Left| {,begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}{ab}&{{b^2} + {x^2}}&{bc}{ca}

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3 and 4 .Determinants and Matrices

hard

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

${a^2}$

${b^2}$

${c^2}$

${x^2}$

Solution

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

Multiply ${C_1},\,{C_2},{C_3}$ by $a,\,\,b,\,c$ respectively and hence divide by $abc$

$\Delta = \frac{1}{{abc}}\,\left| {\,\begin{array}{*{20}{c}}{a({a^2} + {x^2})}&{a{b^2}}&{{c^2}a}\\{{a^2}b}&{b({b^2} + {x^2})}&{b{c^2}}\\{c{a^2}}&{{b^2}c}&{c({c^2} + {x^2})}\end{array}\,} \right|$

Now take out $a, b $ and $c $ common from ${R_1},\,{R_2}$ and ${R_3}$,

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

Now applying ${C_1} \to {C_1} + {C_2} + {C_3}$

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

==> $\Delta = {x^4}({a^2} + {b^2} + {c^2} + {x^2})$

Hence, it is divisible by ${x^2}$.

Standard 12

Mathematics

Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$

medium

The value of $\theta$ lying between $\theta = 0$ and $\theta = \pi /2$ and satisfying the equation : $\left| {\,\begin{array}{*{20}{c}} {1\,\, + \,\,{{\sin }^2}\,\theta }&{{{\cos }^2}\,\theta }&{4\,\sin \,4\,\theta }\\ {{{\sin }^2}\,\theta }&{1\,\, + \,\,{{\cos }^2}\,\theta }&{4\,\sin \,4\,\theta }\\ {{{\sin }^2}\,\theta }&{{{\cos }^2}\,\theta }&{1\,\, + \,\,4\,\sin \,4\,\theta } \end{array}\,} \right|$ $= 0$ are :

normal

By using properties of determinants, show that:

$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$

hard

If $\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&x&{{x^2}}\\{{b^2}}&{ab}&{{a^2}} \end{array}} \right|$ $= 0$ , then :

normal

The determinant $\left| {\begin{array}{*{20}{c}}{{b_1}\, + \,\,{c_1}}&{{c_1}\, + \,\,{a_1}}&{{a_1}\, + \,\,{b_1}}\\{{b_2}\, + \,\,{c_2}}&{{c_2}\, + \,\,{a_2}}&{{a_2}\, + \,\,{b_2}}\\{{b_3}\, + \,\,{c_3}}&{{c_3}\, + \,\,{a_3}}&{{a_3}\, + \,\,{b_3}} \end{array}} \right|$ $=$