Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$
Solution Applying operations $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{1}$ to the given determinant $\Delta$, we have
$\Delta = \left| {\begin{array}{*{20}{c}}
a&{a + b}&{a + b + c} \\
0&a&{2a + b} \\
0&{3a}&{7a + 3b}
\end{array}} \right|$
Now applying $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{2},$ we get
$\Delta=\left|\begin{array}{ccc}
a & a+b & a+b+c \\
0 & a & 2 a+b \\
0 & 0 & a
\end{array}\right|$
Expanding along $C_{1},$ we obtain
$\Delta = a\left| {\begin{array}{*{20}{c}}
a&{2a + b} \\
0&a
\end{array}} \right| + 0 + 0$
$ = a\left( {{a^2} - 0} \right) = a\left( {{a^2}} \right) = {a^3}$
Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}( A )=4$. Let $R _{ i }$ denote the $i ^{\text {th }}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R _{2} \rightarrow 2 R _{2}+5 R _{3}$ on $2 A ,$ then $\operatorname{det}( B )$ is equal to ...... .
Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If
$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$
then value of $\lambda^{2}$ is equal to $.....$
$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = $
$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ - 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$
Then the maximum value of $f(x)$ is equal to $.....$
If $a,b,c$ are positive integers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + x}&{ab}&{ac}\\{ab}&{{b^2} + x}&{bc}\\{ac}&{bc}&{{c^2} + x}\end{array}\,} \right|$ is divisible by