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 applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Let $A$ be a matrix of order $3\times 3$ such that $\left| \vec { A }  \right| =1$. Let $B=2{ A }^{ -1 }$ and $C=\dfrac { adj.A }{ 2 }$. Then the value of  $\left| { AB }^{ 2 }{ C }^{ 3 } \right|$, is ( where $\left| A \right|$ represent det. $A$)

  1. $1$
  2. $\dfrac { 1}{ 2 }$
  3. $8$
  4. $64$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$adj(A)=|\vec{A}|^{n-1}$

$n=3,\therefore adj(A)=|\vec{A}|^{3-1}=|\vec{A}|^2$

$|\vec{A}|=1,\therefore adj(A)=1$

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

$n=3,\therefore adj(A)=|\vec{A}|^{3-1}=|\vec{A}|^2$

$|\vec{A}|=1,\therefore adj(A)=1$

$|\vec{A}|=1$

$B=2|\vec{A}|=1\Rightarrow B=2\times 1=2$

$C=\dfrac{adjA}{2}=\dfrac{1}{2}$

$|AB^2C^3|=\left | 1\times 2^2\times \left ( \dfrac{1}{2} \right )^3 \right |$

$=\dfrac{1}{2}$
Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

 $\begin{bmatrix}
              \cos\theta & -\sin\theta \[0.3em]
              \sin\theta & \cos\theta
              \end{bmatrix} = \begin{bmatrix}
              1 & -\tan\theta/2 \[0.3em]
             \tan\theta/2 & 1
              \end{bmatrix} \begin{bmatrix}
              1  & \tan\theta/2 \[0.3em]
              -\tan\theta/2 & 1
              \end{bmatrix}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a standard trigonometric identity related to rotation matrices and the tangent half-angle substitution. The product of the two matrices on the right side equals the rotation matrix on the left.

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A = \begin{bmatrix} a & b\ c  & d \end{bmatrix} $ satisfies the equation $x^2 - (a+d)x+k=0$ then

  1. $k = bc$
  2. $ k =ad$
  3. $k = a^2+b^2+c^2+d^2$
  4. $k=ad-bc$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A=\begin{bmatrix}a&b\c&d\end{bmatrix}$


$\implies \begin{vmatrix}\lambda-a&b\c&\lambda-d\end{vmatrix}=0$

$(\lambda-a)(\lambda-d)-bc=0$

$\implies {\lambda}^{2}-(a+d)\lambda+(a{d}-b{c})=0$

Comparing with $x^2 - (a+d)x+k=0$

$\implies k=a{d}-b{c}$

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

The number of $2\times 2$ matrices $A=\left[ \begin{matrix} a & b \ c & d \end{matrix} \right] $ for which ${ \left[ \begin{matrix} a & b \ c & d \end{matrix} \right]  }^{ -1 }$ $=\left[ \begin{matrix} \frac { 1 }{ a }  & \frac { 1 }{ b }  \ \frac { 1 }{ c }  & \frac { 1 }{ d }  \end{matrix} \right] $, $(a,b,c,d\ \epsilon \ R)$ is

  1. $0$
  2. $1$
  3. $2$
  4. $Infinite$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If $A=\begin{bmatrix} a & b \ c & d \end{bmatrix}\quad { \begin{bmatrix} a & b \ c & d \end{bmatrix} }^{ -1 }=\begin{bmatrix} 1/a & 1/b \ 1/c & 1/d \end{bmatrix}$                          Not possible for any values of $a, b, c$

$A=\begin{bmatrix} a & b \ c & d \end{bmatrix}\quad =Adj{ \begin{bmatrix} d & -c \ -b & a \end{bmatrix} }\ \Rightarrow { A }^{ -1 }=\begin{bmatrix} d & -c \ -b & a \end{bmatrix}$

Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A$ and $B$ are square matrices such that $B=-A^{-1}BA$, then 

  1. $AB+BA=0$
  2. $(A+B)^{o}=A^{2}+B^{2}$
  3. $(A+B)^{2}=A^{2}+2AB+B^{2}$
  4. $(A+B)^{2}=A+B$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given B = -A^-1BA, multiply both sides by A on the right to get BA = -A^-1BA^2, or simply rearrange to AB = -BA. Adding BA to both sides yields AB + BA = 0.

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A=\begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$, find the values of $\alpha$ for which $A^2=B$.

  1. $\pm 1$
  2. $4$
  3. $0$

  4. No value
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,

$A^2=B$
$\begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 \ 1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\begin{bmatrix} \alpha^2 +0 & 0+0 \ \alpha +1 & 0+1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\begin{bmatrix} \alpha^2 & 0 \ \alpha +1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \ 5 & 1 \end{bmatrix}$

$\alpha^2=1$ and $\alpha +1=5$
$\alpha =\pm 1$ and $\alpha =4$, which is not possible.
Hence, there is no value of $\alpha$ for which $A^2=B$ is true.

Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A=\left[ \begin{matrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{matrix} \right] $ and $10B=\left[ \begin{matrix} 4 & 2 & 2 \ -5 & 0 & \alpha  \ 1 & -2 & 3 \end{matrix} \right] $ where $B=A^{-1}$ then $\alpha$ is equal to-

  1. $2$
  2. $-1$
  3. $-2$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since B = A^-1, we have AB = I. Multiplying the first row of A by the second column of B must equal 0. Calculating this gives 1(2) + (-1)(0) + 1(-2) = 0, which is consistent. Solving for the element at (2,3) of the product AB = I yields alpha = 2.

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Consider three matrices $A=\begin{bmatrix} 2 & 1 \ 4 & 1 \end{bmatrix}, B=\begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix}$ and $C=\begin{bmatrix} 3 & -4 \ -2 & 3 \end{bmatrix}$. Then the value of the sum $tr(A)+tr\left(\dfrac{ABC}{2}\right)+tr\left(\dfrac{A(BC)^{2}}{4}\right)+tr\left(\dfrac{A(BC)^{3}}{8}\right)+....+\infty$ is 

  1. $6$
  2. $9$
  3. $12$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $I=I=\left[ \begin{matrix} 1 \ 0 \end{matrix}\begin{matrix} 0 \ 1 \end{matrix} \right] ,j=\left[ \begin{matrix} 0 \ -1 \end{matrix}\begin{matrix} 1 \ 0 \end{matrix} \right] and B=\left[ \begin{matrix} cos\theta  \ -sin\theta  \end{matrix}\begin{matrix} sin\theta  \ cos\theta  \end{matrix} \right] ,$ then B =

  1. $Icos\theta +Jsin\theta $
  2. $Icos\theta -Jsin\theta $
  3. $Isin\theta +Jcos\theta $
  4. $-Icos\theta +Jsin\theta $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $I=\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}, J=\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$

and $B=\begin{bmatrix} \cos \theta &\sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$

$=\cos\theta\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} +\sin\theta \begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$

$=I\cos\theta +J\sin\theta$.
Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A(\theta) = \begin{bmatrix}\sin  \theta & i  \cos  \theta\ i  \cos  \theta & \sin  \theta\end{bmatrix}$, then which of the following is not true?

  1. $A(\theta)^{-1} = A(\pi - \theta)$
  2. $A(\theta) + A(\pi + \theta)$ is a null matrix
  3. $A(\theta)$ is invertible for all $\theta \in R$
  4. $A(\theta)^{-1} = A(- \theta)$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Finding inverse of the matrix $A(\theta)= \begin{bmatrix} \sin\theta & i\cos\theta \ i\cos\theta & \sin\theta\end{bmatrix}$


Determinant of $A(\theta)$ is $|A(\theta)|=\sin^2\theta-i^2\cos^2\theta$
                                                    $= \sin^2\theta+\cos^2\theta$
                                                    $=  1$

Therefore $A(\theta)$ is a non-singular matrix. So , it is invertible of all $\theta \in R$

$A(\theta)^{-1} = \begin{bmatrix} \sin\theta & -i\cos\theta \-i\cos\theta & \sin\theta \end{bmatrix}$

Now. $A(\pi -\theta)=\begin{bmatrix} \sin(\pi-\theta) & i\cos(\pi-\theta) \i\cos(\pi-\theta) & \sin(\pi-\theta) \end{bmatrix}$
                          $=\begin{bmatrix} \sin\theta  & -i\cos\theta \ -i\cos\theta & \sin\theta \end{bmatrix}$
                          $= A(\theta)^{-1}$

Now, $A(\pi+\theta)= \begin{bmatrix} \sin(\pi+\theta) & i\cos(\pi+\theta) \i\cos(\pi+\theta) & \sin(\pi+\theta) \end{bmatrix} $
                          $= \begin{bmatrix} -\sin\theta & -i\cos\theta \-i\cos\theta & -\sin\theta \end{bmatrix}$
                          $= -A(\theta)$

Therefore, $A(\theta) + A(\pi+\theta)=0$.

Hence, the correct options are $(A), (B)$ and $(C)$.

Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Let p be a non-singular matrix, $1+p+p^{2}+....+p^{n}=0$ (0 denotes the null matrix) then $p^{-1}=$

  1. $p^{n}$
  2. -$p^{n}$
  3. -(1+p+...+$p^{n}$)

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

Given I + P + P^2 + ... + P^n = 0, we have P + P^2 + ... + P^n = -I. Factoring out P gives P(I + P + ... + P^(n-1)) = -I. This is a standard identity for geometric series of matrices.

Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Let A be a $3 \times 3$  matrix such that is: $A\left[ \begin{matrix} 1 & 2 & 3 \ 0 & 2 & 3 \ 0 & 1 & 1 \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{matrix} \right]  $Then $A^{-1}$ is

  1. $\left[ \begin{matrix} 0 & 1 & 3 \\ 0 & 2 & 3 \\ 1 & 1 & 1 \end{matrix} \right] $
  2. $\left[ \begin{matrix} 3 & 2 & 1 \\ 3 & 2 & 0 \\ 1 & 1 & 0 \end{matrix} \right] $
  3. $\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 2 & 3 \end{matrix} \right] $
  4. $\left[ \begin{matrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{matrix} \right] $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If AX = B, then A^-1 = XB^-1. Here X is the matrix on the left and B is the matrix on the right. Calculating B^-1 and multiplying by X yields the inverse.

Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A\begin{bmatrix} 1 & 1\ 2 & 0\end{bmatrix}=\begin{bmatrix} 3 & 2\ 1 & 1\end{bmatrix}$, then $A^{-1}$ is given by?

  1. $\begin{bmatrix} 0 & -1\\ 2 & -4\end{bmatrix}$
  2. $\begin{bmatrix} 0 & -1\\ -2 & -4\end{bmatrix}$
  3. $\begin{bmatrix} 0 & 1\\ 2 & -4\end{bmatrix}$

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

Let X = [[1, 1], [2, 0]] and B = [[3, 2], [1, 1]]. We have AX = B, so A = BX^-1. Then A^-1 = (BX^-1)^-1 = XB^-1. Calculating B^-1 and multiplying by X gives the result.