કિમત મેળવો : left[begin{array}{l}1 2 3end{array}right]left[begin{array

Hindi

3 and 4 .Determinants and Matrices

easy

કિમત મેળવો : $\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{lll}2 & 3 & 4\end{array}\right]$

$\left[ {\begin{array}{*{20}{l}} 2&3&4 \\ 4&6&8 \\ 6&9&{12} \end{array}} \right]$

Solution

$\left[ {\begin{array}{*{20}{l}}
1 \\
2 \\
3
\end{array}} \right]\left[ {\begin{array}{*{20}{l}}
2&3&4
\end{array}} \right] = $ $\left[ {\begin{array}{*{20}{l}}
{1(2)}&{1(3)}&{1(4)} \\
{2(2)}&{2(3)}&{2(4)} \\
{3(2)}&{3(3)}&{3(4)}
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{l}}
2&3&4 \\
4&6&8 \\
6&9&{12}
\end{array}} \right]$

Standard 12

Mathematics

જો $P$ એ એવો $3 \times 3$ વાસ્તવિક શ્રેણિક હોય કે જેથી $P^T=a P+(a-1) I$, જ્યાં $a >1$,તો

hard

(JEE MAIN-2023)

જો $A\, = \,\left[ {\begin{array}{{20}{c}}
1&2&x\\
3&{ – 1}&2
\end{array}} \right]$ અને $B\, = \,\left[ {\begin{array}{{20}{c}}
y\\
x\\
1
\end{array}} \right]$ છે કે જેથી $AB\, = \,\left[ {\begin{array}{*{20}{c}}
6\\
8
\end{array}} \right],$ તો

hard

(JEE MAIN-2014)

જો $A=$ $\left[ {\begin{array}{{20}{l}}
{\frac{2}{3}}&1&{\frac{5}{3}} \\
{\frac{1}{3}}&{\frac{2}{3}}&{\frac{4}{3}} \\
{\frac{7}{3}}&2&{\frac{2}{3}}
\end{array}} \right]$ અને $B=$ $\left[ {\begin{array}{{20}{l}}
{\frac{2}{5}}&{\frac{3}{5}}&1 \\
{\frac{1}{5}}&{\frac{2}{5}}&{\frac{4}{5}} \\
{\frac{7}{5}}&{\frac{6}{5}}&{\frac{2}{5}}
\end{array}} \right],$ હોય, તો $3A-5B$ ની ગણતરી કરો.

medium

શ્રેણિક $\left[ {\begin{array}{*{20}{c}}2&5&{ – 7}\\0&3&{11}\\0&0&9\end{array}} \right]$ એ. . . .. થાય.

easy

જો $A=\left[\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2\end{array}\right]$, $B=\left[\begin{array}{rr}1 & 3 \\ 0 & 2 \\ -1 & 4\end{array}\right]$ અને $C=\left[\begin{array}{rrrr}1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1\end{array}\right],$ અને $A(B C),(A B) C$ શોધો અને દર્શાવો કે $(A B) C=A(B C)$.

medium