यदि A =left[begin{array}{rr}8 & 0 4 & -2 3 & 6end{array}right], B =left[begin{arra

Hindi

3 and 4 .Determinants and Matrices

medium

यदि $A =\left[\begin{array}{rr}8 & 0 \\ 4 & -2 \\ 3 & 6\end{array}\right], B =\left[\begin{array}{cc}2 & -2 \\ 4 & 2 \\ -5 & 1\end{array}\right]$ तथा $2 A +3 X =5 B$ दिया हो तो आव्यूह $X$ ज्ञात कीजिए

$\left[ {\begin{array}{*{20}{c}}
{ 2}&{\frac{{ - 10}}{3}} \\
4&{\frac{{14}}{3}} \\
{\frac{{ - 31}}{3}}&{\frac{{ 7}}{3}}
\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}
{ 2}&{\frac{{ 10}}{3}} \\
4&{\frac{{14}}{3}} \\
{\frac{{ - 31}}{3}}&{\frac{{ 7}}{3}}
\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}
{ - 2}&{\frac{{ 10}}{3}} \\
4&{\frac{{-14}}{3}} \\
{\frac{{ - 31}}{3}}&{\frac{{ 7}}{3}}
\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}
{ - 2}&{\frac{{ - 10}}{3}} \\
4&{\frac{{14}}{3}} \\
{\frac{{ - 31}}{3}}&{\frac{{ - 7}}{3}}
\end{array}} \right]$

Solution

We have $2A+3X=5 B$

or $2A+3X-2A=5B-2A$

or $2 A-2A+3X=5B-2A$ $($ Matrix addition is commutative $)$

or $O+3 X=5B-2 A$ $(-2A$ is the additive inverse of $2A)$

or $3 \mathrm{X}=5 \mathrm{B}-2 \mathrm{A}$ ( $O$ is the additive identity)

or $X=\frac{1}{3}(5 B-2 A)$

or ${\text{X}} = $ $\frac{1}{3}\left( {5\left[ {\begin{array}{*{20}{c}}
2&{ – 2} \\
4&2 \\
{ – 5}&1
\end{array}} \right] – 2\left[ {\begin{array}{*{20}{l}}
8&0 \\
4&{ – 2} \\
3&6
\end{array}} \right]} \right)$ $ = \frac{1}{3}\left( {\left[ {\begin{array}{*{20}{c}}
{10}&{ – 10} \\
{20}&{10} \\
{ – 25}&5
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{ – 16}&0 \\
{ – 8}&4 \\
{ – 6}&{ – 12}
\end{array}} \right]} \right)$

$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
{10 – 16}&{ – 10 + 0} \\
{20 – 8}&{10 + 4} \\
{ – 25 – 6}&{5 – 12}
\end{array}} \right]$

$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
{ – 6}&{ – 10} \\
{12}&{14} \\
{ – 31}&{ – 7}
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
{ – 2}&{\frac{{ – 10}}{3}} \\
4&{\frac{{14}}{3}} \\
{\frac{{ – 31}}{3}}&{\frac{{ – 7}}{3}}
\end{array}} \right]$

Standard 12

Mathematics

यदि आव्यूह $\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\\3&\lambda &5\end{array}} \right]$ एक व्युत्क्रमणीय आव्यूह हो, तो $\lambda $ का मान नहीं हो सकता है

easy

यदि $A =\left[\begin{array}{rrr}1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1\end{array}\right]$ है तो दर्शाइए कि $A ^{3}-23 A -40 I = O$

medium

यदि $A =\left[\begin{array}{rrr}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right], B =\left[\begin{array}{rrr}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right]$ तथा $C =\left[\begin{array}{rrr}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right],$ तो $( A + B )$ तथा $( B – C )$ परिकलित कीजिए। साथ ही सत्यापित कीजिए कि $A +( B – C )=( A + B )- C$

medium

माना $A =\left[ a _{ ij }\right]$, कोटी 3 का वर्ग आव्यूह इस प्रकार है कि सभी $i , j =1,2,3$ के लिये $a _{ ij }=2^{ j }$ ${ }^{-i}$ है। तब आव्यूह $A ^2+ A ^3+\ldots+ A ^{10}$ है :

hard

(JEE MAIN-2022)

$\cos \theta \left[ {\begin{array}{{20}{c}}{\cos \theta }&{\sin \theta }\\{ – \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{{20}{c}}{\sin \theta }&{ – \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right] = $

easy