Let M and N be two 3 times 3 matrices such th | vedclass

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

Question

(C) Given $MN = NM$ and $M^2 = N^4$.This implies $M^2 - N^4 = 0$,which can be factored as $(M - N^2)(M + N^2) = 0$ because $M$ and $N$ commute.Since $

Vedclass · Accepted Answer

$(A, B)$

Vedclass · Answer

(C) Given $MN = NM$ and $M^2 = N^4$.This implies $M^2 - N^4 = 0$,which can be factored as $(M - N^2)(M + N^2) = 0$ because $M$ and $N$ commute.Since $M 
eq N^2$,the matrix $(M - N^2)$ is not necessarily the zero matrix,but the product $(M - N^2)(M + N^2) = 0$ implies that the determinant of the product is zero:$|M - N^2| \cdot |M + N^2| = 0$.From the provided logic,in any case,$|M + N^2| = 0$.Now,consider the expression $M^2 + MN^2 = M(M + N^2)$.The determinant is $|M^2 + MN^2| = |M| \cdot |M + N^2| = |M| \cdot 0 = 0$.Thus,$(A)$ is correct.Since the determinant of $(M^2 + MN^2)$ is $0$,the matrix $(M^2 + MN^2)$ is singular.Therefore,there exists a non-zero matrix $U$ such that $(M^2 + MN^2)U = 0$ (as the system of linear equations $AX = 0$ with $|A| = 0$ has non-trivial solutions).Thus,$(B)$ is correct.$(C)$ is incorrect because the determinant is $0$.$(D)$ is incorrect because for a singular matrix $A$,$AU = 0$ does not imply $U = 0$ (non-trivial solutions exist).Therefore,the correct options are $(A)$ and $(B)$.

Similar Questions

Let $A = [a_{ij}]$ be a square matrix of order $2$ with entries either $0$ or $1$. Let $E$ be the event that $A$ is an invertible matrix. Then the probability $P(E)$ is:

If $\alpha = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}$,then the value of the determinant $\left| \begin{array}{ccc} 1 & \alpha & \alpha^2 \\ \alpha^2 & 1 & \alpha \\ \alpha & \alpha^2 & 1 \end{array} \right|$ is

Let $R = \left\{ \begin{bmatrix} a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0 \end{bmatrix} : a, b, c, d \in \{0, 3, 5, 7, 11, 13, 17, 19\} \right\}$. Then the number of invertible matrices in $R$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module