Maths Matrices

$\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]a_1,b_1,c_{1}t\left\{-1,0,1\right\}$

$\underset{9times}{\underset{?}{a_1+a_2+a_3+....}}=5$

Total = 414

$A=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$

$A^2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill -4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -4\hfill \end{array}\right]=-4\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=-4l$

$N^2=\frac{{\left(2^{20}-1\right)}^2}{25}l$

$M{N}^2=\frac{4}{125}{\left(2^{20}-1\right)}^{3}l$

$=\left[\begin{array}{ccc}\hfill \lambda \hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \lambda \hfill \end{array}\right],\lambda \ne k$

$A\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)$

$A\left(\begin{array}{l}1\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}-1\\ 0\\ 1\end{array}\right)$

$A\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 2\end{array}\right)$

$\Rightarrow A\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

$AB=C\Rightarrow A=C{B}^{-1}$

$\left|A-2l\right|=\left(-4\right)\left(-1\right)+3\left(-1\right)+1\left(-1\right)$

4 – 3 – 1 = 0

$\mathcal{l}\left(-1,0,1\right)+m\left(-1,1,0\right)=\left(-4,3,1\right)\Rightarrow -\mathcal{l}-m=-4$

$m=3,\mathcal{l}=1$

x + y + az = 2………… (i)

3x + y + z = 4 ………… (ii)

x + 2z = 1 ……………. (iii)

x = 1, y = 1, z = 0

(for unique solution $a\ne -3$  

$\left(\alpha ,-1\right),\left(1,\alpha \right)and\left(1,-1\right)$ are collinear.

$\frac{\alpha -1}{1-\alpha }=\frac{-1-\alpha }{1-1}\Rightarrow 1-{\alpha }^2=0\Rightarrow \alpha =\pm 1$                

Sum of absolute value = 2

Apart from upper and lower triangular matrices, we have the unit triangular matrix, strictly-triangular matrix and an atomic-triangular matrix.

A few types of matrices are row and column matrices, singleton, null matrix, square, diagonal, scalar, identity, equal, triangular (both upper and lower), singular and non-singular matrices, symmetric, skew-symmetric, hermitian, and orthogonal matrices. NCERT excercise on matrices chapter covers que

An upper triangular matrix is a square matrix where all the elements above the diagonal are non-zero, and below it is zero. A lower triangular matrix is a square matrix where all the elements above the diagonal are zero.

The elements of a matrix may be real or complex numbers. If all the elements of a matrix are real, then the matrix is called a real matrix.

R₁ =  $\left\{\left(a,1\right)\in N\times N:\left|a-b\right|\le 13\right\}$

 $\left(a,a\right)\in R_{1}as\left|a-a\right|\le 13\left(Reflexive\right)$

$\left(a,b\right)\&\left(b,a\right)\in R_{1}as\left|a-b\right|=\left|b-a\right|\to \left(symmetric\right).$

But it is not necessary that if (a,b) & (b, c)  $\in R,then\left(a,c\right)\in R$

Eg   $-\left(21,10\right)\in R\&\left(10,1\right)\in Rbut\left(21,1\right)\notin R_1$

R₂ =   $\left\{\left(a,b\right)\in N\times N:\left|a-b\right|\ne 13\therefore R_2\to Notequivalencesolutoin.\right\}$

$\left(a,b\right)\&\left(b,a\right)\in R_{2}as\left|a-b\right|=\left|b-a\right|$

But it is not necessary that if (a, b) & (b, c)  $\in R_2$  then (a, c) also $\in R_2$ .

Eg – (21, 1)   $\in R_2\&\left(1,8\right)\in R_{2}but\left(21,8\right)\notin R_2$

$N=M^2+M^4+.....+M^{98}$

$=\left(-{\alpha }^{2}I\right)+{\left(-{\alpha }^{2}I\right)}^2+....+{\left(-{\alpha }^{2}I\right)}^{49}$

$=I\left(-{\alpha }^2+{\alpha }^4-{\alpha }^6+....-{\alpha }^{98}\right)$

N = $-I\left({\alpha }^2-{\alpha }^4+{\alpha }^6.......+{\alpha }^{98}\right)$

$=-I\frac{{\alpha }^2\left(1-{\left(-{\alpha }^2\right)}^{49}\right)}{1-\left(-{\alpha }^2\right)}$

N = $\frac{-I{\alpha }^2\left(1+{\alpha }^{98}\right)}{1+{\alpha }^2}$

Now $\left(I-m^2\right)N=-2I$

$\left(I+{\alpha }^{2}I\right)\frac{\left(-I{\alpha }^2\cdot (1+{\alpha }^{98}\right)}{1+{\alpha }^2}=-2I$

? α¹⁰⁰ + α² = 2

? α = $\pm 1$