Given 3left[begin{array}{ll}x & y z & wend{array}right]=left[begin{array}{cc}x & 6 -1 &a

Hindi

3 and 4 .Determinants and Matrices

medium

Given $3\left[\begin{array}{ll}x & y \\ z & w\end{array}\right]=\left[\begin{array}{cc}x & 6 \\ -1 & 2 w\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right],$ find the values of $x, \,y, \,z$ and $w$.

$x=2$, $y=4$, $z=1,$ $w=3$

Solution

$3\left[\begin{array}{ll}x & y \\ z & w\end{array}\right]=\left[\begin{array}{cc}x & 6 \\ -1 & 2 w\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}3 x & 3 y \\ 3 z & 3 w\end{array}\right]=\left[\begin{array}{cc}x+4 & 6+x+y \\ -1+z+w & 2 w+3\end{array}\right]$

Comparing the corresponding elements of these two matrices, we get :

$3 x=x+4$

$\Rightarrow 2 x=4$

$\Rightarrow x=2$

$3 x=6+x+y$

$\Rightarrow 2 y=6+x=6+2=8$

$\Rightarrow y=4$

$3 w=2 w+3$

$\Rightarrow w=3$

$3 z=-1+z+w$

$\Rightarrow 2 z=-1+w=-1+3=2$

$\Rightarrow z=1$

$\therefore $ $x=2$, $y=4$, $z=1,$ and $w=3$

Standard 12

Mathematics

If $A = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\2&3&0&0\\4&5&6&0\\7&8&9&{10}\end{array}} \right]$, then $A$ is

normal

Let $S=\left\{n \in N \mid\left(\begin{array}{ll}0 & i \\ 1 & 0\end{array}\right)^{n}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) \forall a, b, c, d \in R\right\}$, where $i=\sqrt{-1} .$ Then the number of $2 -$ digit numbers in the set $\mathrm{S}$ is $……$

medium

(JEE MAIN-2021)

Let $A$ and $B$ be $3 \times 3$ real matrices such that $A$ is symmetric matrix and $B$ is skew-symmetric matrix. Then the system of linear equations $\left( A ^{2} B ^{2}- B ^{2} A ^{2}\right) X = O ,$ where $X$ is a $3 \times 1$ column matrix of unknown variables and $O$ is a $3 \times 1$ null matrix, has ……. .

hard

(JEE MAIN-2021)

Let $A = \left( {\begin{array}{{20}{c}}
{\cos \,\alpha }&{ – \sin \,\alpha }\\
{\sin \,\alpha }&{\cos \,\alpha }
\end{array}} \right)$, $\left( {\alpha \in R} \right)$ such that ${A^{32}} = \left( {\begin{array}{{20}{c}}
0&{ – 1}\\
1&0
\end{array}} \right)$. Then a value of $\alpha $ is

hard

(JEE MAIN-2019)

Construct a $3 \times 4$ matrix, whose elements are given by $a_{i j}=\frac{1}{2}|-3 i+j|$.

easy

Given $3\left[\begin{array}{ll}x & y \\ z & w\end{array}\right]=\left[\begin{array}{cc}x & 6 \\ -1 & 2 w\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right],$ find the values of $x, \,y, \,z$ and $w$.

Solution

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Let $A = \left( {\begin{array}{{20}{c}}
{\cos \,\alpha }&{ – \sin \,\alpha }\\
{\sin \,\alpha }&{\cos \,\alpha }
\end{array}} \right)$, $\left( {\alpha \in R} \right)$ such that ${A^{32}} = \left( {\begin{array}{{20}{c}}
0&{ – 1}\\
1&0
\end{array}} \right)$. Then a value of $\alpha $ is