Tag: inverse of a matrix and linear equations

Quizzes (5) Questions (140)

Questions Related to inverse of a matrix and linear equations

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

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

  1. $16$
  2. $64$
  3. $\dfrac{1}{{16}}$

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $|(adjA)^{-1}|=\dfrac{4}{4^3}=\dfrac{1}{16}$

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

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

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$
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 of order $2$, then $det({A}^{-1})$ is equal to

  1. $det(A)$
  2. $\cfrac{1}{det(A)}$
  3. $1$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $|A^{-1}|=\dfrac{1}{|A|}$

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

  1. $C\ (A-B)=A^{-1}B$
  2. $C\ (A-B)=BA^{-1}$
  3. $(A-B)C=A^{-1}B$
  4. $All\ the\ above$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given (A-B)C = BA^-1, we can multiply both sides by A on the right to get (A-B)CA = B. Rearranging or manipulating the equation leads to C(A-B) = A^-1 B.

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

  1. $\displaystyle I-A$
  2. $\displaystyle \frac{1}{2}\left ( I-A \right )$
  3. $\displaystyle I+A$
  4. $\displaystyle \frac{1}{2}\left ( I+A \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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


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

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

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

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