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

. $\mathrm{If}$ $\mathrm{A}$ 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

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 $A = \begin{bmatrix} 2 & 3\ 5 & 1 \end{bmatrix},$ then find $A^{-1}$

  1. $\begin{bmatrix} - \frac {1}{13} & \frac {3}{13}\\ - \frac {5}{13} & - \frac {2}{13} \end{bmatrix}$
  2. $\begin{bmatrix} - \frac {1}{13} & \frac {3}{13}\\ \frac {5}{13} & \frac {2}{13} \end{bmatrix}$
  3. $\begin{bmatrix} - \frac {1}{13} & \frac {3}{13}\\ \frac {5}{13} & - \frac {2}{13} \end{bmatrix}$
  4. $\begin{bmatrix} \frac {1}{13} & \frac {3}{13}\\ \frac {5}{13} & - \frac {2}{13} \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$A=\begin{bmatrix} 2 & 3\\ 5 & 1\end{bmatrix}$

Cofactor matrix of $A=\begin{bmatrix} 1 & -5\\ -3 & 2\end{bmatrix}$

adj. $A=\begin{bmatrix} 1 & -5\\ -3 & 2\end{bmatrix}'=\begin{bmatrix} 1 & -3\\ -5 & 2\end{bmatrix}$

$|A|=2\times 1-(3\times 5)$

$=2-15=-13$

$A^{-1}=\dfrac{adj. A}{|A|}=\dfrac{-1}{13}\begin{bmatrix} 1 & -3\\ -5 & 2\end{bmatrix}$

$=\begin{bmatrix} \dfrac{-1}{13} & \dfrac{3}{13}\\ \dfrac{5}{13} & \dfrac{-2}{13}\end{bmatrix}$.
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 two non singular matrices of the same order such that ${ B }^{ r }=I$, for some positive integer $r>1$, then ${ A }^{ -1 }{ B }^{ r-1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A=$

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

${ A }^{ -1 }{ B }^{ r-1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A$

$=A^{-1}B^rB^{-1}A - A^{-1}B^{-1}A $
$={ A }^{ -1 }I{ B }^{ -1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A       (\because B^{r}=I)$
$={ A }^{ -1 }{ B }^{ -1 }{ A }-{ A }^{ -1 }{ B }^{ -1 }A$
$= O$

Hence, option 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 $\begin{pmatrix}1 & -tan  \theta\ tan  \theta & 1\end{pmatrix} \begin{pmatrix} 1 & tan  \theta\ - tan  \theta & 1\end{pmatrix}^{-1} = \begin{bmatrix} a& -b\ b & a\end{bmatrix}$, then

  1. $a = cos 2 \theta$
  2. $a = 1$
  3. $b = sin 2 \theta$
  4. $b = -1$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
we have 

$ \begin{pmatrix} 1 & tan  \theta\\ - tan  \theta & 1\end{pmatrix}^{-1} = \dfrac{1}{1+tan^2\theta} \begin{pmatrix}1 & -tan  \theta\\ tan  \theta & 1\end{pmatrix} $

$\therefore \begin{bmatrix} a& -b\\ b & a\end{bmatrix}=cos^2\theta \begin{pmatrix}1 & -tan  \theta\\ tan  \theta & 1\end{pmatrix} \begin{pmatrix}1 & -tan  \theta\\ tan  \theta & 1\end{pmatrix}$

We get 

$\begin{pmatrix}cos2\theta & -sin2\theta\\ sin2\theta & cos2\theta \end{pmatrix}$

$\therefore a=cos2\theta, b=sin2\theta$
Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

$A = \begin{bmatrix} 1& 0 & 0\0 &  1& 1\ 0 & -2 & 4\end{bmatrix}, I = \begin{bmatrix}1 & 0 & 0\ 0& 1 & 0\ 0 & 0 & 1\end{bmatrix}$ and $A^{-1} = \left [ \dfrac{1}{6} (A^2 + cA + dI) \right]$

The value of $(c,d)$ is

  1. $(-6, -11)$
  2. $(6, 11)$
  3. $(-6, 11)$
  4. $(6, -11)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $A = \begin{bmatrix} 1& 0 & 0\0 &  1& 1\ 0 & -2 & 4\end{bmatrix}$
The characteristic equation of $A$ is given by 
$|A-\lambda I|=0$
$\begin{vmatrix} 1-\lambda  & 0 & 0 \ 0 & 1-\lambda  & 1 \ 0 & -2 & 4-\lambda  \end{vmatrix}=0$

$\Rightarrow {\lambda}^{3}-6{\lambda}^{2}+11\lambda-6=0$
$\Rightarrow A^{3}-6A^{2}+11A-6=0$    ($\because$ Every square matrix satisfies its characteristic equation )
$\Rightarrow A^{2}-6A+11I-6A^{-1}=0$
$\Rightarrow A^{-1}=\displaystyle \frac{1}{6}(A^{2}-6A+11I)$

Comparing this with $A^{-1} = \left [ \frac{1}{6} (A^2 + cA + dI) \right]$, we get $c=-6, d=11$
$\therefore (c,d) = (-6,11)$

Hence, option 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

Two $n \times n$ square matrices $A$ and $B$ are said to be similar if there exists a non-singular matrix $P$ such that  $P^{-1}A: P=B$
If $A$ and $B$ are two non-singular matrices, then

  1. $A$ is similar to $B$
  2. $AB$ is similar to $BA$
  3. $AB$ is similar to $A^{-1}B$

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

$AB = (B^{-1}B)AB = B^{-1}(BA)B $

$\therefore$ $AB$ is similar to $BA$.

Hence, option 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

Two $n \times n$ square matrices $A$ and $B$ are said to be similar if there exists a non-singular matrix $P$ such that  $P^{-1}A: P=B$
If $A$ and $B$ are similar matrices such that $det :(A) =1$, then

  1. $det \: (B) = 1$
  2. $det\: (A)+det\: (B)=0$
  3. $det \: (B) = -1$

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

As $A$ and $B$ are similar matrices there exists a non-singular matrix $P$ such that

                                      $A=P^{-1}:BP$

$\Rightarrow det : (A) = det: (P^{-1}:BP)$

                $=det: (P^{-1}): det : (B) : det : (P)$

                 $\displaystyle =\frac{1}{det: (P)}det : (B) : (det P)$

                $= det : B$

Thus, $det : (A) = 0 \Leftrightarrow det : (B) = 0 : and : det : (A) =1 \Leftrightarrow  det : (B) = 1$

Hence, option 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

Two $n \times n$ square matrices $A$ and $B$ are said to be similar if there exists a non-singular matrix $P$ such that  $P^{-1}A: P=B$
If $A$ and $B$ are similar and $B$ and $C$ are similar, then

  1. $AB$ and $BC$ are similar
  2. $A$ and $C$ are similar
  3. $A + C$ and $B$ are similar

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

$A = P^{-1}:BP, B = Q^{-1} : C : Q$,

$\Rightarrow      A = P^{-1}(Q^{-1} : C : Q)P = (QP)^{-1}: C : QP$

Thus, $A$ is similar to $C$

Hence, option B.


Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

Which of the given values of $x$ and $y$ make the following pair of matrices equal.
$\displaystyle \begin{bmatrix} 3x+7 & 5 \ y+1 & 2-3x \end{bmatrix}=\begin{bmatrix} 0 & y-2 \ 8 & 4 \end{bmatrix}$

  1. $\displaystyle x=\frac { -1 }{ 3 } ,y=7$

  2. Not possible to find
  3. $\displaystyle y=7,x=\frac { -2 }{ 3 } $
  4. $\displaystyle x=\frac { -1 }{ 3 } ,y=\frac { -2 }{ 3 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle \begin{bmatrix} 3x+7 & 5

\ y+1 & 2-3x \end{bmatrix}=\begin{bmatrix} 0 & y-2 \ 8 & 4

\end{bmatrix}$
Equating the corresponding elements, we get,
$\displaystyle 3x+7=0\Rightarrow x=-\frac { 7 }{ 3 } $
$\displaystyle 5=y-2\Rightarrow y=7$
$\displaystyle y+1=8\Rightarrow y=7$
$\displaystyle 2-3x=4\Rightarrow x=-\frac { 2 }{ 3 } $
We find that on comparing the corresponding elements of the two matrices, we get two different values of $x$, which is not possible.
Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal.

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices


Let $A$ be the set of all $3 \times  3$ symmetric matrices all of whose entries are either $0$ or $1$. Five of these entries are $1$ and four of them are $0$.

The number of matrices $\mathrm{A}$ in $d$ for which the system of linear equations $\mathrm{A}\begin{bmatrix}\mathrm{X}\\mathrm{Y}\\mathrm{Z}\end{bmatrix}=\begin{bmatrix}1\0\0\end{bmatrix}$ has a unique solution, is

  1. less than 4

  2. at least 4 but less than 7

  3. atleast 7 but less than 10

  4. at least 10
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{bmatrix}0 & \mathrm{a} & \mathrm{b}\\mathrm{a} & 0 & \mathrm{c}\\mathrm{b} & \mathrm{c} & 1\end{bmatrix}$
either $\mathrm{b}=0$ or $\mathrm{c}=0\Rightarrow |\mathrm{A}|\neq 0$
$ \Rightarrow  2$ matrices
$\begin{bmatrix}0 & \mathrm{a} & \mathrm{b}\\mathrm{a} & 1 & \mathrm{c}\\mathrm{b} & \mathrm{c} & 0\end{bmatrix}$
either $\mathrm{a}=0$ or $\mathrm{c}=0\Rightarrow |\mathrm{A}|\neq 0 $
$\Rightarrow  2$ matrices
$\begin{bmatrix}1 & \mathrm{a} & \mathrm{b}\\mathrm{a} & \mathrm{o} & \mathrm{c}\\mathrm{b} & \mathrm{c} & \mathrm{o}\end{bmatrix}$
either $\mathrm{a}=0$ or $\mathrm{b}=0\Rightarrow |\mathrm{A}|\neq 0 $
$\Rightarrow  2$ matrices.
$\begin{bmatrix}1 & \mathrm{a} & \mathrm{b}\\mathrm{a} & 1 & \mathrm{c}\\mathrm{b} & \mathrm{c} & 1\end{bmatrix}$
If $\mathrm{a}=\mathrm{b}=0\Rightarrow |\mathrm{A}|=0$
If $\mathrm{a}=\mathrm{c}=0\Rightarrow |\mathrm{A}|=0$
If $\mathrm{b}=\mathrm{c}=0\Rightarrow |\mathrm{A}|=0$
$\Rightarrow $ there will be only 6 matrices.

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

Let $X=\begin{bmatrix} { x } _{ 1 } \ { x } _{ 2 } \ { x } _{ 3 } \end{bmatrix};A=\begin{bmatrix} 1 & -1 & 2 \ 2 & 0 & 1 \ 3 & 2 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 3 \ 1 \ 4 \end{bmatrix}$. If $AX=B$, then $X$ is equal to

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

$X=A^{-1}B$

$A^{-1}=\dfrac{1}{|A|}adj(A)$

Given,

$X=\begin{bmatrix}x _1\\ x _2\\ x _3\end{bmatrix}$

$A=\begin{bmatrix}1 &-1  &2 \\  2& 0 &1 \\  3& 2 &1 \end{bmatrix}$

$B=\begin{bmatrix}3\\ 1\\ 4\end{bmatrix}$

$|A|=\begin{vmatrix}1 &-1  &2 \\ 2 & 0 &1 \\  3&2  &1 \end{vmatrix}$

$=1(0-2)+1(2-3)+2(4-0)=-2-1+8=5$

$C _A=\begin{bmatrix}-2 &1  &4 \\  5& -5 &-5 \\  -1& 3 &2 \end{bmatrix}$

$adj(A)=C _A^T=\begin{bmatrix}-2 &5  &-1 \\  1& -5 &3 \\  4& -5 &2 \end{bmatrix}$

$A^{-1}=\dfrac{1}{5}\begin{bmatrix}-2 &5  &-1 \\  1& -5 &3 \\  4& -5 &2 \end{bmatrix}$

$X=A^{-1}B$

$=\dfrac{1}{5}\begin{bmatrix}-2 &5  &-1 \\  1& -5 &3 \\  4& -5 &2 \end{bmatrix}\begin{bmatrix}3\\ 1\\ 4\end{bmatrix}$

$=\dfrac{1}{5}\begin{bmatrix}-6+5-4\\ 3-5+12\\ 12-5+8\end{bmatrix}$

$=\dfrac{1}{5}\begin{bmatrix}-5\\ 10\\ 15\end{bmatrix}$

$=\begin{bmatrix} \left(\dfrac{-5}{5}\right)\\ \left(\dfrac{10}{5}\right)\\ \left(\dfrac{15}{5}\right)\end{bmatrix}$

$\begin{bmatrix}x _1\\ x _2\\ x _3\end{bmatrix}=\begin{bmatrix}-1\\ 2\\ 3\end{bmatrix}$
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

Which of the given values of $x$ and $y$ make the following pairs of matrices equal?
$\begin{bmatrix}3x + 7 & 5\ y + 1 & 2 - 3x\end{bmatrix}$ and $\begin{bmatrix} 0&y - 2 \ 8 & 4\end{bmatrix}$

  1. $x = -\dfrac {1}{3}, y = 7$
  2. $y = 7, x = -\dfrac {2}{3}$
  3. $x = -\dfrac {1}{3}, 4 = -\dfrac {2}{5}$

  4. Not possible to find
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If the given matrices are equal then we have the following equations,

$3x+7=0$
Or, $ x=-\dfrac{7}{3}.$......(1).

And 
$5=y-2$
Or, $ y=7$.......(2)

And
$y+1=8$
Or, $y=7$.......(3)

And
$ 2-3x=4$
Or, $x=-\dfrac{2}{3}.$....(4)

From (1) and (4) we get contradicting result i.e. why $x$ and $y$ can't be found simultaneously.
So not possible to find $x$ and $y$.

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

Suppose $a _1, :a _2,: ... $ are real numbers, with $a _1\neq 0$. If $a _1, :a _2,:a _3,:...$ are in A.P.  Then,

  1. $A=\begin{bmatrix}a _1&a _2 &a _3 \\a _4 &a _5 &a _6 \\a _5 &a _6 &a _7 \end{bmatrix}$ is singular
  2. the system of equations $a _1x+a _2y+a _3z=0, \: a _4x+a _5y+a _6z=0,\:a _7x+a _8y+a _9z=0$ has infinite number of solutions
  3. $B=\begin{bmatrix}a _1&ia _2 \\ ia _2 & a _1\end{bmatrix}$ is non singular

  4. none of these
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation
Given ${ a } _{ 1 },{ a } _{ 2 }...$are in AP
${ a } _{ 1 }\neq 0$
Option A$=\begin{bmatrix} { a } _{ 1 } & { a } _{ 2 } & { a } _{ 3 } \\ { a } _{ 4 } & { a } _{ 5 } & { a } _{ 6 } \\ { a } _{ 7 } & { a } _{ 8 } & { a } _{ 9 } \end{bmatrix}$
$=\begin{bmatrix} { a } _{ 1 } & { a } _{ 1 }+d & { a } _{ 1 }+2d \\ { a } _{ 1 }+3d & { a } _{ 1 }+4d & { a } _{ 1 }+5d \\ { a } _{ 1 }+4d & { a } _{ 1 }+5d & { a } _{ 1 }+6d \end{bmatrix}\quad { C } _{ 1 }\rightarrow { C } _{ 2 }-{ C } _{ 2 }$
$=\begin{bmatrix} d & { a } _{ 1 }+d & { a } _{ 1 }+2d \\ d & { a } _{ 1 }+4d & { a } _{ 1 }+5d \\ d & { a } _{ 1 }+5d & { a } _{ 1 }+6d \end{bmatrix}{ C } _{ 2 }\rightarrow { C } _{ 3 }-{ C } _{ 2 }$
$=\begin{bmatrix} d & d & { a } _{ 1 }+2d \\ d & d & { a } _{ 1 }+5d \\ d & d & { a } _{ 1 }+6d \end{bmatrix}$
${ C } _{ 1 }\& { C } _{ 2 }$ are same $\Rightarrow \triangle =0$
The matrix is singular-correct
Option B - co efficients matrix$=\begin{bmatrix} { a } _{ 1 } & { a } _{ 2 } & { a } _{ 3 } \\ { a } _{ 4 } & { a } _{ 5 } & { a } _{ 6 } \\ { a } _{ 7 } & { a } _{ 8 } & { a } _{ 9 } \end{bmatrix}$
They are in AP
$\Rightarrow \triangle =0$
$\therefore $They have infinite solutions
Option B is also correct
Option C $B=\begin{bmatrix} { a } _{ 1 } & i{ a } _{ 2 } \\ i{ a } _{ 2 } & { a } _{ 1 } \end{bmatrix}$
$\Rightarrow \triangle ={ a } _{ 1 }^{ 2 }-\left( { i }^{ 2 }{ a } _{ 2 }^{ 2 } \right) $
$={ a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 }$
$={ a } _{ 1 }^{ 2 }+{ \left( { a } _{ 1 }+d \right)  }^{ 2 }$
$=2{ a } _{ 1 }^{ 2 }+{ d }^{ 2 }+2{ a } _{ 2 }d>0$
Non singular
Option A,B, C are true