Multiplicative inverse of a matrix - class-XII
| Description: multiplicative inverse of a matrix | |
| Number of Questions: 64 | |
| Created by: Neema Pandya | |
| Tags: maths matrix determinants business maths matrices applications of matrices and determinants |
Find the inverse f the following matrices by using transformation method.
If $A=\begin{bmatrix} \cos { x } & \sin { x } \ -\sin { x } & \cos { x } \end{bmatrix}$ and $A(AdjA)=k\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ then the value of $k$ is
If A be square matrix of order n and k is a scalar, then adj (KA) is:
If $A=\left[ \begin{matrix} 2 & -3 \ -4 & 7 \end{matrix} \right] $, then ${2A}^{-1}=$
If AB = AC then
$A=\begin{bmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos\theta\end{bmatrix}$ and $AB=BA=I$, then B is equal to
$A=\begin{bmatrix} 2&2&1\0&1&4\0&2&6\end{bmatrix}$, $B=\begin{bmatrix} 2&2&1\0&1&4\0&0&1\end{bmatrix}$
A= $\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$.
$A=\begin{bmatrix} 2&2&1\4&5&6\6&8&9\end{bmatrix}$, $B=\begin{bmatrix} 2&2&1\0&1&4\0&2&6\end{bmatrix}$
Multiply the fourth row by $3$.
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$
$\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$
A= $\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$.
In echelon form, which of the following is incorrect?
The system $\begin{pmatrix} 1 & -1 & 2 \ 3 & 5 & -3 \ 2 & 6 & a \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}=\begin{pmatrix} 3 \ b \ 2 \end{pmatrix}$ has no solution, if
Let $A$ be a matrix of order $3\times 3$ such that $\left| \vec { A } \right| =1$. Let $B=2{ A }^{ -1 }$ and $C=\dfrac { adj.A }{ 2 }$. Then the value of $\left| { AB }^{ 2 }{ C }^{ 3 } \right|$, is ( where $\left| A \right|$ represent det. $A$)
$\begin{bmatrix}
\cos\theta & -\sin\theta \[0.3em]
\sin\theta & \cos\theta
\end{bmatrix} = \begin{bmatrix}
1 & -\tan\theta/2 \[0.3em]
\tan\theta/2 & 1
\end{bmatrix} \begin{bmatrix}
1 & \tan\theta/2 \[0.3em]
-\tan\theta/2 & 1
\end{bmatrix}$
If $A = \begin{bmatrix} a & b\ c & d \end{bmatrix} $ satisfies the equation $x^2 - (a+d)x+k=0$ then
The number of $2\times 2$ matrices $A=\left[ \begin{matrix} a & b \ c & d \end{matrix} \right] $ for which ${ \left[ \begin{matrix} a & b \ c & d \end{matrix} \right] }^{ -1 }$ $=\left[ \begin{matrix} \frac { 1 }{ a } & \frac { 1 }{ b } \ \frac { 1 }{ c } & \frac { 1 }{ d } \end{matrix} \right] $, $(a,b,c,d\ \epsilon \ R)$ is
Let A=$\left( {\begin{array}{{20}{c}}{ - 5}&{ - 8}&{ - 7}\3&5&4\2&3&3\end{array}} \right),B = \left( {\begin{array}{{20}{c}}x\y\z\end{array}} \right)$. If AB is scalar $\left( { \ne 0} \right)$ multiple of B, then x+y=
If $A = \left[ {\begin{array}{*{20}{c}}1&2\3&4\end{array}} \right]$, then $8A^{-4}$ is equal to
If $A$ and $B$ are square matrices such that $B=-A^{-1}BA$, then
If $A$ is a $2\times 2$ matrix such that $A^{2}-4A+3I=0$, then the inverse of $A+3I$ is equal to
If $A^{-1} = \alpha I + \beta I$ where $\alpha, \beta \in R$, then $\alpha + \beta$ is equal to (where $A^{-1}$ denotes inverse of matrix $A$)-
If $A=\begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$, find the values of $\alpha$ for which $A^2=B$.
If $A=\left[ \begin{matrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{matrix} \right] $ and $10B=\left[ \begin{matrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end{matrix} \right] $ where $B=A^{-1}$ then $\alpha$ is equal to-
The inverse of the matrix $\left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \ { 3 } & { 3 } & { 0 } \ { 5 } & { 2 } & { - 1 } \end{array} \right]$ is
If $A=\left[ \begin{matrix} 1 & 0 & -1 \ 3 & 4 & 5 \ 0 & 6 & 7 \end{matrix} \right]$ and $A^{-1}=[\alpha _{ij}] _{3\times 3}$ then $\alpha _{23}=$
Let $P=\begin{bmatrix} \cos { \dfrac { \pi }{ 9 } } & \sin { \dfrac { \pi }{ 9 } } \ -\sin { \dfrac { \pi }{ 9 } } & \cos { \dfrac { \pi }{ 9 } } \end{bmatrix}$ and $\alpha,\ \beta,\ \gamma$ be non-zero real numbers such that $\alpha P^{6}+\beta P^{3}+\gamma 1$ is the zero matrix. Then, $(\alpha^{2}+\beta^{2}+\gamma^{2})^{(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)}$ is
Inverse of $\begin{bmatrix} -1 & 5 \ -3 & 2 \end{bmatrix}$ is
Consider three matrices $A=\begin{bmatrix} 2 & 1 \ 4 & 1 \end{bmatrix}, B=\begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix}$ and $C=\begin{bmatrix} 3 & -4 \ -2 & 3 \end{bmatrix}$. Then the value of the sum $tr(A)+tr\left(\dfrac{ABC}{2}\right)+tr\left(\dfrac{A(BC)^{2}}{4}\right)+tr\left(\dfrac{A(BC)^{3}}{8}\right)+....+\infty$ is
If A is a 2 X 2 matrix such that $A^2009 + A^2008$= I, then : $(A^2008)^-1$=
If $I=I=\left[ \begin{matrix} 1 \ 0 \end{matrix}\begin{matrix} 0 \ 1 \end{matrix} \right] ,j=\left[ \begin{matrix} 0 \ -1 \end{matrix}\begin{matrix} 1 \ 0 \end{matrix} \right] and B=\left[ \begin{matrix} cos\theta \ -sin\theta \end{matrix}\begin{matrix} sin\theta \ cos\theta \end{matrix} \right] ,$ then B =
If $A(\theta) = \begin{bmatrix}\sin \theta & i \cos \theta\ i \cos \theta & \sin \theta\end{bmatrix}$, then which of the following is not true?
Write the following transformation in matrix form
$\quad x _1 = \displaystyle\frac{\sqrt 3}{2}y _1 + \displaystyle\frac{1}{2}y _2; \quad x _2 = -\displaystyle\frac{1}{2}y _1 + \displaystyle\frac{\sqrt 3}{2}y _2$.
Hence find the transformation in matrix form which expresses $y _1, y _2$ in terms of $x _1, x _2$.
Let p be a non-singular matrix, $1+p+p^{2}+....+p^{n}=0$ (0 denotes the null matrix) then $p^{-1}=$
Let A be a $3 \times 3$ matrix such that is: $A\left[ \begin{matrix} 1 & 2 & 3 \ 0 & 2 & 3 \ 0 & 1 & 1 \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{matrix} \right] $Then $A^{-1}$ is
Use the method of elementary row transformation to compute the inverse of
$\quad \begin{bmatrix} 1 & 2 & 5 \ 2 & 3 & 1 \ -1 & 1 & 1\end{bmatrix}$
If $
A=\left[ \begin{array}{ll}{x} & {1} \ {1} & {0}\end{array}\right]
$ and $
A^{2}=I
$, $
A^{-1}
$ is equal to ...............
A is an involuntary matrix given by $A=\begin{bmatrix} 0 & 1 & -1\ 4 & -3 & 4\ 3 & -3 & 4\end{bmatrix}$ then the inverse of $\dfrac{A}{2}$ will be?
If $A\begin{bmatrix} 1 & 1\ 2 & 0\end{bmatrix}=\begin{bmatrix} 3 & 2\ 1 & 1\end{bmatrix}$, then $A^{-1}$ is given by?
If $A=\left[ \begin{matrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{matrix} \right] $, then value of $A^{-1}$ is equal to
If A and B are any $2\times2$ matrices, then det. (A+B) =0 implies
If $A^2-A+1=0$, then the inverse of A is?
Let $\begin{bmatrix} 1 & 1\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 3\ 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & n-1\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\ 0 & 1\end{bmatrix}$
If $A=\begin{bmatrix} 1 & n\ 0 & 1\end{bmatrix}$ then $A^{-1}=?$
If $\displaystyle A=\begin{bmatrix} 0 & 0 & 1\ 0 & 1&0 \ 1& 0 & 0\end{bmatrix}$, then $A^{-1}$ is.
Let $A=\begin{bmatrix} 1 & -1 & -1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}$ and $10B=\begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end{bmatrix}$, if $B$ is the inverse of matrix $A$, then $\alpha $ is
If $\begin{bmatrix} 1 & 2 \ 3 & -5 \end{bmatrix}$, then ${A}^{-1}$ is equal to
If you switch the first row with the fourth row, what will the new first row be?
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$
Which of the following is the new row that results when you add rows $1$ and $3$?
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$
Use a transformation matrix to find the image of $D(-7,6)$ after a rotation of $180^0$ counterclockwise around the origin.
$A=\begin{bmatrix} 1&-2&3\7&-8&9\4&-5&6\end{bmatrix}$ the new matrix formed by adding $\ 2^{nd}\ row \ to \ 1^{st} $ row will be
A=$\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$
The new matrix formed after interchanging $2^{nd}$ and $3^{rd}$rows will be
For a matrix $A \begin{pmatrix} 1& 0 & 0\ 2 & 1 & 0\ 3 & 2 & 1\end{pmatrix}$, if $U _{1}, U _{2}$ and $U _{3}$ are $3\times 1$ column matrices satisfying $AU _{1} = \begin{pmatrix}1\ 0 \ 0
\end{pmatrix}, AU _{2} \begin{pmatrix}2\3 \ 0
\end{pmatrix}, AU _{3} = \begin{pmatrix}2\ 3\ 1
\end{pmatrix}$ and $U$ is $3\times 3$ matrix whose columns are $U _{1}, U _{2}$ and $U _{3}$
Then sum of the elements of $U^{-1}$ is
The inverse of a diagonal matrix is a :
Inverse of $A = \begin{bmatrix} 1& 3\ 2 & -2\end{bmatrix} $ is equal to?A
If a matrix A is such that $3{A^3} + 2{A^2} + 5A + I = 0$ , then $A^{-1}$ is equal to
If $A$ is a non zero square matrix of order $n$ with $det\left( I+A \right) \neq 0$, and ${A}^{3}=0$, where $I,O$ are unit and null matrices of order $n\times n$ respectively, then ${ \left( I+A \right) }^{ -1 }=$
If $A=\begin{bmatrix} 3 & -2 \ 5 & 8 \end{bmatrix}$, then $A^{-1}=$
If the matrix $\begin{bmatrix} 0 & 2\beta & \Upsilon \ \alpha & \beta & -\Upsilon \ \alpha & -\beta & \Upsilon \end{bmatrix}$is orthogonal, then
The inverse of the $\begin{bmatrix}- 1 & 5\ - 3 & 2\end{bmatrix}$ is
The inverse of the matrix $\begin{bmatrix} 5 & -2 \ 3 & 1 \end{bmatrix}$ is
What is the inverse of the matrix
$A=\begin{bmatrix} \cos { \theta } & \sin { \theta } & 0 \ -\sin { \theta } & \cos { \theta } & 0 \ 0 & 0 & 1 \end{bmatrix}$ ?