Tag: properties of inverses of matrices

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Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

Matrices $A$ and $B$ will be inverse of each other only if

  1. $AB=BA$
  2. $AB=0,BA=I$
  3. $AB=BA=0$
  4. $AB=BA=I$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that if $A$ is a square of order $m$, and if there exists another square matrix $B$ of the same order $m$, such that $AB=I$, then $B$ is said to be the inverse of $A$. 

In this case, it is clear that $A$ is the inverse of $B$.
Thus , matrices $A$ and $B$ will be inverses of each other only if $AB=BA=I.$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants


 Let A$= \left[\begin{array}{lll}1 & 0 & 0\2 & 1 & 0\3 & 2 & 1\end{array}\right]$.If $\mathrm{u _1}$ and $\mathrm{u} _{2}$ are column matrices such that  $\mathrm{Au _{1}}=\left[\begin{array}{l}1\0\0\end{array}\right]$ and  $\mathrm{Au _{2}}=\left[\begin{array}{l}0\1\0\end{array}\right]$ then $\mathrm{u _{1}+u _{2}}$ is equal to:

  1. $\left[\begin{array}{l}

    -1\\

    1\\

    0

    \end{array}\right]$
  2. $\left[\begin{array}{l}

    -1\\

    1\\

    -1

    \end{array}\right]$
  3. $\left[\begin{array}{l}

    -1\\

    -1\\

    0

    \end{array}\right]$
  4. $\left[\begin{array}{l}

    1\\

    -1\\

    -1

    \end{array}\right]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given:Matrices are 
$A=\left[ \begin{matrix} 1 & 0  & 0 \\ 2 & 1  & 0 \\ 3 & 2  & 1 \end{matrix} \right]$

$A{u} _{1}=\left[ \begin{matrix} 1  \\  0 \\ 0 \end{matrix} \right]$ and 
$A{u} _{2}=\left[ \begin{matrix} 0  \\  1 \\ 0 \end{matrix} \right]$

To find:Matric ${u} _{1}+{u} _{2}$

Since both $A{u} _{1}$ and $A{u} _{2}$ are given, hence adding them, we get

$A{u} _{1}+A{u} _{2}=\left[ \begin{matrix} 1  \\  0 \\ 0 \end{matrix} \right]+\left[ \begin{matrix} 0  \\  1 \\ 0 \end{matrix} \right]$

$A\left({u} _{1}+{u} _{2}\right)=\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix} \right]$

Since,$A$ is a non-singular matrix,we have
$\left|A\right|\neq\,0$

Hence multiplying both sides by ${A}^{-1}$ from RHS we get

${A}^{-1}A\left({u} _{1}+{u} _{2}\right)={A}^{-1}\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix}\right]$

${u} _{1}+{u} _{2}={\left[ \begin{matrix} 1 & 0  & 0 \\ 2 & 1  & 0 \\ 3 & 2  & 1 \end{matrix} \right]}^{-1}\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix}\right]$     ..........$(1)$

Now, $\left|A\right|=\left|\begin{matrix} 1 & 0  & 0 \\ 2 & 1  & 0 \\ 3 & 2  & 1 \end{matrix} \right|$

$=1\left| \begin{matrix} 1 & 0   \\ 2 & 1  \end{matrix} \right|-0+0$(by expanding the determinant along row $1$)

$\Rightarrow\,\left|A\right|=1$

Now, co-factor matrix of $A$ (i.e., the matrix in which every element is replaced by corresponding co-factor)

$=\left[\begin{matrix} \left| \begin{matrix} 1 & 0   \\ 2 & 1  \end{matrix} \right| & -\left| \begin{matrix} 2 & 0   \\ 3 & 1  \end{matrix} \right|  & \left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & 0   \\ 2 & 1  \end{matrix} \right| & \left| \begin{matrix} 1 & 0   \\ 3 & 1  \end{matrix} \right|  & -\left| \begin{matrix} 1 & 0   \\ 3 & 2 \end{matrix} \right| \\ \left| \begin{matrix} 0 & 0   \\ 1 & 0  \end{matrix} \right| & -\left| \begin{matrix} 1 & 0   \\ 2 & 0  \end{matrix} \right|  & \left| \begin{matrix} 1 & 0   \\ 2 & 1  \end{matrix} \right| \end{matrix} \right]$

$=\left[\begin{matrix} 1 & -2 & 1 \\ 0 & 1  & -2 \\ 0 & 0 & 1 \end{matrix} \right]$
$\therefore\,adj\left(A\right)={\left[\begin{matrix} 1 & -2 & 1 \\ 0 & 1  & -2 \\ 0 & 0 & 1 \end{matrix} \right]}^{T}=\left[\begin{matrix} 1 & 0 & 0 \\ -2 & 1  & 0 \\ 1 & -2 & 1 \end{matrix} \right]$

$\Rightarrow\,{A}^{-1}=\dfrac{adj\left(A\right)}{\left|A\right|}$

$=\left[\begin{matrix} 1 & 0  & 0 \\ -2 & 1  & 0 \\ 1 & -2  & 1 \end{matrix} \right|\,\,\,\because\left|A\right|=1$

From eqn$(1)$ we get

${u} _{1}+{u} _{2}={\left[\begin{matrix} 1 & 0  & 0 \\ 2 & 1  & 0 \\ 3 & 2  & 1 \end{matrix} \right]}^{-1}\times\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix}\right]$ 

$=\left[\begin{matrix} 1 & 0  & 0 \\ -2 & 1  & 0 \\ 1 & -2  & 1 \end{matrix} \right]\times\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix}\right]$ 

$=\left[ \begin{matrix} 1+0+0  \\  -2+1+0 \\ 1-2+0 \end{matrix}\right]$ 

$\therefore\,{u} _{1}+{u} _{2}=\left[ \begin{matrix} 1  \\  -1 \\ -1 \end{matrix}\right]$ 
Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If a $3\times 3$ matrix $A$ has its inverse equal to $A$, then ${A}^{2}$ is equal to

  1. $\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
  2. $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}$
  3. $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
  4. $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $A^{-1}=A$

$AA^{-1}=I$
$\implies A.A=I$
$\implies A^{2}=I=\begin{bmatrix}1&0&0\0&1&0\0&0&1\end{bmatrix}$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A$ is an $3\times 3$ non -singular matrix that $AA'=A'A$ and $B=A^{-1}A'$,then $BB'$ equal ?

  1. $I+B^{-1}$
  2. $(B^{-1})$
  3. $I+B$
  4. $I$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given A is non-singular and AA' = A'A, we have B = A^-1 A'. Then BB' = (A^-1 A')(A^-1 A')'. Since (A^-1 A')' = (A')'(A^-1)' = A(A^-1)' = A(A')^-1, the expression simplifies based on the commutativity of A and A'.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

 ${( -A )}^{ -1 }$ is always equal to (where $A$ is $nth$ order square matrix)

  1. ${ (-1) }^{ n }{ A }^{ -1 }$
  2. ${ -A }^{ -1 }$
  3. ${( -1) }^{ n-1 }{ A }^{ -1 }$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that if $A^{-1}$ exist then $(cA)^{-1}=\dfrac{1}{c}A^{-1}$ where c is  a constant
Hence $(-A)^{-1}=\dfrac{1}{-1}A^{-1}=-A^{-1}$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A\left( \alpha ,\beta  \right) =\left[ \begin{matrix} \cos { \alpha  }  & \sin { \alpha  }  & 0 \ -\sin { \alpha  }  & \cos { \alpha  }  & 0 \ 0 & 0 & { e }^{ \beta  } \end{matrix} \right]$, then $A{ \left( \alpha ,\beta  \right)  }^{ -1 }$ is equal to 

  1. $A{ \left( -\alpha ,-\beta \right) }$
  2. $A{ \left( -\alpha ,\beta \right) }$
  3. $A{ \left(\alpha ,-\beta \right) }$
  4. $A{ \left(\alpha ,\beta \right) }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The matrix A(alpha, beta) is a rotation matrix combined with a scaling factor e^beta. The inverse of a rotation matrix is obtained by replacing alpha with -alpha, and the inverse of the scaling factor e^beta is e^-beta.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

Let $a, b, c$ are non real number satisfying equation $x^{5}=1$ and $S$ be the set of all non-invertible matrices of the from $\begin{bmatrix} 1 & a & b \ w & 1 & c \ { w }^{ 2 } & w & 1 \end{bmatrix}$ where $w={ e }^{ \dfrac { 12\pi  }{ 5 }  }$. The number of distinct matrices in set $S$ is 

  1. $1$
  2. $28$
  3. $32$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The matrix is non-invertible if its determinant is zero. Given the structure and the roots of unity, the determinant evaluates to zero only under specific conditions related to the values of a, b, and c.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If A is an invertible matrix, then det $\displaystyle :\left ( A^{-1} \right )$ is equal to

  1. $\displaystyle \:det\left ( A \right )$
  2. $\displaystyle \:\frac{1}{det\left ( A \right )}$
  3. $1$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that $|A^{n}|=|A|^{n}$  n be any integer
$\Rightarrow |A^{-1}|=|A|^{-1}=\displaystyle \frac{1}{|A|}$

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $\displaystyle [A]\neq 0 $ then which of the following is not true?

  1. $\displaystyle (A^{2})^{-1}= (A^{-1})^{2}$
  2. $\displaystyle (A')^{-1}= (A^{-1})^{'}$
  3. $\displaystyle A^{-1}= \left | A \right |^{-1}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know, $(A^{n})^{-1}=(A^{-1})^{n}$
So, $(A^{2})^{-1}=(A^{-1})^{2}$
Hence, option A is correct.

We know that inverse of transpose of matrix is equal to transpose of inverse of matrix
$(A^{-1})' =(A')^{-1}$
Hence, option B is correct

For option C,
In the LHS, there is a matrix and in RHS , its a determinant i.e. a single value.
So, option C is incorrect.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

Which of the following matrix is inverse of itself

  1. $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
  2. $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
  3. $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}$
  4. $\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Inverse of unit matrix also unit matrix.

Ans: B