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The determinant  $\begin{vmatrix} b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y \end{vmatrix}$  equals

  1. $\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}$
  2. $2\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}$
  3. $3\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}$
  4. None of these
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1 Answer

1 vote

Answer: $\mathbf B$

Explanation:

Let $\mathrm {M} = \begin {bmatrix} \mathrm {b+c} & \mathrm {c+ a} & \mathrm {a + b} \\ \mathrm {q+r} & \mathrm {r+ p} & \mathrm {p+q} \\ \mathrm {y + z} & \mathrm {z+x} & \mathrm {x+y}  \end {bmatrix}$

Apply  $\mathrm {C_1 \rightarrow C_1 + C_2 + C_3}$

$=\begin{bmatrix} \mathrm {2(a+b+c)} & \mathrm{c+a} & \mathrm {a + b} \\ \mathrm{2(p+q+r)} & \mathrm{r+p} & \mathrm {p+q} \\ \mathrm{2(x+y+z)} & \mathrm{z+x} & \mathrm{x+y} \end{bmatrix}$

$=2\begin{bmatrix} \mathrm {(a+b+c)} & \mathrm{c+a} & \mathrm {a + b} \\ \mathrm{(p+q+r)} & \mathrm{r+p} & \mathrm {p+q} \\ \mathrm{(x+y+z)} & \mathrm{z+x} & \mathrm{x+y} \end{bmatrix}$

Apply $\mathrm {C_2} \rightarrow \mathrm{C_2-C_1}$ & $\mathrm {C_2} \rightarrow \mathrm{C_3-C_1}$

$=2\begin{bmatrix} \mathrm {(a+b+c)} & \mathrm{-b} & \mathrm {-c} \\ \mathrm{(p+q+r)} & \mathrm{-q} & \mathrm {-r} \\ \mathrm{(x+y+z)} & \mathrm{-y} & \mathrm{-z} \end{bmatrix}$

Apply $\mathrm {C_1 \rightarrow C_1+C_2+C_3}$, we get:

$=2\begin{bmatrix} \mathrm {a} & \mathrm{-b} & \mathrm {-c} \\ \mathrm{p} & \mathrm{-q} & \mathrm {-r} \\ \mathrm{x} & \mathrm{-y} & \mathrm{-z} \end{bmatrix}$

Taking out $-1$ common from $\mathrm {C_2}$ and $\mathrm {C_3}$, we get:

$=2(-1)(-1)\begin{bmatrix} \mathrm {(a+b+c)} & \mathrm{b} & \mathrm {c} \\ \mathrm{(p+q+r)} & \mathrm{q} & \mathrm {r} \\ \mathrm{(x+y+z)} & \mathrm{y} & \mathrm{z} \end{bmatrix}$

$=2\begin{bmatrix} \mathrm {(a+b+c)} & \mathrm{b} & \mathrm {c} \\ \mathrm{(p+q+r)} & \mathrm{q} & \mathrm {r} \\ \mathrm{(x+y+z)} & \mathrm{y} & \mathrm{z} \end{bmatrix}$

$\therefore$ is the correct option.


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