समीकरण $\left[ {\begin{array}{*{20}{c}}1&0&1\\{ – 1}&1&0\\0&{ – 1}&1\end{array}} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}1\\1\\2\end{array} \right]$  का हल है, $(x,y,z)$=

यदि $\left[ {\begin{array}{*{20}{c}}3&1\\4&1\end{array}} \right]X = \left[ {\begin{array}{*{20}{c}}5&{ – 1}\\2&3\end{array}} \right]$ है, तो $X =$

मान लीजिए कि $A =\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B =\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C =\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right],$ तो निम्नलिखित ज्ञात कीजिए:

$3 A – C$

यदि $A = \left[ {\begin{array}{*{20}{c}}{\,\,\,\cos \alpha }&{\sin \alpha }\\{ – \sin \alpha }&{\cos \alpha }\end{array}} \right]$, तो ${A^2} = $

यदि $A =\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ इस प्रकार है कि $A ^{2}= I ,$ तो