If x is a positive integer, then Delta = left| {,begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x +

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3 and 4 .Determinants and Matrices

hard

If $x$ is a positive integer, then $\Delta = \left| {\,\begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x + 2)!}\\{(x + 1)!}&{(x + 2)!}&{(x + 3)!}\\{(x + 2)!}&{(x + 3)!}&{(x + 4)!}\end{array}\,} \right|$ is equal to

$2(x!)(x + 1)!$

$2(x!)(x + 1)!(x + 2)!$

$2(x!)(x + 3)!$

None of these

Solution

(b) $\Delta = \left| {\begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x + 2)\,!}\\{(x + 1)!}&{(x + 2)!}&{(x + 3)\,!}\\{(x + 2)!}&{(x + 3)!}&{(x + 4)\,!}\end{array}} \right|$

= $x\,!\,(x + 1)\,!\,(x + 2)\,!\,\left| {\,\begin{array}{*{20}{c}}1&{(x + 1)}&{(x + 2)\,(x + 1)}\\1&{(x + 2)}&{(x + 3)\,(x + 2)}\\1&{(x + 3)}&{(x + 4)\,(x + 3)}\end{array}\,} \right|$

Applying ${R_1} \to {R_2} – {R_1},{R_2} \to ({R_3} – {R_2})$ we get

$ = x\,!(x + 1)\,!(x + 2)\,!$ $\,\left| {\,\begin{array}{*{20}{c}}0&1&{2(x + 2)}\\0&1&{2(x + 3)}\\1&{(x + 3)}&{(x + 4)\,(x + 3)}\end{array}\,} \right|$

$ = 2\,x!(x + 1)!(x + 2)!$,(on simplification).

Trick: Put $x = 1$ and then match the alternate.

Standard 12

Mathematics

$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $

hard

(IIT-1980)

If the system of equations $ax + y + z = 0$, $x + by + z = 0$ and $x + y + cz = 0 $, where $a,b,c \ne 1,$ has a non trivial solution, then the value of $\frac{1}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}}$is

medium

The number of real values of $x$ satisfying $\left| {\,\begin{array}{*{20}{c}} x&{3x + 2}&{2x – 1}\\{2x – 1}&{4x}&{3x + 1}\\{7x – 2}&{17x + 6}&{12x – 1}\end{array}\,} \right|$ $= 0$ is

normal

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = $