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

If $A, B, C$ are the angles of a triangle and $\left| {\begin{array}{*{20}{c}}1&1&1\\{1 + \sin A}&{1 + \sin B}&{1 + \sin C}\\{\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right| = 0$,then the triangle is

The value of $\left| \begin{array}{ccc} 265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181 \end{array} \right|$ is equal to

The roots of the equation $\left| \begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5x^2 \end{array} \right| = 0$ are

If the area of a triangle is $4$ sq. units whose vertices are $(k, 0), (4, 0)$ and $(0, 2)$,then the value of $k$ is . . . . . . .

$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{matrix} \right| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module