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

English

Gujarati

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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

Prove that

$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$

medium

If $a+x=b+y=c+z+1,$ where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to

hard

(JEE MAIN-2020)

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

If $a,b,c$ are distinct and rational numbers then $\left| {\begin{array}{*{20}{c}}
{\left( {{a^2} + {b^2} + {c^2}} \right)}&{ab + bc + ca}&{ab + bc + ca}\\
{ab + bc + ca}&{\left( {{a^2} + {b^2} + {c^2}} \right)}&{\left( {bc + ca + ab} \right)}\\
{ab + bc + ca}&{\left( {ab + bc + ca} \right)}&{\left( {{a^2} + {b^2} + {c^2}} \right)}
\end{array}} \right|$ is always

normal

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