Maths Matrices

$A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=l+B$

$B^2=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]*\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill ab\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

B³ = 0

$\therefore A^n={\left(1+B\right)}^n{=}^n{C}_{0}l{+}^n{C}_{1}B{+}^n{C}_2{B}^2{+}^n{C}_3{B}^3+....$

On comparing we get na = 48, nb = 96 and na + $\frac{n\left(n-1\right)}{2}ab=2160$

$\Rightarrow a=4,n=12andb=8$

$\Rightarrow n+a+b=24$

Here,

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

we get A² = A and similarly, for

$B=A-I=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill -1\hfill \end{array}\right]$            

We get B² = -B $\Rightarrow$  B³ = B

$\therefore A^n+{\left(\omega B\right)}^n=A+{\left(\omega B\right)}^{n}forn\in N$

For ${\omega }^n$  to be unity n shall be multiple of 3 and for Bⁿ to be B. n shell be 3, 5, 7, ….9

$\therefore n=\left\{3,9,15,.....99\right\}$     

Number of elements = 17.

$A+B=\left[\begin{array}{cc}\hfill \beta +1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill \alpha \hfill \end{array}\right]$

${\left(A+B\right)}^2=\left[\begin{array}{cc}\hfill {\left(\beta +1\right)}^2\hfill & \hfill 0\hfill \\ \hfill 3\left(\beta +1\right)+3\alpha \hfill & \hfill {\alpha }^2\hfill \end{array}\right]$

$\therefore \left[\begin{array}{cc}\hfill 1\hfill & \hfill -\alpha +1\hfill \\ \hfill 2\alpha +4\hfill & \hfill {\alpha }^2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill {\left(\beta +1\right)}^2\hfill & \hfill 0\hfill \\ \hfill 3\left(\alpha +\beta +1\right)\hfill & \hfill {\alpha }^2\hfill \end{array}\right]$

= 1 = 1

$B^2=\left[\begin{array}{cc}\hfill \beta \hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill \beta \hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

$\beta =0,\alpha =-1={\alpha }_2$

$\left|{\alpha }_1-{\alpha }_2\right|=\left|1-\left(-1\right)\right|=2$

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

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

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 questions related to this topic

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.