$\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}\,} \right| = $

If $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x – z}&{x – y}\\{y – z}&{z – x}&{y – x}\\{z – y}&{z – x}&{x + y}\end{array}\,} \right| = k\,xyz$, then the value of $k $ is

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$

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 $ALWAYS TRUE$ ?

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

By using properties of determinants, show that:

$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$