The value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{b + c}&{c + a}&{a + b}\\{b + c - a}&{c + a - b}&{a + b - c}\end{array}\,} \right|$ is

  • A

    $abc$

  • B

    $a + b + c$

  • C

    $ab + bc + ca$

  • D

    $0$

Similar Questions

The determinant $\left| {\begin{array}{*{20}{c}}{^x{C_1}}&{^x{C_2}}&{^x{C_3}}\\ {^y{C_1}}&{^y{C_2}}&{^y{C_3}}\\{^z{C_1}}&{^z{C_2}}&{^z{C_3}}\end{array}} \right|$ $=$

If $\omega $ is the cube root of unity, then $\left| {\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\\omega &{{\omega ^2}}&1\\{{\omega ^2}}&1&\omega \end{array}} \right|$=

Let $M$ be a $3 \times 3$ invertible matrix with real entries and let $I$ denote the $3 \times 3$ identity matrix. If $M ^{-1}=\operatorname{adj}(\operatorname{adj} M )$, then which of the following statement is/are $ALWAYS TRUE$ ?

$(A)$ $M=I$   $(B)$ $\operatorname{det} M =1$   $(C)$ $M ^2= I$  $(D)$ $(\operatorname{adj} M)^2=I$

  • [IIT 2020]

By using properties of determinants, show that:

$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

Show that $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$