Determinant left| {begin{array}{*{20}{c}}{^x{C₁}}&{^x{C₂}}&{^x{C₃}} {^y{C₁}}&{^y

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3 and 4 .Determinants and Matrices

normal

The determinant $\left| {\begin{array}{*{20}{c}}{^x{C_1}}&{^x{C_2}}&{^x{C_3}}\\ {^y{C_1}}&{^y{C_2}}&{^y{C_3}}\\{^z{C_1}}&{^z{C_2}}&{^z{C_3}}\end{array}} \right|$ $=$

$\frac{1}{3} \,xyz (x + y) (y + z) (z + x)$

$\frac{1}{4} \,xyz (x + y - z) (y + z - x)$

$\frac{1}{12} \, xyz (x - y) (y - z) (z - x)$

none

Solution

$\left| {\,\begin{array}{*{20}{c}}x&{\frac{{x(x – 1)}}{2}}&{\frac{{x(x – 1)(x – 2)}}{6}}\\y&{\frac{{y(y – 1)}}{2}}&{\frac{{y(y – 1)(y – 2)}}{6}}\\z&{\frac{{z(z – 1)}}{2}}&{\frac{{z(z – 1)(z – 2)}}{6}}\end{array}\,} \right|$ $=$ $\frac{{xyz}}{{12}}\;\;\left| {\,\begin{array}{*{20}{c}}1&x&{{x^2}}\\1&y&{{y^2}}\\1&z&{{z^2}}\end{array}\,} \right|$

$R_1 \rightarrow R_1 – R_2 \, and \, R_2 \rightarrow R_2 – R_3$

Standard 12

Mathematics

For a non – zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is

normal

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.

hard

Show that $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$

easy

Using properties of determinants, prove this:

$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

medium

Let $P=\left[a_{\|}\right]$be $a \times 3$ matrix and let $Q=\left[b_1\right]$, where $b_1=2^{1+j} a_{\|}$for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$ , then the determinant of the matrix $Q$ is

hard

(IIT-2012)