Let $a-2 b+c=1$

If $f(x)=\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|,$ then

If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.

If $a, b $ and $ c$ are non zero numbers, then $\Delta = \left| {\,\begin{array}{*{20}{c}}{{b^2}{c^2}}&{bc}&{b + c}\\{{c^2}{a^2}}&{ca}&{c + a}\\{{a^2}{b^2}}&{ab}&{a + b}\end{array}\,} \right|$ is equal to

By using properties of determinants, show that:

$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3 x+k)$

If $a, b, c > 0 \, and \, x, y, z \in R$ , then the determinant $\left|{\begin{array}{*{20}{c}}{{{\left( {{a^x}\, + \,\,{a^{ – x}}} \right)}^2}}&{{{\left( {{a^x}\, – \,\,{a^{ – x}}} \right)}^2}}&1\\{{{\left( {{b^y}\, + \,\,{b^{ – y}}} \right)}^2}}&{{{\left( {{b^y}\, – \,\,{b^{ – y}}} \right)}^2}}&1\\{{{\left( {{c^z}\, + \,\,{c^{ – z}}} \right)}^2}}&{{{\left( {{c^z}\, – \,\,{c^{ – z}}} \right)}^2}}&1\end{array}} \right|$ $=$