Matrix Algebra and Multiplication Properties - Class XII

For A = [[1,1,1],[1,1,1],[1,1,1]], A^2 = 3A, A^3 = 9A = 3^2 * A. By induction, A^n = 3^(n-1) * A. This matches the given matrix.

If A is a column vector [1, 2, 3]^T, then A^T is the row vector [1, 2, 3]. The product AA^T is a 3x3 matrix: [[11, 12, 13], [21, 22, 23], [31, 32, 3*3]] = [[1, 2, 3], [2, 4, 6], [3, 6, 9]].

$A$ is a square matrix such that $A^2 = A$ and $|A| \neq 0$
$\Rightarrow A^{-1}A^2=A^{-1}A=I$
$\Rightarrow A=I$
$\therefore A=A'=I$
Hence, option A.

It is not necessary that, if $AB=0$ then $A=0$ or $B=0$

Singular matrix is square matrix whose determinant is equal to Zero.

By performing the matrix multiplication BAB and comparing it with the inverse of A (A^-1), one can verify the identity. Given the definitions of A and B, the statement is mathematically true.

We know that from the propeties of inverse matrices,

The inverse of a matrix A is (1/det(A)) * adj(A). For the given matrix, the determinant is 1(2-6) - 0 + 1(0-2) = -4 - 2 = -6. Calculating the adjugate matrix and dividing by -6 yields the correct inverse.

Given $A=\begin{bmatrix} i & 0  \ 0 & -1 \end{bmatrix}$ where $i^{2}=-1$

A is a rotation matrix R(theta). A^n = R(n*theta). Here, A^2017 = R(2017 * 2pi / 19). Since 2017 = 19 * 106 + 3, A^2017 = R(3 * 2pi / 19) = A^3.

$\displaystyle :\left ( A+B \right )^{2}=(A+B)(A+B)$
$= A^{2}+AB+BA+B^{2}$
If $AB=BA$ then 
$\left ( A+B \right )^{2}=A^{2}+2AB+B^{2}$

$\begin{bmatrix} \cos { \theta  }  & -\sin { \theta  }  \ \sin { \theta  }  & \cos { \theta  }  \end{bmatrix}\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}=\begin{bmatrix} a\cos { \theta  }  & -b\sin { \theta  }  \ a\sin { \theta  }  & b\cos { \theta  }  \end{bmatrix}$   ...(1)

It is well known fact that, Multiplication of matrices is distributive with respect to the matrix addition. i.e $ A(B+C)=AB+AC$

To check if the given matrix is the inverse, multiply it with the original matrix. The product of a matrix and its true inverse must yield the identity matrix. Since the resulting product is not the identity matrix, the statement is false.

Let $\quad A = \begin{bmatrix}1 & -1 & 1 \ -3 & 2 & -1

\ -2 & 1 & 0\end{bmatrix}$ and $B = \begin{bmatrix}1&2

& 3\2&4&6 \ 1&2 &3\end{bmatrix}$

$\therefore\quad

AB = \begin{bmatrix}1 & -1 & 1 \ -3 & 2 & -1 \ -2

& 1 & 0\end{bmatrix}\times\begin{bmatrix}1&2 &

3\2&4&6 \ 1&2 &3\end{bmatrix}$

$\quad                   = \begin{bmatrix}0&0&0 \ 0&0&0 \ 0&0&0\end{bmatrix} = O$

$\therefore \quad AB = O$
But neither $A = O$ nor $B = O$.

$A=\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ and $B=\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}$

$AA^{-1} = I\Rightarrow |AA^{-1}| = |I| = 1$
$\Rightarrow |A|\cdot |A^{-1}| = 1\Rightarrow |A^{-1}| = \dfrac {1}{|A|}$.

$AA^{-1} = I\Rightarrow |AA^{-1}| = |I|\Rightarrow |A|\cdot |A^{-1}| = 1\Rightarrow |A^{-1}| = \dfrac {1}{|A|}$.
$|A| = \begin{vmatrix}-2 & 3\ 1 & 1\end{vmatrix} = (-2 - 3) = (-5) \Rightarrow |A^{-1}| = \dfrac {-1}{5}$.

If two Square matrices $A$ and $B$ follow property of anti-commutation, then 

(i) is false.

$|A| = \begin{vmatrix}2 & -1\ 1 & 3\end{vmatrix} = (6 + 1) = 7\neq 0$
$M _{11} = 3, M _{12} = 1, M _{21} = -1$ and $M _{22} = 2$

The inverse of a product of two invertible matrices is the product of their inverses in reverse order: (AB)^-1 = B^-1 * A^-1.

Binomial theorem is applicable if and only if $AB=BA$.

$\displaystyle A1=\frac { 1 }{ 2 } \left( AI+IA \right) =\frac { 1 }{ 2 } \left( A+A \right) =A$

$AB^2 =(AB)B = 0B =0$  

Let $\displaystyle B=I+A+{ A }^{ 2 }+{ A }^{ 3 }+...+{ A }^{ k-1 }$

$\displaystyle A \ast B=\frac{1}{2}(AB'+A'B)$

$A$ and $B$ commute each other then $AB=BA$
${ \left( AB \right)  }^{ n }={ \left( BA \right)  }^{ n }\ \Rightarrow { \left( AB \right)  }^{ n }={ B }^{ n }{ A }^{ n }={ A }^{ n }{ B }^{ n }={ \left( BA \right)  }^{ n }$

Since $\left| A \right| \neq 0$, therefore ${ A }^{ -1 }$ exists.

Let A be a skew-symmeteic matric of order $n.$

We have 

Given, $A=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix},B=\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}$

For a diagonal matrix D = diag(d1, d2, ..., dn), the inverse is D^-1 = diag(1/d1, 1/d2, ..., 1/dn), provided all di are non-zero.

Taking $AB=A$

Given, $A=\begin{pmatrix}
8 & -6 & 2\
-6 & 7 & -4\
2 & -4 & \lambda
\end{pmatrix}$ is a singular matrix
So, det A=0
$\therefore $ BY operation of matrix (s),
$det A=8(7 \lambda-16)+6[-6 \lambda + 8]+2[24-14]$
$=56 \lambda - 128 -36 \lambda +48 +20$
$=20 \lambda - 60$
So, $det A = 0= 20 \lambda -60$
$\lambda =3$

Given the matrix equality, x=1 and y^3=8, which implies y=2. The matrix to invert is [[1, 2], [2, 0]]. The determinant is (10) - (22) = -4. The inverse is (1/det) * [[0, -2], [-2, 1]], which simplifies to -1/4 * [[0, -2], [-2, 1]].

$\left | A \right |= \cos ^{2}\alpha + \sin ^{2}\alpha = 1$
$A^{T}= \begin{bmatrix}\cos\alpha    & \sin \alpha \ -\sin \alpha  & \cos\alpha \end{bmatrix}\neq A$
$A^{-1}= \frac{1}{1}\begin{bmatrix}\cos\alpha    & \sin \alpha \ -\sin \alpha  & \cos\alpha \end{bmatrix}\neq -A$
$\begin{bmatrix}A^{T}= A^{-1}\end{bmatrix}$

For any given square matrix , $ A{A}^{-1} $ is always equal to Identity Matrix $ I $

So, $ A{A}^{-1} = \begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} $

A square matrix that is not invertible is called singular matrix in which its determinant is 0.
$\displaystyle \left[\begin{matrix} 2 & 3 \ 4 & 6 \end{matrix}  \right]$ is a non invertible matrix.
$\because$ its determinant is    $12-12 =0$
Option C is correct.

Since matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is commutative with the matrix $\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$

For two $m \times  n$ and $n \times m$ matrices $A$ and $B$

If $A$ is of order $3$ then,$A^-1$ is also of order $3$.

Solution:

$A=AB-BA$
$AI+BA=AB$
$A(I+B)A=AB$
$\left| I+B \right| \left| A \right| =\left| A \right| \left| B \right| $
$\left| B \right| =\left| I+B \right| \longrightarrow 1$
$A=AB-BA$
$BA=AB-A$
$BA=A(B-I)$
$\left| B \right| \left| A \right| =\left| A \right| \left| B-I \right| $
$\left| B \right| =\left| B-I \right| \longrightarrow 2$
From equation 1 and 2
$\left| B-I \right| =\left| B+I \right| $

$A^2B =A(AB) =A(BA) =(AB)A$

Let ${ D } _{ 1 }=\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix},{ D } _{ 2 }=\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}$

$\begin{bmatrix}2 &1 \ 7 &4 \end{bmatrix}$A$\begin{bmatrix}-3 &2 \ 5 &-3 \end{bmatrix}=\begin{bmatrix}1 &0 \ 0&1 \end{bmatrix}$

$B=(I-A)(I+A)^{-1}$