Using property of determinants and without expanding, Prove that left|begin{array}{lll}x &

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3 and 4 .Determinants and Matrices

easy

Using the property of determinants and without expanding, Prove that

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

Option A

Option B

Option C

Option D

Solution

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

Clearly, the two determinants have two identical columns. Thus,

$=0+0=0$

Standard 12

Mathematics

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

hard

Solve the equations $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0, a \neq 0$

medium

$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$ is divisor of

hard

$\left| {\,\begin{array}{{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ then $K = $

medium

The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right|$ $= 11$ is

normal

Using the property of determinants and without expanding, Prove that $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$