Mathematics

Linear Algebra

449 Questions

Linear algebra involves the study of matrices, vectors, and linear transformations. Common topics include finding the rank of a matrix, calculating eigenvalues and eigenvectors, and performing LU decomposition. These advanced mathematical concepts are frequently tested in engineering, statistics, and civil service examinations.

Matrix rank calculationLU decompositionEigenvalues and eigenvectorsMatrix invertibilityDeterminant properties

Linear Algebra Questions

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


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

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

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

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

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 p be a nonsingular matrix, and $I + p + p^2 + ..... + p^n = 0$, then find $p^{-1}$.

  1. $I$
  2. $p^{n+1}$
  3. $p^n$
  4. $\left( p^{n+1} - I\right) \left( p-I\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have $I + p + p^2 + ..... + p^n = O$  ----------$(1)$
Since p is a nonsingular matrix, p is invertible.
Multiplying both sides of (1) by $p^{-1}$, we get
$p^{-1} + I + Ip + ..... + p^{n - 1} I = O. p^{-1}$
or $ p^{-1} + I (1 + p + ...... + p^{n-1}) = O$
or $ p^{-1} = - I (I + p + p^2 + ..... + p^{n - 1}) = - I(-p^n) = p^n$

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 satisfy $AB = B^{-1}$, where $ B\quad =\quad \begin{bmatrix} 2 & -1 \ 2 & 0 \end{bmatrix}$, then find without finding $A^{-1}$, the matrix X satisfying $A^{-1}XA = ?$

  1. $B$
  2. $B^2$
  3. $A$

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

Given, $A^{-1} XA = B$


$AA^{-1} XA = AB$

$IXA =AB$

$XAB =AB^2$ 

$XAB =I$ since $\left[AB=B^{-1}\Rightarrow AB^2=I\right]$

$XAB^2 = B$

$XI = B$

$\therefore X = B$

Hence option $'A'$ is the answer.

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 $A$ and $B$ are symmetric matrices and $AB=BA$, then ${ A }^{ -1 }B$ is a

  1. Symmetric matrix

  2. Skew-symmetric matrix

  3. Identity matrix

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

We have $AB=BA=B'A'=\left( AB \right) '$

$\Rightarrow AB$ is symmetric.
Also, $AB{ A }^{ -1 }=BA{ A }^{ -1 }=B\quad \quad \left( \because AB=BA \right) $
$\Rightarrow { A }^{ -1 }AB{ A }^{ -1 }={ A }^{ -1 }B\Rightarrow B{ A }^{ -1 }={ A }^{ -1 }B$
Therefore, $\left( { A }^{ -1 }B \right) '=\left( B{ A }^{ -1 } \right) =\left( { A }^{ -1 } \right) 'B'=A'B$   $[\because { A }^{ -1 }$ and $B$ are symmetric $]$
Thus, the matrix ${ A }^{ -1 }B$ is symmetric.