The solution(s) of the equation left| begin{m | vedclass

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

Question

(A) Given the determinant $\Delta = \left| \begin{matrix} x & a & b \ a & x & a \ b & b & x \end{matrix} ight| = 0$. Applying $C_1 ightarrow C_1

Vedclass · Accepted Answer

$x = -(a + b)$

Vedclass · Answer

(A) Given the determinant $\Delta = \left| \begin{matrix} x & a & b \ a & x & a \ b & b & x \end{matrix} ight| = 0$. Applying $C_1 ightarrow C_1 + C_2 + C_3$,we get: $\left| \begin{matrix} x+a+b & a & b \ x+a+b & x & a \ x+a+b & b & x \end{matrix} ight| = 0$. Taking $(x+a+b)$ common from $C_1$: $(x+a+b) \left| \begin{matrix} 1 & a & b \ 1 & x & a \ 1 & b & x \end{matrix} ight| = 0$. Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$: $(x+a+b) \left| \begin{matrix} 1 & a & b \ 0 & x-a & a-b \ 0 & b-a & x-b \end{matrix} ight| = 0$. Expanding along $C_1$: $(x+a+b) [(x-a)(x-b) - (a-b)(b-a)] = 0$. $(x+a+b) [x^2 - bx - ax + ab + (a-b)^2] = 0$. $(x+a+b) [x^2 - (a+b)x + ab + a^2 - 2ab + b^2] = 0$. $(x+a+b) [x^2 - (a+b)x + a^2 - ab + b^2] = 0$. Thus,$x = -(a+b)$ is a solution. The quadratic part $x^2 - (a+b)x + a^2 - ab + b^2 = 0$ yields complex roots unless $a=b=x$. If $a=b$,the determinant becomes $\left| \begin{matrix} x & a & a \ a & x & a \ a & a & x \end{matrix} ight| = (x-a)^2(x+2a) = 0$,giving $x=a$ or $x=-2a$. Since the question implies general solutions,$x=-(a+b)$ is the primary root.

The solution$(s)$ of the equation $\left| \begin{matrix} x & a & b \\ a & x & a \\ b & b & x \end{matrix} \right| = 0$ is/are:

Similar Questions

Find the values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$.

The roots of the equation $\left| \begin{matrix} 1+x & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+x \end{matrix} \right| = 0$ are

If $a x^{4}+b x^{3}+c x^{2}+d x+e = \left|\begin{array}{ccc}x^{3}+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x\end{array}\right|$,then $e$ is equal to

If $a, b, c$ are real,then the value of the determinant $\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac}\\{ab}&{{b^2} + 1}&{bc}\\{ac}&{bc}&{{c^2} + 1}\end{array}}\right| = 1$ if

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det A$ =

Vedclass Test Series

Exam Paper Generator

Online Exam Module