Value of left| {,begin{array}{*{20}{c}}{265}&{240}&{219}{240}&{225}&{198}{219}&{

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3 and 4 .Determinants and Matrices

easy

The value of $\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right|$ is equal to

$0$

$679$

$779$

$1000$

Solution

(a) $\left| {{\mkern 1mu} \begin{array}{*{20}{c}}
{265}&{240}&{219}\\
{240}&{225}&{198}\\
{219}&{198}&{181}
\end{array}{\mkern 1mu} } \right|$ $ = \left| {{\mkern 1mu} \begin{array}{*{20}{c}}
{25}&{21}&{219}\\
{15}&{27}&{198}\\
{21}&{17}&{181}
\end{array}{\mkern 1mu} } \right|$

$=\left| {\,\begin{array}{*{20}{c}}4&{21}&9\\{ – 12}&{27}&{ – 72}\\4&{17}&{11}\end{array}\,} \right|$

{Applying ${C_1} \to {C_1} – {C_2};\,{C_3} \to {C_3} – 10{C_2} $}

$= \,\left| {\,\begin{array}{*{20}{c}}4&{21}&9\\{ – 12}&{27}&{ – 72}\\0&{ – 4}&2\end{array}\,} \right|$ {Applying ${R_3} \to {R_3} – {R_1}$}

$=4\,\left| {\,\begin{array}{*{20}{c}}1&{21}&9\\{ – 3}&{27}&{ – 72}\\0&{ – 4}&2\end{array}\,} \right| = 4\,\left| {\,\begin{array}{*{20}{c}}1&{21}&9\\0&{90}&{ – 45}\\0&{ – 4}&2\end{array}\,} \right|$

by ${R_2} \to 3{R_1} + {R_2}$ $=4(90 × 2 – 45 × 4)=0.$

Standard 12

Mathematics

Suppose $a_1, a_2, …….$ real numbers, with $a_1 \ne 0$. If $a_1, a_2, a_3, ……….$ are in $A.P$. then

normal

Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

hard

If ${a_1},{a_2},{a_3},……..,{a_n},……$ are in G.P. and ${a_i} > 0$ for each $i$, then the value of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 2}}}&{\log {a_{n + 4}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 8}}}&{\log {a_{n + 10}}}\\{\log {a_{n + 12}}}&{\log {a_{n + 14}}}&{\log {a_{n + 16}}}\end{array}} \right|$ is equal to

medium

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{31}&{37}&{92}\\{31}&{58}&{71}\\{31}&{105}&{24}\end{array}\,} \right|$ is

easy

If $\left| {\begin{array}{*{20}{c}} {a – b}&{b – c}&{c – a} \\ {b – c}&{c – a}&{a – b} \\ {c – a + 1}&{a – b}&{b – c} \end{array}} \right| = 0$ ,$\left( {a,b,c \in R – \left\{ 0 \right\}} \right),$ then