If -9 is a root of the equation left| begin{a | vedclass

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

Question

(A) Given the determinant equation: $\left| \begin{array}{ccc} x & 3 & 7 \ 2 & x & 2 \ 7 & 6 & x \end{array} ight| = 0$. Applying the row operatio

Vedclass · Accepted Answer

$2, 7$

Vedclass · Answer

(A) Given the determinant equation: $\left| \begin{array}{ccc} x & 3 & 7 \ 2 & x & 2 \ 7 & 6 & x \end{array} ight| = 0$. Applying the row operation $R_1 	o R_1 + R_2 + R_3$,we get: $\left| \begin{array}{ccc} x+9 & x+9 & x+9 \ 2 & x & 2 \ 7 & 6 & x \end{array} ight| = 0$. Taking $(x+9)$ as a common factor from $R_1$: $(x+9) \left| \begin{array}{ccc} 1 & 1 & 1 \ 2 & x & 2 \ 7 & 6 & x \end{array} ight| = 0$. Expanding the determinant: $(x+9) [1(x^2 - 12) - 1(2x - 14) + 1(12 - 7x)] = 0$. $(x+9) [x^2 - 12 - 2x + 14 + 12 - 7x] = 0$. $(x+9) (x^2 - 9x + 14) = 0$. Factoring the quadratic expression: $(x+9) (x-2) (x-7) = 0$. Thus,the roots are $x = -9, 2, 7$. The other two roots are $2$ and $7$.

If $-9$ is a root of the equation $\left| \begin{array}{ccc} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array} \right| = 0$,then the other two roots are:

Similar Questions

The area of the triangle formed by the points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$ is:

The solutions of $\operatorname{det}(A-\lambda I_2)=0$ are $4$ and $8$,where $A=\begin{bmatrix} 2 & 3 \\ x & y \end{bmatrix}$. Then:

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$ and $|2A|^3 = 2^{21}$,where $\alpha, \beta \in \mathbb{Z}$. Then a value of $\alpha$ is:

Evaluate $\Delta = \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$

Find the area of the triangle with vertices $(a \cos \theta, b \sin \theta)$,$(-a \sin \theta, b \cos \theta)$,and $(-a \cos \theta, -b \sin \theta)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module