If a neq p, b neq q, c neq r and left|begin{a | vedclass

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

Question

(C) Given the determinant equation: $\left|\begin{array}{ccc}p & b & c \ p+a & q+b & 2c \ a & b & r\end{array}ight|=0$. Applying the property of d

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given the determinant equation: $\left|\begin{array}{ccc}p & b & c \ p+a & q+b & 2c \ a & b & r\end{array}ight|=0$. Applying the property of determinants to split the second row: $\left|\begin{array}{ccc}p & b & c \ p & q & c \ a & b & r\end{array}ight| + \left|\begin{array}{ccc}p & b & c \ a & b & c \ a & b & r\end{array}ight| = 0$. Expanding the first determinant: $p(qr - bc) - b(ar - ac) + c(ab - aq) = pqr - pbc - abr + abc + abc - acq = pqr - pbc - abr - acq + 2abc = 0$. Thus,$pqr - pbc - abr - acq = -2abc$. Now,consider the expression $E = \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c}$. This can be rewritten as $E = (1 + \frac{a}{p-a}) + (1 + \frac{b}{q-b}) + (1 + \frac{c}{r-c}) = 3 + \frac{a}{p-a} + \frac{b}{q-b} + \frac{c}{r-c}$. Alternatively,dividing the equation $pqr - pbc - abr - acq = -2abc$ by $(p-a)(q-b)(r-c)$ leads to the result $2$.

If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2c \\ a & b & r\end{array}\right|=0$,then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to :

Similar Questions

$\left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|=$ . . . . . . .

The value of the determinant $\left| {\begin{array}{*{20}{c}}1&1&1\\{\cos (nx)}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin (nx)}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}} \right|$ is independent of:

$\left| \begin{array}{ccc} a - b & b - c & c - a \\ x - y & y - z & z - x \\ p - q & q - r & r - p \end{array} \right| = $

Evaluate the determinant: $\left| {\begin{array}{*{20}{c}}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}} \right|$

Let $P = [a_{ij}]$ be a $3 \times 3$ matrix and let $Q = [b_{ij}]$,where $b_{ij} = 2^{i+j} a_{ij}$ for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$,then the determinant of the matrix $Q$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module