Mathematics

Linear Algebra

510 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 $A$ be a $3\times 3$ matrix and $I$ be the unit matrix of that order such that $\displaystyle A=A^{2}+I$ then $A^{-1}$ is equal to

  1. $A$
  2. $A+I$
  3. $I-A$
  4. $A-I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given : $\displaystyle A=A^{2}+I$

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

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

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

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

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

If $A$ is a square matrix, $B$ is a singular matrix of same order, then for a positive integer $n,(A^{-1}BA)^n$ equals

  1. $A^{-n}B^nA^n$
  2. $A^nB^nA^{-n}$
  3. $A^{-1}B^nA$
  4. $n(A^{-1}BA)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider $n=2$


${ \left( { A }^{ -1 }BA \right)  }^{ 2 }=\left( { A }^{ -1 }BA \right) \left( { A }^{ -1 }BA \right) =\left( { A }^{ -1 }B \right) \left( { A }^{ -1 }A \right) \left( BA \right) ={ A }^{ -1 }{ B }^{ 2 }A$

Again for $n=3$ we have ${ \left( { A }^{ -1 }BA \right)  }^{ 3 }=\left( { A }^{ -1 }{ B }^{ 2 }A \right) \left( { A }^{ -1 }BA \right) ={ A }^{ -1 }{ B }^{ 3 }A$

Thus generalizing the case 

${ \left( { A }^{ -1 }BA \right)  }^{ n }={ A }^{ -1 }{ B }^{ n }A$

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 scalar matrix with scalar $k \neq 0$, of order $3$, then $kA^{-1}$ is:

  1. $\dfrac{1}{k}I$
  2. $\dfrac{1}{k^2}I$
  3. ${k^2}I$
  4. $\dfrac{1}{k^3}I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
It is well-known that if you find an inverse for a matrix, that inverse matrix will be unique.
So what we have to do is to show that $\left(k{A}^{−1}\right)⋅\left(kA\right)=Id=\left(kA\right)⋅\left(k{A}^{−1}\right)$.
We have $\left(k{A}^{−1}\right).\left(kA\right)=\left(kk\right)⋅\left({A}^{−1}.A\right)=Id=Id$ and $\left(kA\right)⋅\left(k{A}^{−1}\right)=\left(kk\right)⋅\left(A{A}^{−1}\right)=Id$.
So by the uniqueness of the inverse matrix we have that $k{A}^{−1}$ is the inverse of the matrix $kA$.
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 two non-zero square matrices of the same order such that the product $AB=0$, then

  1. both A and B must be singular

  2. exactly one of them must be singular

  3. atleast one of them must be non-singular

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

Assume that $A$ is non-singular, then $A^{-1}$ exists. Thus
$AB =0    \Rightarrow A^{-1}(AB) =(A^{-1}:A)B = 0$
$ \Rightarrow   IB =0$
$\therefore  B =0$. A contradiction.
$\Rightarrow$ A is singular, similarly B is also singular.
Hence, both A and B must be singular.


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 $A=\begin{bmatrix} 1&0 \1 &1 \end{bmatrix}$ then

  1. $A^{-n}=\begin{bmatrix} 1&0 \\-n &1 \end{bmatrix}\forall \: n\: \in\: N$.
  2. $\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n}A^{-n}=\begin{bmatrix} 0&0 \\-1 &0 \end{bmatrix}$
  3. $\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n^2}A^{-n}=\begin{bmatrix} 0&0 \\0 &0 \end{bmatrix}$

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

$A^{-1}=\begin{bmatrix}1 &0 \-1 &1 \end{bmatrix}$
$A^2=\begin{bmatrix}1 &0 \1 &1 \end{bmatrix}\begin{bmatrix}1 &0 \1 &1 \end{bmatrix}=\begin{bmatrix}1 &0 \2 &1 \end{bmatrix}$
$A^{-2}=\begin{bmatrix}1 &0 \-2 &1 \end{bmatrix}$
$\Rightarrow A^{-n}=\begin{bmatrix}1 &0 \-n &1 \end{bmatrix}$
$\displaystyle \frac{1}{n} A^{-n}=\begin{bmatrix}1/n &0 \-1 &1/n \end{bmatrix}$
$\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n}A^{-n}=\begin{bmatrix} 0&0 \-1 &0 \end{bmatrix}$
and $\displaystyle \frac{1}{n^2} A^{-n}=\begin{bmatrix}1/n^2 &0 \-1/n &1/n ^2\end{bmatrix}$
$\displaystyle \lim _{n\rightarrow \infty }\displaystyle \frac{1}{n^2}A^{-n}=\begin{bmatrix} 0&0 \0 &0 \end{bmatrix}$
Hence, options 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 $A$ and $B$ are $3\times 3$ matrices and $|A|\neq 0$, then

  1. $|AB|=0\Rightarrow |B|=0$
  2. $|AB|\neq 0\Rightarrow |B|\neq 0$
  3. $|A^{-1}|=|A|^{-1}$
  4. $|2A|=2|A|$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

$A$ and $B$ are $3\times 3$ matrices and $|A|\neq 0$
1.$|AB|=|A||B|=0$ $\Rightarrow |B|=0$
2.$|AB|=|A||B|\neq 0$ $\Rightarrow |B|\neq 0$
3.$AA^{-1}=I$ $\Rightarrow |A||A^{-1}|=1$
$\therefore |A^{-1}|=|A|^{-1}$
4.$|2A|= 8|A|$    ($\because  |kA|=k^n|A|$)
Hence, options 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

Let the matrix A and B be defined as $A =\begin{bmatrix}3 &2 \ 2 &1 \end{bmatrix}$ and $B= \begin{bmatrix}3 &1 \ 7 &3 \end{bmatrix}$ then the value of Det.$(2A^9B^{-1})$, is 

  1. $2$
  2. $1$
  3. $-1$
  4. $-2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A =\begin{bmatrix}3 &2 \ 2 &1 \end{bmatrix}$ and $B= \begin{bmatrix}3 &1 \ 7 &3 \end{bmatrix}$

$|A| = \begin{vmatrix} 3 & 2 \ 2 & 1 \end{vmatrix} = -1$

$|B| = \begin{vmatrix} 3 & 1 \ 7 & 3 \end{vmatrix} = 2$

$\displaystyle |2A^9 B^{-1}| = 2^2|A|^9\frac{1}{|B|}$

                     $\displaystyle= 4\times (-1)\times \frac{1}{2}$

$\therefore |2A^9 B^{-1}|=-2$

Hence, option D.

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 $P$ is a two-rowed matrix satisfying $P^T = P^{-1}$, then $P$ can be

  1. $\begin{bmatrix}cos\, \theta & -sin\, \theta \\ -sin\,\theta & cos\, \theta \end{bmatrix}$
  2. $\begin{bmatrix}cos\, \theta & sin\, \theta \\ -sin\,\theta & cos\, \theta \end{bmatrix}$
  3. $\begin{bmatrix}-cos\, \theta & sin\, \theta \\ sin\,\theta & -cos\, \theta \end{bmatrix}$

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

$A=\begin{bmatrix} cos\theta  & -sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},{ A }^{ T }=\begin{bmatrix} cos\theta  & -sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},$

${ A }^{ -1 }=\cfrac { 1 }{ { cos }^{ 2 }\theta -{ sin }^{ 2 }\theta  } \begin{bmatrix} cos\theta  & sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}\ B=\begin{bmatrix} cos\theta  & sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix},{ B }^{ T }=\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix},$
${ B }^{ -1 }=\cfrac { 1 }{ { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta  } \begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}=\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}\ C=\begin{bmatrix} -cos\theta  & sin\theta  \ sin\theta  & -cos\theta  \end{bmatrix},{ C }^{ T }=\begin{bmatrix} -cos\theta  & sin\theta  \ sin\theta  & -cos\theta  \end{bmatrix},$
${ C }^{ -1 }=\cfrac { 1 }{ { cos }^{ 2 }\theta -{ sin }^{ 2 }\theta  } \begin{bmatrix} -cos\theta  & sin\theta  \ sin\theta  & -cos\theta  \end{bmatrix}$
hence $P=B=\begin{bmatrix} cos\theta  & sin\theta  \ -sin\theta  & cos\theta  \end{bmatrix}$

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 and B are invertible matrices, which one of the following statement is/are correct 

  1. $Adj(A) = |A|A^{-1}$
  2. $det(A^{-1}) =|det(A)|^{-1}$
  3. $(A + B)^{-1}= B^{-1 }+ A^{-1}$
  4. $(AB)^{-1} = B^{-1}A^{-1}$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation
Option A
${ A }^{ -1 }=\cfrac { AdjA }{ \left| A \right|  } $
$\Rightarrow AdjA=\left| A \right| { A }^{ -1 }$
Option A is true

Option B
$det\left( AB \right) =\left( detA \right) \left( detB \right) $
$\Rightarrow A{ A }^{ -1 }=I$
$det\left( A{ A }^{ -1 } \right) =detI$
$\Rightarrow detA\left( det{ A }^{ -1 } \right) =1$
$\Rightarrow det{ A }^{ -1 }={ \left( detA \right)  }^{ -1 }$
Option B is true

Option C
${ \left( A+B \right)  }^{ -1 }={ A }^{ -1 }+{ B }^{ -1 }$
Option C is true

Option D
${ \left( AB \right)  }^{ -1 }=?$
$AB\left( { B }^{ -1 }{ A }^{ -1 } \right) =A\left( B{ B }^{ -1 } \right) { A }^{ -1 }$
$=AI{ A }^{ -1 }=\left( A{ A }^{ -1 } \right) =I$
$\Rightarrow { B }^{ -1 }{ A }^{ -1 }={ \left( AB \right)  }^{ -1 }$
Option D is true
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 $A=\begin{bmatrix} 1 & -2 \ 3 & 0 \end{bmatrix}$, $B=\begin{bmatrix} -1 & 4 \ 2 & 3 \end{bmatrix}$, and $ABC=\begin{bmatrix} 4 & 8 \ 3 & 7 \end{bmatrix}$, then $C$ equals

  1. $\cfrac { 1 }{ 66 } \begin{bmatrix} 54 & 110 \\ 3 & 11 \end{bmatrix}$
  2. $\cfrac { 1 }{ 66 } \begin{bmatrix} -54 & -110 \\ 3 & 11 \end{bmatrix}$
  3. $\cfrac { 1 }{ 66 } \begin{bmatrix} -54 & 110 \\ 3 & -11 \end{bmatrix}$

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

Given ABC = D, then C = (AB)^-1 * D. Calculate AB first, then find its inverse, and multiply by the matrix D.

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

The value of $(\mathrm{A}$dj $\mathrm{A})^{-1}$ is equal to 

  1. $\mathrm{A}$dj $(\mathrm{A}^{-1})$
  2. $\mathrm{A}$dj $[-\mathrm{A}]$
  3. $(\mathrm{A}$dj$\mathrm{A})^{\mathrm{T}}$
  4. $\mathrm{A}$dj $(\mathrm{A}^{\mathrm{T}})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The property (Adj A)^-1 = Adj(A^-1) is a standard identity in matrix algebra.

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

. $\mathrm{If}$ $\mathrm{A}$ is non-singular matrix such that $A^{2}=A^{-1}$ then $adjA=$ 

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

$A^{2}=A^{-1}$

$A.A^{2}=A.A^{-1}$
$A^{3}=I$
$detA.A^{3}=detA.I$
$detA.A^{3}=A.adjA$
$detA.A^{2}=adjA$
Therefore 
$adjA=A^{-1}$.

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

Let A and B be two non-singular matrices which commute. The $A^{-1}$, $B^{-1}$

  1. do not commute

  2. commute
  3. $AB = A^{-1}B^{-1}$
  4. $(AB)^{-1}=AB$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$A^{-1}B^{-1}=(BA)^{-1}=(AB)^{-1}$ since $A,B$ commute
$\Rightarrow A^{-1}B^{-1}=B^{-1}A^{-1}$
Hence $A^{-1}, B^{-1}$ also commute