Let M = begin{bmatrix} a & b b & c end{bmatr | vedclass

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

Question

(B) The determinant of $M$ is $|M| = ac - b^2$.$(A)$ The first column is $\begin{bmatrix} a \ b \end{bmatrix}$ and the transpose of the second row is

Vedclass · Accepted Answer

$(C, D)$

Vedclass · Answer

(B) The determinant of $M$ is $|M| = ac - b^2$.$(A)$ The first column is $\begin{bmatrix} a \ b \end{bmatrix}$ and the transpose of the second row is $\begin{bmatrix} b \ c \end{bmatrix}$. If they are equal, then $a=b$ and $b=c$, so $a=b=c$. Then $M = \begin{bmatrix} a & a \ a & a \end{bmatrix}$, which has $|M| = a^2 - a^2 = 0$. Thus, $M$ is not invertible. $(A)$ is incorrect.$(B)$ The second row is $[b, c]$ and the transpose of the first column is $\begin{bmatrix} a \ b \end{bmatrix}$. If they are equal, then $b=a$ and $c=b$, so $a=b=c$. This leads to the same matrix as $(A)$, which is not invertible. $(B)$ is incorrect.$(C)$ If $M$ is a diagonal matrix, $M = \begin{bmatrix} a & 0 \ 0 & c \end{bmatrix}$. The determinant is $|M| = ac$. Since $a, c 
eq 0$, $|M| 
eq 0$. Thus, $M$ is invertible. $(C)$ is correct.$(D)$ For $M$ to be invertible, $|M| 
eq 0$, which means $ac - b^2 
eq 0$, or $ac 
eq b^2$. Since $b$ is an integer, $b^2$ is a perfect square. If $ac$ is not a perfect square, then $ac 
eq b^2$ is guaranteed. $(D)$ is correct.Therefore, $(C, D)$ are correct.

Similar Questions

Let $A = \left| \begin{array}{cc} 2 & e^{i \pi} \\ -1 & i^{2012} \end{array} \right|$,$C = \left. \frac{d}{dx} \left( \frac{1}{x} \right) \right|_{x=1}$,and $D = \int_{e^2}^{1} \frac{dx}{x}$. If the sum of two roots of the equation $Ax^3 + Bx^2 + Cx - D = 0$ is equal to zero,then $B$ is equal to:

If $M$ and $N$ are square matrices of order $3$, then which one of the following statements is not true?

Let $A$ be a non-zero periodic matrix with period $4$ and $A^{12} + B = I$,where $I$ is the identity matrix and $B$ is any square matrix of the same order as $A$. The matrix product $AB$ is equal to:

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 $A(\theta)=\begin{bmatrix} i \sin \theta & \cos \theta \\ \cos \theta & i \sin \theta \end{bmatrix}$ is a matrix,where $i=\sqrt{-1}$,then which of the following is not true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module