Tag: inverse of a matrix and linear equations

Questions Related to inverse of a matrix and linear equations

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 

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

  1. $1$

  2. $28$

  3. $32$

  4. $4$

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Correct Option: A

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

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

  1. $\displaystyle :det\left ( A \right )$

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

  3. $1$

  4. none of these

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Correct Option: B
Explanation:

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

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

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

  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

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Correct Option: C
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.

Which of the following matrix is inverse of itself

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

  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. $\begin{bmatrix} 0 & 1 & 0 \ 1 & 1 & 1 \ 0 & 1 & 0 \end{bmatrix}$

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Correct Option: B
Explanation:

Inverse of unit matrix also unit matrix.

Ans: B

For two suitable order matrices $A, B$; correct statement is-

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

  1. ${(AB)}^{-1}={A}^{-1}{B}^{-1}$

  2. ${(AB)}^{-1}={B}^{-1}{A}^{-1}$

  3. ${(AB)}^{-1}={(BA)}^{-1}$

  4. none of these

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Correct Option: B
Explanation:

It is fundamental concept that, $(AB)^{-1}=B^{-1}A^{-1}$

If A is a $3 \times 3$ matrix such that $\left| A \right| = 4\ than\ \left| {{{\left( {adjA} \right)}^{ - 1}}} \right| = $

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

  1. $16$

  2. $64$

  3. $\dfrac{1}{{16}}$

  4. None

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Correct Option: A
Explanation:

We know, $A^{-1}=\dfrac{adjA}{|A|}$

Multiplying above equation with A both sides,
$AA^{-1}=\dfrac{A\times adjA}{|A|}\Rightarrow|A|=A\times adjA$
Multiplying with $(adjA)^{-1}$ both sides ,
$|A|\times(adjA)^{-1}=A\times adjA \times(adjA)^{-1}\Rightarrow|A|\times(adjA)^{-1}=A$
Taking determinant both sides,$||A|\times(adjA)^{-1}|=|A|\Rightarrow ||A||\times |(adjA)^{-1}|=|A|\Rightarrow |A|^n\times |(adjA)^{-1}|=|A|$
Where n is the order of matrix A, i.e. $n=3$ and $|A|=4$
Thus, $|(adjA)^{-1}|=\dfrac{4}{4^3}=\dfrac{1}{16}$

If the matrices $A, B, (A+B)$ are non singular then ${[A{(A+B)}^{-1}B]}^{-1}$ is equal to-

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

  1. $A+B$

  2. ${A}^{-1}+{B}^{-1}$

  3. $A{(A+B)}^{-1}$

  4. None

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Correct Option: B
Explanation:
$ (A(A+B)^{-1}B)^{-1}$

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

$ = [(AA^{-1}+AB^{-1})B]^{-1}$

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

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

If $A$ is an invertible matrix of order $2$, then $det({A}^{-1})$ is equal to

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

  1. $det(A)$

  2. $\cfrac{1}{det(A)}$

  3. $1$

  4. $0$

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Correct Option: B
Explanation:

We know that 

$AA^{-1}=I$
Taking determinant both sides
$|AA^{-1}|=|I|$
$|A||A^{-1}|=|I|$       $[\because |AB|=|A||B|]$
$|A||A^{-1}|=1$         $[\because |I|=1]$
$|A^{-1}|=\dfrac{1}{|A|}$
Since $|A|\neq0$
Hence, $|A^{-1}|=\dfrac{1}{|A|}$

Let $A,B$ and $C$ be square matrices of order $3\ \times 3$. If $A$ invertible $(A-B)C=BA^{-1}$, then

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

  1. $C\ (A-B)=A^{-1}B$

  2. $C\ (A-B)=BA^{-1}$

  3. $(A-B)C=A^{-1}B$

  4. $All\ the\ above$

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Correct Option: A

A square non-singular matrix A satisfies $\displaystyle A^{2}-A+2I=0$, then $\displaystyle A^{-1}=$

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

  1. $\displaystyle I-A$

  2. $\displaystyle \frac{1}{2}\left ( I-A \right )$

  3. $\displaystyle I+A$

  4. $\displaystyle \frac{1}{2}\left ( I+A \right )$

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Correct Option: B
Explanation:

Given, $\displaystyle A^{2}-A+2I=0$


$\Rightarrow A^{2}A^{-1}-AA^{-1}+2IA^{-1}=0$

$\Rightarrow A-I+2A^{-1}=0$

$\Rightarrow 2A^{-1}=I-A$

$\Rightarrow A^{-1}=\displaystyle \frac{1}{2}(I-A)$