Multiplicative inverse of a matrix - class-XII

$A(AdjA)=\begin{bmatrix}c^{2}+s^{2}&cs-cs\-cs+cs&c^{2}+s^{2}\end{bmatrix}=\begin{bmatrix}1&0\0&1\end{bmatrix}\implies k=1$

For an n x n matrix A and scalar k, the property of the adjugate is adj(kA) = k^(n-1) * adj(A).

$A=\left[ { \begin{array} { *{ 20 }{ c } }2 & { -3 } \ { -4 } & 7 \end{array} } \right]  \ \left| { A-\lambda I } \right| =0 \ \left| { \begin{array} { *{ 20 }{ c } }{ 2-\lambda  } & { -3 } \ { -4 } & { 7-\lambda  } \end{array} } \right| =0 \ \left( { 2-\lambda  } \right) \left( { 7-\lambda  } \right) -12=0 \ \left( { \lambda -7 } \right) \left( { \lambda -2 } \right) -12=0 \ { \lambda ^{ 2 } }-9\lambda +2=0 \ { A^{ 2 } }-9A+2I=0 \ A-9I+2{ A^{ -1 } }=0 \ 2{ A^{ -1 } }=9I-0 \ 2{ A^{ -1 } }=9I-A$

Value of B and C depends on the value of A. if A is non-singular, i.e non zero, then B=C. 
But if A=0, then both terms AB and AC becomes zero, so B and C need not be necessarily equal.

Given, $A=\begin{bmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos\theta \end{bmatrix}$ and $AB=BA=I$
$\Rightarrow B=A^{-1}I=A^{-1}$
$=\displaystyle\frac{1}{\cos^2\theta +\sin^2\theta}\begin{bmatrix} \cos\theta & \sin\theta \ -\sin\theta & \cos\theta\end{bmatrix}$
$\Rightarrow B=\begin{bmatrix}\cos\theta & \sin\theta \ -\sin\theta & \cos\theta\end{bmatrix}$

Fourth row = $0,0,6,1$

multiplying by $3$

$(0\times 3 , 0\times 3 , 6\times 3 , 1\times 3)$

$(0,0,18,3)$

Echelon form of matrix has the following characteristics:

1. The first non-zero entry of every row is 1.

2. The first non-zero entry in every row is one position right to the previous row.

3. The row with all zero elements will be below the rows having a non-zero element.

Therefore, according to the question statement in option D is incorrect.

Hence, the answer is option D.

Given 
$\begin{pmatrix} 1 & -1 & 2 \ 3 & 5 & -3 \ 2 & 6 & a \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} 3 \ b \ 2 \end{pmatrix}$
The augment matrix of given system is
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 3 & 5 & -3 & | & b \ 2 & 6 & a & | & 2 \end{bmatrix}$
${ R } _{ 2 }\rightarrow { R } _{ 2 }-3{ R } _{ 1 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 2 & 6 & a & | & 2 \end{bmatrix}$
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 1 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 0 & 8 & a-4 & | & -4 \end{bmatrix}$
${ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 2 }$ gives
$\left[ A|B \right] =\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 8 & -9 & | & b-9 \ 0 & 0 & a+5 & | & 5-b \end{bmatrix}$
We know that
For no solution
Rank A < Rank $\left[ A|B \right] $
$\therefore$ $a+5=0$ and $5-b\ne 0$
$a=-5, b\ne 5$

This is a standard trigonometric identity related to rotation matrices and the tangent half-angle substitution. The product of the two matrices on the right side equals the rotation matrix on the left.

$A=\begin{bmatrix}a&b\c&d\end{bmatrix}$

If $A=\begin{bmatrix} a & b \ c & d \end{bmatrix}\quad { \begin{bmatrix} a & b \ c & d \end{bmatrix} }^{ -1 }=\begin{bmatrix} 1/a & 1/b \ 1/c & 1/d \end{bmatrix}$                          Not possible for any values of $a, b, c$

Using the Cayley-Hamilton theorem for matrix A = [[1, 2], [3, 4]], the characteristic equation is A^2 - 5A - 2I = 0. By manipulating this equation, one can express higher powers of A in terms of A and I, eventually leading to the expression 27I - 145A^-1.

Given B = -A^-1BA, multiply both sides by A on the right to get BA = -A^-1BA^2, or simply rearrange to AB = -BA. Adding BA to both sides yields AB + BA = 0.

From A^2 - 4A + 3I = 0, we can write A^2 - 4A = -3I. Factoring gives A(A - 4I) = -3I, so A(4I - A) = 3I. The inverse of (A + 3I) is found by manipulating the characteristic equation; the result is 1/24 A - 7/24 I.

We have,

Since B = A^-1, we have AB = I. Multiplying the first row of A by the second column of B must equal 0. Calculating this gives 1(2) + (-1)(0) + 1(-2) = 0, which is consistent. Solving for the element at (2,3) of the product AB = I yields alpha = 2.

The inverse of a matrix M is (1/det(M)) * adj(M). The determinant of the given matrix is 1(3*-1 - 0) = -3. Calculating the adjugate matrix and multiplying by -1/3 yields the correct option.

The element alpha_23 of the inverse matrix A^-1 is given by (-1)^(2+3) times the minor of the element at row 3, column 2 of matrix A, divided by the determinant of A. Computing the determinant of A and the appropriate cofactor yields 2/5.

Matrix P represents a rotation by pi/9. Raising P to the 6th and 3rd powers corresponds to rotations by 6pi/9 and 3pi/9 respectively. By analyzing the characteristic equation or evaluating the matrix powers, the coefficients alpha, beta, and gamma make the expression simplify such that the exponent product evaluates to 1.

For a 2x2 matrix [[a, b], [c, d]], the inverse is (1/(ad-bc)) * [[d, -b], [-c, a]]. Here, det = (-1)(2) - (5)(-3) = -2 + 15 = 13. The inverse is (1/13) * [[2, -5], [3, -1]].

The given series involves traces of matrix products that form a geometric progression of matrices. By computing BC and evaluating the traces step-by-step, the sum converges to a total trace value of 6.

Given A^2009 + A^2008 = I, we can factor as A^2008(A + I) = I. This implies (A^2008)^-1 = A + I.

Finding inverse of the matrix $A(\theta)= \begin{bmatrix} \sin\theta & i\cos\theta \ i\cos\theta & \sin\theta\end{bmatrix}$

$ \displaystyle  { x } _{ 1 }=\frac { \sqrt { 3 }  }{ 2 } { y } _{ 1 }+\frac { 1 }{ 2 } { y } _{ 2 }  $ and $\displaystyle { x } _{ 2 }=\frac { -1 }{ 2 } { y } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } { y } _{ 2 } $ 
We observe $ \displaystyle \frac { \sqrt { 3 }  }{ 2 } { x } _{ 1 }-\frac { 1 }{ 2 } { x } _{ 2 }=\frac { 3 }{ 4 } { y } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } .\frac { 1 }{ 2 } { y } _{ 2 }+\frac { 1 }{ 4 } { y } _{ 1 }-\frac { \sqrt { 3 }  }{ 2 } \frac { 1 }{ 2 } { y } _{ 2 } $
$ \displaystyle \Rightarrow \frac { \sqrt { 3 }  }{ 2 } { x } _{ 1 }-\frac { 1 }{ 2 } { x } _{ 2 }={ y } _{ 1 } $
Similarly $ \displaystyle \frac { 1 }{ 2 } { x } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } { x } _{ 2 }=\frac { 1 }{ 4 } { y } _{ 2 }+\frac { 3 }{ 4 } { y } _{ 2 }={ y } _{ 2 } $
$ \displaystyle \therefore { y } _{ 1 }=\frac { \sqrt { 3 }  }{ 2 } { x } _{ 1 }-\frac { 1 }{ 2 } { x } _{ 2 };{ y } _{ 2 }=\frac { 1 }{ 2 } { x } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } { x } _{ 2 }  $ 

Given I + P + P^2 + ... + P^n = 0, we have P + P^2 + ... + P^n = -I. Factoring out P gives P(I + P + ... + P^(n-1)) = -I. This is a standard identity for geometric series of matrices.

If AX = B, then A^-1 = XB^-1. Here X is the matrix on the left and B is the matrix on the right. Calculating B^-1 and multiplying by X yields the inverse.

Let $\quad A = \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}$

$\Rightarrow \quad Write \space A A^{-1}= I$

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

$\quad \begin{matrix}R _{21}(-2)\ \mbox{}\ R _{31}(1)\end{matrix}\begin{bmatrix}1 & 0 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \ -2 & 1 & 0 \ 1 & 0 & 1\end{bmatrix}$

$\quad \begin{matrix}R _2(-1) \ \mbox{} \ R _3(1/3)\end{matrix}\begin{bmatrix}1 & 2 & 5 \ 0 & 1 & 9 \ 0 & 1 & 2\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \ 2 & -1 & 0 \ \displaystyle\frac{1}{3} & 0 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R _{12}(-2) \ \mbox{} \ R _{32}(-1)\end{matrix}\begin{bmatrix}1 & 0 & -13 \ 0 & 1 & 9 \ 0 & 0 & -7\end{bmatrix} A^{-1}= \begin{bmatrix}-3 & 2 & 0 \ 2 & -1 & 0 \ -\displaystyle\frac{5}{3} & 1 & \displaystyle\frac{1}{3}\end{bmatrix}$

$\quad \begin{matrix}R _3(-1/7)\ \mbox{}\end{matrix}\begin{bmatrix}1 & 0 & -13 \ 0 & 1 & 9 \ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}-3 & 2 & 0 \ 2 & -1 & 0 \ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

$\quad \begin{matrix}R _{13}(13) \ \mbox{~} \ R _{23}(-9)\end{matrix}\begin{bmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

Hence, $\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$

An involuntary matrix satisfies A^2 = I, which means A = A^-1. If we want the inverse of (A/2), we use the property (kA)^-1 = (1/k)A^-1. Thus, (A/2)^-1 = 2A^-1 = 2A.

Let X = [[1, 1], [2, 0]] and B = [[3, 2], [1, 1]]. We have AX = B, so A = BX^-1. Then A^-1 = (BX^-1)^-1 = XB^-1. Calculating B^-1 and multiplying by X gives the result.

Calculating A^2 shows that A^2 = I. If A^2 = I, then A = A^-1.

The determinant of a sum of matrices is not generally related to the sum of their determinants or the individual determinants being zero.

Given the matrix equation A^2 - A + I = 0 (where 1 represents the identity matrix I), we can rearrange it to isolate I: I = A - A^2 = A(I - A), or multiply through by A^-1. Multiplying A^2 - A + I = 0 by A^-1 gives A - I + A^-1 = 0, which rearranges to A^-1 = I - A.

$\begin{bmatrix} 1 & 1\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 3\ 0 & 1\end{bmatrix}..\begin{bmatrix} 1 & n-1\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\ 0 & 1\end{bmatrix}$
$\Rightarrow \dfrac{n(n-1)}{2}=78\Rightarrow n=13$
$A=\begin{bmatrix} 1 & 13\ 0 & 1\end{bmatrix}$
so $A^{-1}=\begin{bmatrix} 1 & -13\ 0 & 1\end{bmatrix}$.

We have, $A=\begin{bmatrix} 0 & 0&1\ 0 &1 &0\ 1&0 &0\end{bmatrix}$
$\Rightarrow |A|=0(0-0)-0(0-0)+1(0-1)$
$\Rightarrow |A|=-1$
and cofactors of A are
$A _{11}=0, A _{12}=0, A _{13}=-1,$
$A _{21}=0, A _{22}=-1, A _{23}=0,$
$A _{31}=-1, A _{32}=0, A _{33}=0$
$\therefore A^{-1}=\displaystyle\frac{adj(A)}{|A|}$
$=-\displaystyle\frac{1}{1}\begin{bmatrix} 0 & 0 & -1\0 & -1 &0\ -1 &0 &0\end{bmatrix}$

Since, $B$ is the inverse of $A$.
ie, $B=10{ A }^{ -1 }$
$\therefore \left( 10 \right) { A }^{ -1 }=\begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{bmatrix}$
$\therefore \left( 10 \right) { A }^{ -1 }\cdot A=\begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{bmatrix}A$
$\Rightarrow 10I=\begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 10 & 0 & 0 \ 0 & 10 & 0 \ 0 & 0 & 10 \end{bmatrix}=\begin{bmatrix} 10 & 0 & 0 \ -5+\alpha  & 5+\alpha  & -5+\alpha  \ 0 & 0 & 10 \end{bmatrix}$
$\Rightarrow 5+\alpha =10$
$\Rightarrow \alpha =5$

Since $A=\begin{bmatrix} 1 & 2 \ 3 & -5 \end{bmatrix}$

The transformation matrix for rotation  is $\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}$

Let $U _{i} = \begin{pmatrix}a _{i}\b _{i} \c _{i} \end{pmatrix} i = 1, 2, 3$
$AU _{1} = \begin{pmatrix}a _{1}\2a _{1} + b _{1} \ 3a _{1} + 2b _{1} + c _{1}
\end{pmatrix} = \begin{pmatrix}1\0 \ 0
\end{pmatrix}$, So $a _{1} = 1, b _{1} = -2, c _{1} = 1$
$AU _{2} = \begin{pmatrix}a _{2}\2a _{2} + b^{2} \ 3a _{2} + 2b _{2} + c _{2}
\end{pmatrix} = \begin{pmatrix}2\ 3\ 0
\end{pmatrix}$,
So, $a _{2} = 2, b _{2} = -1, c _{2} = -4$. Similarly, $a _{3} = 2, b _{3} = -1, c _{3} = -3$
So, $U = \begin{pmatrix} 1& 2 & 2\ -2 & -1 & -1\ 1 & -4 & -3\end{pmatrix}$. So, sum of elements of $U^{-1}$ is zero.

A diagonal matrix has elements only in it's diagonal.
So the inverse will also have all non zero elements in the diagonal.
So, it will be symmetric and will also be a diagonal matrix.
Hence, option A and C are correct

$3A^3+2A^2+5A+I=0$
$3A^3+2A^2+5A+AA^{-1}=0$
$A^{-1}=-3A^2-2A-5$

$A=\left[{\begin{array}{cc}3&2\5&8\end{array}}\right]$

$\begin{array}{l} A=\left[ \begin{array}{l} 5, , , , , -2 \ 3, , , , , , , , , 1 \end{array} \right]  \ \left| A \right| =5+6=11\ne 0 \ so,, A, , is, , , non-\sin  gular, ,, { A^{ -1 } }, , is, , exist \ so,m, { A _{ 11 } }=1,, , , , , { A _{ 12 } }=-3,, , , , { A _{ 21 } }=2,, , , , , , { A _{ 22 } }=5 \ A=\left( { \begin{array} { *{ 20 }{ c } }1 & { -3 } \ 2 & 5 \end{array} } \right) \Rightarrow AdjA=\left( { \begin{array} { *{ 20 }{ c } }1 & 2 \ { -3 } & 5 \end{array} } \right)  \ { A^{ -1 } }=\frac { 1 }{ { \left| A \right|  } } adjA, , , , \Rightarrow , , , , \frac { 1 }{ { 11 } } \left( { \begin{array} { *{ 20 }{ c } }1 & 2 \ { -3 } & 5 \end{array} } \right)  \end{array}$