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

If A and B are any $2\times2$ matrices, then det. (A+B) =0 implies

  1. None of these

  2. det A=0 and det B=0

  3. det A=0 or det B=0

  4. det A=0 + det B=0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The determinant of a sum of matrices is not generally related to the sum of their determinants or the individual determinants being zero.

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

Let $\begin{bmatrix} 1 & 1\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 3\ 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & n-1\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\ 0 & 1\end{bmatrix}$
If $A=\begin{bmatrix} 1 & n\ 0 & 1\end{bmatrix}$ then $A^{-1}=?$

  1. $\begin{bmatrix} 1 & 12\\ 0 & 1\end{bmatrix}$
  2. $\begin{bmatrix} 1 & -13\\ 0 & 1\end{bmatrix}$
  3. $\begin{bmatrix} 1 & -12\\ 0 & 1\end{bmatrix}$
  4. $\begin{bmatrix} 1 & 0\\ -13 & 1\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{bmatrix} 1 & 1\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 3\ 0 & 1\end{bmatrix}..\begin{bmatrix} 1 & n-1\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 78\ 0 & 1\end{bmatrix}$
$\Rightarrow \dfrac{n(n-1)}{2}=78\Rightarrow n=13$
$A=\begin{bmatrix} 1 & 13\ 0 & 1\end{bmatrix}$
so $A^{-1}=\begin{bmatrix} 1 & -13\ 0 & 1\end{bmatrix}$.

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

If $\displaystyle A=\begin{bmatrix} 0 & 0 & 1\ 0 & 1&0 \ 1& 0 & 0\end{bmatrix}$, then $A^{-1}$ is.

  1. $-A$
  2. $A$
  3. $1$

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

We have, $A=\begin{bmatrix} 0 & 0&1\ 0 &1 &0\ 1&0 &0\end{bmatrix}$
$\Rightarrow |A|=0(0-0)-0(0-0)+1(0-1)$
$\Rightarrow |A|=-1$
and cofactors of A are
$A _{11}=0, A _{12}=0, A _{13}=-1,$
$A _{21}=0, A _{22}=-1, A _{23}=0,$
$A _{31}=-1, A _{32}=0, A _{33}=0$
$\therefore A^{-1}=\displaystyle\frac{adj(A)}{|A|}$
$=-\displaystyle\frac{1}{1}\begin{bmatrix} 0 & 0 & -1\0 & -1 &0\ -1 &0 &0\end{bmatrix}$

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

If $\begin{bmatrix} 1 & 2 \ 3 & -5 \end{bmatrix}$, then ${A}^{-1}$ is equal to

  1. $\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$
  2. $\begin{bmatrix} -\cfrac { 5 }{ 11 } & -\cfrac { 2 }{ 11 } \\ -\cfrac { 3 }{ 11 } & -\cfrac { 1 }{ 11 } \end{bmatrix}$
  3. $\begin{bmatrix} \cfrac { 5 }{ 11 } & \cfrac { 2 }{ 11 } \\ \cfrac { 3 }{ 11 } & \cfrac { 1 }{ 11 } \end{bmatrix}$
  4. $\begin{bmatrix} 5 & 2 \\ 3 & -1 \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since $A=\begin{bmatrix} 1 & 2 \ 3 & -5 \end{bmatrix}$


$\therefore \left| A \right| =\begin{bmatrix} 1 & 2 \ 3 & -5 \end{bmatrix}=-5-6=-11$

and $adj(A)=\begin{bmatrix} -5 & -2 \ -3 & 1 \end{bmatrix}$

$\therefore { A }^{ -1 }=\cfrac { 1 }{ \left| A \right|  } adj(A)$

$=-\cfrac { 1 }{ 11 } \begin{bmatrix} -5 & -2 \ -3 & 1 \end{bmatrix}=\cfrac { 1 }{ 11 } \begin{bmatrix} 5 & 2 \ 3 & -1 \end{bmatrix}$

$\quad =\begin{bmatrix} \cfrac { 5 }{ 11 }  & \cfrac { 2 }{ 11 }  \ \cfrac { 3 }{ 11 }  & -\cfrac { 1 }{ 11 }  \end{bmatrix}$

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

If you switch the first row with the fourth row, what will the new first row be?
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$

  1. $3, 4, 2, 11$
  2. $9, 1, 0, 0$
  3. $0, 1, 0, 2$
  4. $0, 0, 6, 1$
  5. $0, 2, 0, 3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given Matrix$=\begin{bmatrix} 3 & 4 & 2 & 11 \\ 9 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 6 & 1 \end{bmatrix}$
First and fourth row are interchanged
New matrix obtained $=\begin{bmatrix} 0 & 0 & 6 & 1 \\ 9 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 3 & 4 & 2 & 1 \end{bmatrix}$
New first row$=\begin{bmatrix} 0 & 0 & 6 & 1 \end{bmatrix}$ Option D
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Which of the following is the new row that results when you add rows $1$ and $3$?
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$

  1. $6, 8, 4, 22$
  2. $3, 5, 2, 13$
  3. $3, 4, 2, 11$
  4. $3, 4, 8, 12$
  5. $4, 5, 3, 12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given Matrix$=\begin{bmatrix} 3 & 4 & 2 & 11 \\ 9 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 6 & 1 \end{bmatrix}$
Row $1=\begin{bmatrix} 3 & 4 & 2 & 11 \end{bmatrix}$
Row $3=\begin{bmatrix} 0 & 1 & 0 & 2 \end{bmatrix}$
Sum$=\begin{bmatrix} 3+0 & 4+1 & 2+0 & 11+2 \end{bmatrix}$
$\begin{bmatrix} 3 & 5 & 2 & 13 \end{bmatrix}$
$\therefore $Option $2$ is correct
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

$A=\begin{bmatrix} 1&-2&3\7&-8&9\4&-5&6\end{bmatrix}$ the new matrix formed by adding $\ 2^{nd}\ row \ to \ 1^{st} $ row  will be

  1. $\begin{bmatrix}8&-10&12\\7&-8&9\\4&-5&6\end{bmatrix}$
  2. $\begin{bmatrix} 6&6&6\\7&8&9\\4&5&6\end{bmatrix}$
  3. $\begin{bmatrix} 1&2&3\\7&8&9\\11&-13&14\end{bmatrix}$
  4. $\begin{bmatrix} 1&-2&3\\7&8&-29\\4&-2&6\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $A=\begin{bmatrix} 1 & -2 & 3 \\ 7 & -8 & 9 \\ 4 & -5 & 6 \end{bmatrix}$
Now$\Rightarrow { R } _{ 1 }\rightarrow { R } _{ 1 }+{ R } _{ 2 }$
Resultant matrix$=\begin{bmatrix} 1+7 & -2-8 & 3+9 \\ 7 & -8 & 9 \\ 4 & -5 & 6 \end{bmatrix}$
$=\begin{bmatrix} 8 & -12 & 12 \\ 7 & -8 & 9 \\ 4 & -5 & 6 \end{bmatrix}$
Option A is correct
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

 A=$\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$
The new matrix formed  after interchanging $2^{nd}$ and $3^{rd}$rows  will be 

  1. $-\begin{bmatrix} 1&2&3\\4&5&6\\7&8&9\end{bmatrix}$
  2. $\begin{bmatrix} 4&5&6\\1&2&3\\7&8&9\end{bmatrix}$
  3. $-\begin{bmatrix} 1&2&3\\7&8&9\\4&5&6\end{bmatrix}$
  4. $\begin{bmatrix} 1&2&3\\7&8&9\\4&5&6\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
Now second and third rows are interchanged $\Rightarrow $all elements are interchanged
Resultant matrix$=\begin{bmatrix} 1 & 2 & 3 \\ 7 & 8 & 9 \\ 4 & 5 & 6 \end{bmatrix}$
Option D is correct
Multiple choice business maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

For a matrix $A \begin{pmatrix} 1& 0 & 0\ 2 & 1 & 0\ 3 & 2 & 1\end{pmatrix}$, if $U _{1}, U _{2}$ and $U _{3}$ are $3\times 1$ column matrices satisfying $AU _{1} = \begin{pmatrix}1\ 0 \ 0
\end{pmatrix}, AU _{2} \begin{pmatrix}2\3 \ 0
\end{pmatrix}, AU _{3} = \begin{pmatrix}2\ 3\ 1
\end{pmatrix}$ and $U$ is $3\times 3$ matrix whose columns are $U _{1}, U _{2}$ and $U _{3}$
Then sum of the elements of $U^{-1}$ is

  1. $6$
  2. $0 (zero)$
  3. $1$
  4. $2/3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $U _{i} = \begin{pmatrix}a _{i}\b _{i} \c _{i} \end{pmatrix} i = 1, 2, 3$
$AU _{1} = \begin{pmatrix}a _{1}\2a _{1} + b _{1} \ 3a _{1} + 2b _{1} + c _{1}
\end{pmatrix} = \begin{pmatrix}1\0 \ 0
\end{pmatrix}$, So $a _{1} = 1, b _{1} = -2, c _{1} = 1$
$AU _{2} = \begin{pmatrix}a _{2}\2a _{2} + b^{2} \ 3a _{2} + 2b _{2} + c _{2}
\end{pmatrix} = \begin{pmatrix}2\ 3\ 0
\end{pmatrix}$,
So, $a _{2} = 2, b _{2} = -1, c _{2} = -4$. Similarly, $a _{3} = 2, b _{3} = -1, c _{3} = -3$
So, $U = \begin{pmatrix} 1& 2 & 2\ -2 & -1 & -1\ 1 & -4 & -3\end{pmatrix}$. So, sum of elements of $U^{-1}$ is zero.

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

If $A$ is a non zero square matrix of order $n$ with $det\left( I+A \right) \neq 0$, and ${A}^{3}=0$, where $I,O$ are unit and null matrices of order $n\times n$ respectively, then ${ \left( I+A \right)  }^{ -1 }=$

  1. $I-A+{ A }^{ 2 }$
  2. $I+A+{ A }^{ 2 }$
  3. $I+{ A }^{ 2 }$
  4. $I+A$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$det(I+A)\neq 0$
$A^3=0$   where $0$ is null matrix, $I$ is the identity matrix
$A^3+I=I$ [adding $I$ on both sides]
$(A+I)(A^2-IA+I^2)=I$ [by the formula of $a^3+b^3$]
$(A+I)(A^2-A+I)=I$
$(I+A)(I+A)^{-1}=I$ [by the rule of inverse matrix]
hence $(I+A)^{-1}=(A^2-A+I)$
Ans: $I-A+A^2$
Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $A=\begin{bmatrix} 3 & -2 \ 5 & 8 \end{bmatrix}$, then $A^{-1}=$

  1. $\frac{1}{30}\begin{bmatrix} 8 & 2 \\ -5 & 3 \end{bmatrix}$
  2. $\frac{1}{34}\begin{bmatrix} 8 & 2 \\ -5 & 3 \end{bmatrix}$
  3. $-\frac{1}{34}\begin{bmatrix} -8 & -2 \\ -5 & 3 \end{bmatrix}$

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

$A=\left[{\begin{array}{cc}3&2\5&8\end{array}}\right]$

$A^{-1}=\cfrac{1}{ad-bc}\left[{\begin{array}{cc}d&-b\-c&a\end{array}}\right]$  (determinant)
$=\cfrac{1}{3\times 8-(-2\times 5)}\left[{\begin{array}{cc}8&2\-5&3\end{array}}\right]$  
$=\cfrac{1}{34}\left[{\begin{array}{cc}8&2\-5&3\end{array}}\right]$  

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

If the matrix $\begin{bmatrix} 0 & 2\beta & \Upsilon \ \alpha & \beta & -\Upsilon \ \alpha & -\beta & \Upsilon \end{bmatrix}$is orthogonal, then

  1. $\alpha = \pm\dfrac{1}{\sqrt{2}}$
  2. $\beta = \pm\dfrac{1}{\sqrt{6}}$
  3. $\gamma = \pm\dfrac{1}{\sqrt{3}}$

  4. all of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\begin{bmatrix} 0 & 2\beta  & \gamma  \\ \alpha  & \beta  & -\gamma  \\ \alpha  & -\beta  & \gamma  \end{bmatrix}$
for orthogonal matrix we have 
$A.A^{T}=I$
$\begin{bmatrix} 0 & 2\beta  & \gamma  \\ \alpha  & \beta  & -\gamma  \\ \alpha  & -\beta  & \gamma  \end{bmatrix}\begin{bmatrix} 0 & \alpha  & \alpha  \\ 2\beta  & \beta  & -\beta  \\ \gamma  & -\gamma  & \gamma  \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 0+4{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } & 0+2{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } & 0-2{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } \\ 0+2{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }-{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } \\ 0-2{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }-{ \beta  }^{ 2 }-{ \gamma  }^{ 2 } & { \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$4\beta^{2}+\gamma^{2}=1, 2\beta^{2}-\gamma^{2}=0$, 
$4\left(\dfrac{\gamma^{2}}{2}\right)+\gamma^{2}=1$        $\beta^{2}=\dfrac{r^{2}}{2}$
$r^{2}[3]=1$
$r=\pm \dfrac{1}{\sqrt{3}}$
$2\beta^{2}-\gamma^{2}=0, \alpha^{2}+\beta^{2}+\gamma^{2}+\gamma^{2}=1, \alpha^{2}-\beta^{2}-\gamma^{2}=0$
$\beta^{2}=\dfrac{\gamma^{2}}{2},  \alpha^{2}+\dfrac{\gamma^{2}}{2}+\dfrac{\gamma^{2}}{1}=1$
$\alpha^{2}+\dfrac{3\gamma^{2}}{2}=1$
$\beta^{2}=\dfrac{1}{6}\alpha^{2}+\dfrac{3}{2}\times \dfrac{1}{3}=1$
$\beta=\pm \dfrac{1}{\sqrt{6}}$            $\alpha=\pm \dfrac{1}{\sqrt{2}}$
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A=\begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}$ then $A^n=\begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \ 3^{n-1} & 3^{n-1} & 3^{n-1} \ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}$ , $n \in N$

  1. True

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

For A = [[1,1,1],[1,1,1],[1,1,1]], A^2 = 3A, A^3 = 9A = 3^2 * A. By induction, A^n = 3^(n-1) * A. This matches the given matrix.