Let A, B be two 3 times 3 matrices and C be a | vedclass

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

Question

(C) Let $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix

Vedclass · Accepted Answer

Both Statement $I$ and Statement $II$ are true

Vedclass · Answer

(C) Let $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$, and $C = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Then $AB = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$, so $AB-C = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$.Since $\det(AB-C) = -1 
eq 0$, $AB-C$ is non-singular and $D = (AB-C)^{-1} = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix} = -C$.For Statement $I$: $\det(BA) = \det(0) = 0$. $\det(BA-C) = \det(-C) = -1$. $\det(BDA) = \det(-CBA) = \det(0) = 0$. Thus $0 = (-1)(0)$ is true.For Statement $II$: $ABD = AB(-C) = -AB = 0$. $DAB = (-C)AB = -AB = 0$. Thus $ABD = DAB$ is true.Both statements are true.

Similar Questions

If one of the cube roots of $1$ be $\omega$, then $\left|\begin{array}{ccc}1 & 1+\omega^2 & \omega^2 \\ 1-i & -1 & \omega^2-1 \\ -i & -1+\omega & -1\end{array}\right|=$

If $\begin{bmatrix} x & 4 & -1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$,then $x=$

If $A$ is a skew-symmetric matrix and $n$ is a positive integer,then $A^n$ is

Let $M$ be a $3 \times 3$ invertible matrix with real entries and let $I$ denote the $3 \times 3$ identity matrix. If $M^{-1} = \operatorname{adj}(\operatorname{adj} M)$,then which of the following statement$(s)$ is/are $ALWAYS \text{ } TRUE$?

Which of the following is not true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module