Suppose a_1, a_2, dots are real numbers,with | vedclass

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

Question

(D) Let the common difference of the $A.P.$ be $d$. Then $a_n = a_1 + (n-1)d$. For matrix $A$,$|A| = \begin{vmatrix} a_1 & a_2 & a_3 \ a_4 & a_5 & a_

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) Let the common difference of the $A.P.$ be $d$. Then $a_n = a_1 + (n-1)d$. For matrix $A$,$|A| = \begin{vmatrix} a_1 & a_2 & a_3 \ a_4 & a_5 & a_6 \ a_5 & a_6 & a_7 \end{vmatrix}$. Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_2$,we get $R_2 = (3d, 3d, 3d)$ and $R_3 = (d, d, d)$. Since $R_2 = 3R_3$,the rows are linearly dependent,so $|A| = 0$. Thus,$A$ is singular. For the system of equations,the determinant of the coefficient matrix is $|C| = \begin{vmatrix} a_1 & a_2 & a_3 \ a_4 & a_5 & a_6 \ a_7 & a_8 & a_9 \end{vmatrix}$. Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_2$,we get $R_2 = (3d, 3d, 3d)$ and $R_3 = (3d, 3d, 3d)$. Since $R_2 = R_3$,$|C| = 0$. Thus,the system has infinite solutions. For matrix $B$,$|B| = a_1^2 - (i a_2)^2 = a_1^2 + a_2^2$. Since $a_1, a_2$ are real and $a_1 
eq 0$,$|B| > 0$,so $B$ is non-singular. Therefore,all statements are correct.

Suppose $a_1, a_2, \dots$ are real numbers,with $a_1 \neq 0$. If $a_1, a_2, a_3, \dots$ are in $A.P.$,then:

Similar Questions

Let matrix $A = \begin{bmatrix} 5 & -3 & 0 \\ -3 & 5 & 0 \\ 0 & 0 & 2 \end{bmatrix}$,$X$ be a non-zero matrix of order $3 \times 1$,and $c$ be a real number. If $A^2 X = cAX$,then the number of distinct values of $c$ is:

Let $[A]_{3 \times 3}$ be a non-singular matrix such that $A^{-1}=\frac{1}{3}(A^2-5A+7I)$. Then $17A^8-85A^7+119A^6-51A^5-19A^4+95A^3-133A^2+58A+I=$

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

If matrix $A = [a_{ij}]_{3 \times 3}$ and $B = [b_{ij}]_{3 \times 3}$,where $a_{ij} + a_{ji} = 0$ and $b_{ij} - b_{ji} = 0$ for all $i, j$,then $A^4B^3$ is:

The value of $\theta$ lying between $-\frac{\pi}{4}$ and $\frac{\pi}{2}$ and $0 \le A \le \frac{\pi}{2}$ satisfying the equation $\begin{vmatrix} 1 + \sin^2 A & \cos^2 A & 2 \sin 4\theta \\ \sin^2 A & 1 + \cos^2 A & 2 \sin 4\theta \\ \sin^2 A & \cos^2 A & 1 + 2 \sin 4\theta \end{vmatrix} = 0$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module