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 business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

 In matrices $AB = O$ does not necessarily mean that 

  1. $A=0$
  2. $B=0$
  3. Both $ A = 0$ and $B=0$

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $\quad A = \begin{bmatrix}1 & -1 & 1 \ -3 & 2 & -1

\ -2 & 1 & 0\end{bmatrix}$ and $B = \begin{bmatrix}1&2

& 3\2&4&6 \ 1&2 &3\end{bmatrix}$

$\therefore\quad

AB = \begin{bmatrix}1 & -1 & 1 \ -3 & 2 & -1 \ -2

& 1 & 0\end{bmatrix}\times\begin{bmatrix}1&2 &

3\2&4&6 \ 1&2 &3\end{bmatrix}$

$\quad                   = \begin{bmatrix}0&0&0 \ 0&0&0 \ 0&0&0\end{bmatrix} = O$

$\therefore \quad AB = O$
But neither $A = O$ nor $B = O$.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If matrix $A = [a _{ij}] _{2\times 2}$, where $a _{ij} = \left{\begin{matrix} 1,& \ \text{if}\ &i\neq j \ 0, & \ \text{if}\ & i + j\end{matrix}\right.$, then $A^{2}$ is equal to

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

$A=\begin{bmatrix}  a _{11}& a _{12} \  a _{21}&  a _{22}\end{bmatrix}$


$a _{11}:1\,\,\,a _{12}:0$

$a _{21}:0\,\,\,a _{22}:1$

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

$A^2=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \ 0 &  0\end{bmatrix}=\begin{bmatrix}  1& 0 \  0&  1\end{bmatrix}=I$

$\boxed{Hence\,A^2=I}$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A = \begin{bmatrix} -2& 3\ 1 & 1\end{bmatrix}$ then $|A^{-1}| = ?$

  1. $-5$
  2. $\dfrac {-1}{5}$
  3. $\dfrac {1}{25}$
  4. $25$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$AA^{-1} = I\Rightarrow |AA^{-1}| = |I|\Rightarrow |A|\cdot |A^{-1}| = 1\Rightarrow |A^{-1}| = \dfrac {1}{|A|}$.
$|A| = \begin{vmatrix}-2 & 3\ 1 & 1\end{vmatrix} = (-2 - 3) = (-5) \Rightarrow |A^{-1}| = \dfrac {-1}{5}$.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Let $A$ and $B$ be two $2 \times 2$ matrices. Consider the statements
          $(i)$ $AB =0 \Rightarrow A = 0 :or :B = 0$
         $ (ii)$ $AB =I \Rightarrow A =B^{-1}$
          $(iii)$ $(A + B)^2 = A^2 + 2AB + B^2$

  1. $(i)$ is false, $(ii)$ and $(iii)$ are true
  2. $(i)$ and $(iii)$ are false, $(ii)$ is true
  3. $(i)$ and $(ii)$ are false, $(iii)$ is true
  4. $(ii)$ and $(iii)$ are false, $(i)$ is true
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(i) is false.


If $A=\begin{bmatrix}0& 1\0 & -1\end{bmatrix}$ and $B =\begin{bmatrix}1& 1\0 & 0\end{bmatrix}$, then

 $AB=\begin{bmatrix}0& 0\0 & 0\end{bmatrix}=0$

Thus,  $AB = 0$ but neither $A = 0$ nor $B = 0$

(iii) is false since matrix multiplication is not commutative.

$(A + B)^2 = A^2 + AB + BA + B^2$

(ii) is true as the product $AB$ is an identity matrix, if and only $B$ is inverse of the matrix $A$.

Hence, option B.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

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

  1. $\begin{bmatrix}\dfrac {3}{7}
    &\dfrac {-1}{7} \\
    \dfrac {1}{7} & \dfrac {2}{7}
    \end{bmatrix}$
  2. $\begin{bmatrix}\dfrac {3}{7}
    &\dfrac {1}{7} \\
    \dfrac {-1}{7} & \dfrac {2}{7}
    \end{bmatrix}$
  3. $\begin{bmatrix}\dfrac {3}{7}
    &\dfrac {1}{7} \\
    \dfrac {1}{7} & \dfrac {2}{7}
    \end{bmatrix}$

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

$|A| = \begin{vmatrix}2 & -1\ 1 & 3\end{vmatrix} = (6 + 1) = 7\neq 0$
$M _{11} = 3, M _{12} = 1, M _{21} = -1$ and $M _{22} = 2$


$\therefore C _{11} = 3, C _{12} = -1, C _{21} = 1$ and $C _{22} = 2$

$\Rightarrow Adj\ A = \begin{bmatrix} 3& -1\ 1 & 2\end{bmatrix}^T = \begin{bmatrix} 3& 1\ -1 & 2\end{bmatrix}$

$\Rightarrow A^{-1} = \dfrac {1}{|A|} (adj\ A) = \dfrac {1}{7} \begin{bmatrix}3 & 1\ -1 & 2\end{bmatrix} = \begin{bmatrix}\dfrac {3}{7} & \dfrac {1}{7}\ -\dfrac {1}{7} & \dfrac {2}{7}\end{bmatrix}$.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A$ and $B$ are two square matrices of the same order and $m$ is a positive integer, then
$(A + B)^m =$ $^mC _0A^m +$ $^mC _1 A^{m -1} B + ^mC _2A^{m-2} B^2 + ... +$ $^mC _{m- 1} AB^{m-1}+$ $^mC _m B^m$ if

  1. $AB =BA$
  2. $AB + BA =0$
  3. $A^m = 0, \:B^m = 0$

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

Binomial theorem is applicable if and only if $AB=BA$.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Let $A, : B : and : C$ be $2\times 2$ matrices with entries from the set of real numbers. Define $\ast $ as follows: $\displaystyle A\ast B=\frac{1}{2}(AB + BA)$, then

  1. $A\ast B=B\ast A$
  2. $A\ast A=A^2$
  3. $A\ast (B+C)=A\ast B+A\ast C$
  4. $A\ast I=A$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

$\displaystyle A*1=\frac { 1 }{ 2 } \left( AI+IA \right) =\frac { 1 }{ 2 } \left( A+A \right) =A$

$\therefore A*I=A,\therefore$ (a)holds.
$\displaystyle A*A=\frac { 1 }{ 2 } \left( AA+AA \right) =\frac { 1 }{ 2 } \left( { A }^{ 2 }+{ A }^{ 2 } \right) ={ A }^{ 2 }$
$\therefore$ (b) holds.
$\displaystyle \left( c \right) A*B=\frac { 1 }{ 2 } \left( AB+BA \right) B*A=\frac { 1 }{ 2 } \left( BA+AB \right) =\frac { 1 }{ 2 } \left( AB+BA \right) $
$[\because $ addition is commutative$]$
$\therefore A*B=B*A,\therefore$ (c)holds
$\displaystyle A*\left( B+C \right) =\frac { 1 }{ 2 } \left( A\left( B+C \right) +\left( B+C \right) A \right) =\frac { 1 }{ 2 } \left( AB+CA+BA+CA \right)$
$\displaystyle  =\frac { 1 }{ 2 } \left( AB+BA \right) +\frac { 1 }{ 2 } \left( AC+CA \right) =A*B+A*C$
$\therefore$ (d) holds.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Let $A, : B : and : C$ be $2\times 2$ matrices with entries from the set of real numbers. Define $\ast $ as follows:
  $\displaystyle A \ast B=\frac{1}{2}(AB\,'+A'B)$. Which of the given is true?

  1. $A\ast B= B \ast A$
  2. $A\ast A=A^2$
  3. $A\ast (B+C)=A\ast B+A \ast C$
  4. $A\ast I =A+A'$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$\displaystyle A \ast B=\frac{1}{2}(AB'+A'B)$

1) $\displaystyle B \ast A=\frac{1}{2}(BA'+B'A)=\frac{1}{2}(AB'+A'B)=A \ast B$

2)$\displaystyle A \ast A=\frac{1}{2}(AA'+A'A)$

3)$\displaystyle A \ast (B+C)=\frac{1}{2}(A(B+C)'+A'(B+C))$

                             $=\displaystyle\frac{1}{2}(AB'+A'B)+\frac{1}{2}(AC'+A'C)$

                             $=A\ast B+A \ast C$

4)$\displaystyle A \ast I=\frac{1}{2}(AI'+A'I)=\frac{1}{2}(A+A')$

Hence, options A and C.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Say true or false:

Let A, B be two matrices such that they commute, then $(AB)^n = A^nB^n$.

  1. True

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

$A$ and $B$ commute each other then $AB=BA$
${ \left( AB \right)  }^{ n }={ \left( BA \right)  }^{ n }\ \Rightarrow { \left( AB \right)  }^{ n }={ B }^{ n }{ A }^{ n }={ A }^{ n }{ B }^{ n }={ \left( BA \right)  }^{ n }$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A$ is a non-singular matrix, then 

  1. ${ A }^{ -1 }$ is symmetric if $A$ is symmetric
  2. ${ A }^{ -1 }$ is skew-symmetric if $A$ is symmetric
  3. $\left| { A }^{ -1 } \right| =\left| A \right| $
  4. $\left| { A }^{ -1 } \right| ={ \left| A \right| }^{ -1 }$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

Since $\left| A \right| \neq 0$, therefore ${ A }^{ -1 }$ exists.

Now, $A{ A }^{ -1 }=I={ A }^{ -1 }A$
$\Rightarrow \left( A{ A }^{ -1 } \right) '=I'=\left( { A }^{ -1 }A \right) '\Rightarrow \left( { A }^{ -1 } \right) 'A'=I=A'\left( { A }^{ -1 } \right) '\quad \quad \quad \left( \because A'=A \right) $
$\Rightarrow \left( { A }^{ -1 } \right) 'A=I=A\left( { A }^{ -1 } \right) '\Rightarrow { A }^{ -1 }=\left( { A }^{ -1 } \right) '\Rightarrow { A }^{ -1 }$ is symmetric
Also, since $\left| A \right| \neq 0,\therefore { A }^{ -1 }$ exists such that
$A{ A }^{ -1 }=I={ A }^{ -1 }A\Rightarrow \left| A{ A }^{ -1 } \right| =\left| I \right| $
$\Rightarrow \left| A \right| \left| { A }^{ -1 } \right| =1\quad \quad \left( \because \left| AB \right| =\left| A \right| \left| B \right|  \right) $
$\displaystyle \Rightarrow \left| { A }^{ -1 } \right| =\frac { 1 }{ \left| A \right|  } $

Multiple choice maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $AB=A$ and $BA=B$, where $A$ and $B$ are square matrices, then 

  1. ${ B }^{ 2 }=B$ and ${ A }^{ 2 }=A$
  2. ${ B }^{ 2 }=A$ and ${ A }^{ 2 }=B$
  3. $AB=BA$

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

We have 

$AB=A\Rightarrow A\left( BA \right) =A\quad$, subsitute $BA=B$ 
${$ $A(BA) =(AB)A$ $}$
$\Rightarrow \left( AB \right) A=A$
Subsitute $AB = A$
$\Rightarrow AA=A\quad \quad \left[ \therefore AB=A \right] \ \Rightarrow { A }^{ 2 }=A$
Again $BA=B$

$\Rightarrow B\left( AB \right) =B\quad \quad \left[ \because AB=A \right] $
$\Rightarrow \left( BA \right) B=B$
$\Rightarrow BB=B$
$\Rightarrow { B }^{ 2 }=B$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}$, $B=\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}$ then ${(A+B)}^{2}$ equals

  1. ${A}^{2}+{B}^{2}$
  2. ${A}^{2}+{B}^{2}+2AB$
  3. ${A}^{2}+{B}^{2}+AB-BA$

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

Given, $A=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix},B=\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}$


$ A+B=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}+\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}=\begin{bmatrix} 0 & 1-i \ i+1 & 0 \end{bmatrix}$

$ { \left( A+B \right)  }^{ 2 }=\begin{bmatrix} 0 & 1-i \ i+1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1-i \ i+1 & 0 \end{bmatrix}=\begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}$

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

$ { B }^{ 2 }=\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}\begin{bmatrix} 0 & -i \ i & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

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