If A = begin{bmatrix} 0 & x & 16 x & 5 & 7 | vedclass

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

Question

(A) matrix $A$ is singular if its determinant $|A| = 0$.Given $A = \begin{bmatrix} 0 & x & 16 \ x & 5 & 7 \ 0 & 9 & x \end{bmatrix}$.Expanding the d

Vedclass · Accepted Answer

$0, 12, -12$

Vedclass · Answer

(A) matrix $A$ is singular if its determinant $|A| = 0$.Given $A = \begin{bmatrix} 0 & x & 16 \ x & 5 & 7 \ 0 & 9 & x \end{bmatrix}$.Expanding the determinant along the first column:$|A| = 0(5x - 63) - x(x^2 - 0) + 0(7x - 80) = 0$.$-x(x^2) = 0$.Wait,let us re-evaluate the expansion along the first column correctly:$|A| = 0 \cdot \begin{vmatrix} 5 & 7 \ 9 & x \end{vmatrix} - x \cdot \begin{vmatrix} x & 16 \ 9 & x \end{vmatrix} + 0 \cdot \begin{vmatrix} x & 16 \ 5 & 7 \end{vmatrix} = 0$.$-x(x^2 - 144) = 0$.$-x(x - 12)(x + 12) = 0$.Therefore,the possible values of $x$ are $0, 12, -12$.

If $A = \begin{bmatrix} 0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x \end{bmatrix}$ is a singular matrix,then the possible values of $x$ are:

Similar Questions

Evaluate the determinant: $\left| \begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array} \right|$

The system of linear equations $x + \lambda y - z = 0, \lambda x - y - z = 0, x + y - \lambda z = 0$ has a non-trivial solution for:

If $\omega$ is the cube root of unity,then $\left| \begin{array}{ccc} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{array} \right| = $

Given the system of equations $a(x + y + z) = x$,$b(x + y + z) = y$,$c(x + y + z) = z$ where $a, b, c$ are non-zero real numbers. If the real numbers $x, y, z$ are such that $xyz \neq 0$,then $(a + b + c)$ is equal to-

If $x^4+y^4+z^4=0$ then,$\left|\begin{array}{ccc}1 & xy & yz \\ zx & 1 & xy \\ yz & zx & 1\end{array}\right|=$ . . . . . . . $(\because x, y, z \in \mathbb{R})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module