 2I = 0$.

Let $ABC = I$. Then $tr(ABC + BCA + CAB)$ is (where the order of matrices $A, B, C$ is $3 \times 3$ and $tr(A)$ is the sum of the diagonal elements in $A$).

$A$ and $B$ are two square matrices such that $A^2B = BA$. If $(AB)^{10} = A^K B^{10}$,then $k$ is:

The number of ordered pairs $(x, y)$ for which $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2 \end{bmatrix}$ is a singular and symmetric matrix is

Vedclass Test Series

Exam Paper Generator

Online Exam Module