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 business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

If $A=\begin{bmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{bmatrix}$ is singular matrix and $x\in [-5, -2]$ then x=?$

  1. $0$
  2. $-2$
  3. $-4$
  4. $0, -4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $A=\begin{bmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{bmatrix}$ is a singular matrix.


$\therefore |A|=0$

$\begin{vmatrix}3&-1+x&2\3&-1&x+2\x+3&-1&2\end{vmatrix}=0$

$\Rightarrow 3(-2+x+2)-(-1+x)(6-(x+3)(x+2))+2(-3+x+3)=0$

$\Rightarrow 3x-(x-1)(6-x^2-5x-6)+2x=0$

$\Rightarrow 5x-(x-1)(-x^2-5x)=0$

$\Rightarrow 5x-(-x^3-5x^2+x^2+5x)=0$

$\Rightarrow 5x+x^3+4x^2-5x=0$

$\Rightarrow x^3+4x^2=0$

$\Rightarrow x^2(x+4)=0$

$\Rightarrow x=0:and\;x=-4$

It is given that $x\in [-5,-2]$

$\therefore x=-4$

Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

If $A=\begin{bmatrix} 0 & x & 16 \ x & 5 & 7 \ 0 & 9 & x \end{bmatrix}$ is singular, then the possible values of $x$ are

  1. $0, \pm 1$
  2. $0, \pm 12$
  3. $0, \pm 5$
  4. $0, \pm 4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} { { Sin } }gular\, \, means \ \left| A \right| =0 \ \left( { \begin{array} { *{ 20 }{ c } }0 & x & { 16 } \ x & 5 & 7 \ 0 & 9 & x \end{array} } \right) =0 \ 0\left( { 51-63 } \right) -x\left( { { x^{ 2 } } } \right) +16\left( { 9x } \right) =0 \ -{ x^{ 3 } }+144x=0 \ { x^{ 3` } }-144x=0 \ x\left( { { x^{ 2 } }-144 } \right) =0 \ x=0 \ { x^{ 2 } }-144=0 \ { x^{ 2 } }=144 \ x=\pm 12 \ x=0,\pm 12 \end{array}$


Hence, this is the answer.

Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

If $\omega\neq 1$ is a cube root of unity, then


$A=\begin{bmatrix}1+2\omega ^{100}+\omega ^{200}&\omega ^2  &1  \1 &1+\omega ^{101}+2\omega ^{202}  &\omega  \\omega  & \omega ^2 &2+ \omega ^{100}+2\omega ^{200}\end{bmatrix}$

  1. $A$ is singular
  2. $|A|=0$
  3. $A$ is symmetric

  4. none of these
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation
$A=\begin{vmatrix} 1+{ \omega  }^{ 100 }+{ \omega  }^{ 200 } & { \omega  }^{ 2 } & 1 \\ 1 & 1+{ \omega  }^{ 101 }+{ \omega  }^{ 202 } & \omega  \\ \omega  & { \omega  }^{ 2 } & 2+{ \omega  }^{ 100 }+{ \omega  }^{ 200 } \end{vmatrix}$

$1+\omega +{ \omega  }^{ 2 }=0$

${ \omega  }^{ 100 }={ \omega  }^{ 99 }\times \omega =\omega $

${ \omega  }^{ 200 }={ \omega  }^{ 198 }\times { \omega  }^{ 2 }={ \omega  }^{ 2 }$

${ \omega  }^{ 101 }={ \omega  }^{ 2 }$${ \omega  }^{ 202 }={ \omega  }^{ 4 }=\omega $

$A=\begin{vmatrix} 1+2{ \omega  }+{ \omega  }^{ 2 } & { \omega  }^{ 2 } & 1 \\ 1 & 1+{ \omega  }^{ 2 }+{ 2\omega  } & \omega  \\ \omega  & { \omega  }^{ 2 } & 2+{ \omega  }+2{ \omega  }^{ 2 } \end{vmatrix}$

$A=\begin{vmatrix} { \omega  } & { \omega  }^{ 2 } & 1 \\ 1 & { \omega  } & \omega  \\ \omega  & { \omega  }^{ 2 } & -{ \omega  } \end{vmatrix}$

$A=\omega \begin{vmatrix} { \omega  } & { \omega  } & 1 \\ 1 & { 1 } & \omega  \\ \omega  & { \omega  } & -{ \omega  } \end{vmatrix}$

${ C } _{ 1 }\& { C } _{ 2 }$are same

$\left| A \right| =0$

$A\rightarrow $singular

$\triangle =0$
Option A and B
Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

Let $A$ be a square matrix of order $n \times  n$. A constant $\lambda $ is said to be characteristic root of $A$ if there exists a $n \times  1$ matrix $X$ such that  $AX=\lambda X$

Let $P$ be a non-singular matrix, then which of the following matrices have the same characteristic roots.

  1. $A$ and $AP$
  2. $A$ and $PA$
  3. $A$ and $P^{-1} \:AP$

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

Since $X\neq 0$ is such that $(A-\lambda I)X=0,:|A-\lambda 
I|=0\Leftrightarrow A-\lambda I$ is singular. If $A-\lambda I$ is non-singular the then equation $(A-\lambda I)X=0\Rightarrow X=0$
If $\lambda= 0$, we get $|A|=0\Rightarrow A$ is singular.
We have $A^2 :X=A(AX) =A(\lambda X) = \lambda(AX)$
$=\lambda^2X$,
$A^3 :X=A(A^2X) =A(\lambda^2 X)$
$=\lambda^2(AX)=\lambda^2(\lambda X)=\lambda^3 X$
Continuing in this way, we obtain
$A" X=\lambda^": X: \forall : n\in N$
Also,$|P^{-1}:AP -\lambda I|=|P^{-1}(A -\lambda I)P|$
$=|P^{-1}|: |A -\lambda I|: |P|=|A -\lambda I|$

Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

Let $A$ and $B$ be two non-null square matrices. If the product $AB$ is a null matrix, then

  1. $A$ is singular
  2. $B$ is singular
  3. $A$ is non-singular
  4. $B$ is non-singular
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Let $B$ be non-singular, then ${ B }^{ -1 }$ exists.

Now, $AB=0($ given$) \Rightarrow \left( AB \right) { B }^{ -1 }=0{ B }^{ -1 }$
$($ post multiplying both sides by ${ B }^{ -1 })$
$\Rightarrow A\left( B{ B }^{ -1 } \right) =0$     $($ by associativity $)$
$\Rightarrow A{ I } _{ n }=0\quad \quad \quad \left( \because B{ B }^{ -1 }={ I } _{ n } \right) $
$\Rightarrow A=0$
But $A$ is a non-null matrix.
Hence $B$ is a singular matrix.
Similarly, it can be shown that $A$ is a singular matrix.

Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

Let $A=\begin{bmatrix}x+\lambda& x&x\x &x+\lambda&x\x&x&x+\lambda  \end{bmatrix}$, then $A^{-1}$ exists if

  1. $x\neq 0$
  2. $\lambda \neq 0$
  3. $3x+\lambda \neq 0, \lambda \neq 0$
  4. $x\neq 0, \lambda \neq 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$A=\left[ \begin{matrix} x+\lambda  & x & x \ x & x+\lambda  & x \ x & x & x+\lambda  \end{matrix} \right] $
${ A }^{ -1 }$ exists if $\left| A \right| \neq 0$
$\left| A \right| =\left| \begin{matrix} x+\lambda  & x & x \ x & x+\lambda  & x \ x & x & x+\lambda  \end{matrix} \right| \neq 0$
${ R } _{ 1 }\rightarrow { R } _{ 1 }+{ R } _{ 2 }+{ R } _{ 3 }$
$\left( 3x+\lambda  \right) \left| \begin{matrix} 1 & 1 & 1 \ x & x+\lambda  & x \ x & x & x+\lambda  \end{matrix} \right| \neq 0$
${ C } _{ 2 }\rightarrow { C } _{ 2 }-{ C } _{ 1 }$ and ${ C } _{ 3 }\rightarrow { C } _{ 3 }-{ C } _{ 1 }$
$\left( 3x+\lambda  \right) \left| \begin{matrix} 1 & 0 & 0 \ x & \lambda  & 0 \ x & 0 & \lambda  \end{matrix} \right| \neq 0$
$\left( 3x+\lambda  \right) { \left[ { \lambda  }^{ 2 } \right]  }\neq 0$
$3x+\lambda \neq 0,\lambda \neq 0$

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

If $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \neq 0 , x = c y + b z , y = a z + c x$ and $z = b x + a y ,$ then $a ^ { 2 } + b ^ { 2 } + c ^ { 2 } + 2 a b c =$

  1. 2
  2. $a + b + c$

  3. 1
  4. $ab + bc + ca$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given the system of equations, this is a classic problem where the determinant of the coefficient matrix must be zero for non-trivial solutions. Solving the system leads to the identity a^2 + b^2 + c^2 + 2abc = 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$= \left[\begin{array}{lll}1 & 0 & 0\2 & 1 & 0\3 & 2 & 1\end{array}\right]$.If $\mathrm{u _1}$ and $\mathrm{u} _{2}$ are column matrices such that  $\mathrm{Au _{1}}=\left[\begin{array}{l}1\0\0\end{array}\right]$ and  $\mathrm{Au _{2}}=\left[\begin{array}{l}0\1\0\end{array}\right]$ then $\mathrm{u _{1}+u _{2}}$ is equal to:

  1. $\left[\begin{array}{l}

    -1\\

    1\\

    0

    \end{array}\right]$
  2. $\left[\begin{array}{l}

    -1\\

    1\\

    -1

    \end{array}\right]$
  3. $\left[\begin{array}{l}

    -1\\

    -1\\

    0

    \end{array}\right]$
  4. $\left[\begin{array}{l}

    1\\

    -1\\

    -1

    \end{array}\right]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given:Matrices are 
$A=\left[ \begin{matrix} 1 & 0  & 0 \\ 2 & 1  & 0 \\ 3 & 2  & 1 \end{matrix} \right]$

$A{u} _{1}=\left[ \begin{matrix} 1  \\  0 \\ 0 \end{matrix} \right]$ and 
$A{u} _{2}=\left[ \begin{matrix} 0  \\  1 \\ 0 \end{matrix} \right]$

To find:Matric ${u} _{1}+{u} _{2}$

Since both $A{u} _{1}$ and $A{u} _{2}$ are given, hence adding them, we get

$A{u} _{1}+A{u} _{2}=\left[ \begin{matrix} 1  \\  0 \\ 0 \end{matrix} \right]+\left[ \begin{matrix} 0  \\  1 \\ 0 \end{matrix} \right]$

$A\left({u} _{1}+{u} _{2}\right)=\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix} \right]$

Since,$A$ is a non-singular matrix,we have
$\left|A\right|\neq\,0$

Hence multiplying both sides by ${A}^{-1}$ from RHS we get

${A}^{-1}A\left({u} _{1}+{u} _{2}\right)={A}^{-1}\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix}\right]$

${u} _{1}+{u} _{2}={\left[ \begin{matrix} 1 & 0  & 0 \\ 2 & 1  & 0 \\ 3 & 2  & 1 \end{matrix} \right]}^{-1}\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix}\right]$     ..........$(1)$

Now, $\left|A\right|=\left|\begin{matrix} 1 & 0  & 0 \\ 2 & 1  & 0 \\ 3 & 2  & 1 \end{matrix} \right|$

$=1\left| \begin{matrix} 1 & 0   \\ 2 & 1  \end{matrix} \right|-0+0$(by expanding the determinant along row $1$)

$\Rightarrow\,\left|A\right|=1$

Now, co-factor matrix of $A$ (i.e., the matrix in which every element is replaced by corresponding co-factor)

$=\left[\begin{matrix} \left| \begin{matrix} 1 & 0   \\ 2 & 1  \end{matrix} \right| & -\left| \begin{matrix} 2 & 0   \\ 3 & 1  \end{matrix} \right|  & \left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & 0   \\ 2 & 1  \end{matrix} \right| & \left| \begin{matrix} 1 & 0   \\ 3 & 1  \end{matrix} \right|  & -\left| \begin{matrix} 1 & 0   \\ 3 & 2 \end{matrix} \right| \\ \left| \begin{matrix} 0 & 0   \\ 1 & 0  \end{matrix} \right| & -\left| \begin{matrix} 1 & 0   \\ 2 & 0  \end{matrix} \right|  & \left| \begin{matrix} 1 & 0   \\ 2 & 1  \end{matrix} \right| \end{matrix} \right]$

$=\left[\begin{matrix} 1 & -2 & 1 \\ 0 & 1  & -2 \\ 0 & 0 & 1 \end{matrix} \right]$
$\therefore\,adj\left(A\right)={\left[\begin{matrix} 1 & -2 & 1 \\ 0 & 1  & -2 \\ 0 & 0 & 1 \end{matrix} \right]}^{T}=\left[\begin{matrix} 1 & 0 & 0 \\ -2 & 1  & 0 \\ 1 & -2 & 1 \end{matrix} \right]$

$\Rightarrow\,{A}^{-1}=\dfrac{adj\left(A\right)}{\left|A\right|}$

$=\left[\begin{matrix} 1 & 0  & 0 \\ -2 & 1  & 0 \\ 1 & -2  & 1 \end{matrix} \right|\,\,\,\because\left|A\right|=1$

From eqn$(1)$ we get

${u} _{1}+{u} _{2}={\left[\begin{matrix} 1 & 0  & 0 \\ 2 & 1  & 0 \\ 3 & 2  & 1 \end{matrix} \right]}^{-1}\times\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix}\right]$ 

$=\left[\begin{matrix} 1 & 0  & 0 \\ -2 & 1  & 0 \\ 1 & -2  & 1 \end{matrix} \right]\times\left[ \begin{matrix} 1  \\  1 \\ 0 \end{matrix}\right]$ 

$=\left[ \begin{matrix} 1+0+0  \\  -2+1+0 \\ 1-2+0 \end{matrix}\right]$ 

$\therefore\,{u} _{1}+{u} _{2}=\left[ \begin{matrix} 1  \\  -1 \\ -1 \end{matrix}\right]$ 
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 an $3\times 3$ non -singular matrix that $AA'=A'A$ and $B=A^{-1}A'$,then $BB'$ equal ?

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

Given A is non-singular and AA' = A'A, we have B = A^-1 A'. Then BB' = (A^-1 A')(A^-1 A')'. Since (A^-1 A')' = (A')'(A^-1)' = A(A^-1)' = A(A')^-1, the expression simplifies based on the commutativity of A and 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

 ${( -A )}^{ -1 }$ is always equal to (where $A$ is $nth$ order square matrix)

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

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

We know that if $A^{-1}$ exist then $(cA)^{-1}=\dfrac{1}{c}A^{-1}$ where c is  a constant
Hence $(-A)^{-1}=\dfrac{1}{-1}A^{-1}=-A^{-1}$

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

If $A\left( \alpha ,\beta  \right) =\left[ \begin{matrix} \cos { \alpha  }  & \sin { \alpha  }  & 0 \ -\sin { \alpha  }  & \cos { \alpha  }  & 0 \ 0 & 0 & { e }^{ \beta  } \end{matrix} \right]$, then $A{ \left( \alpha ,\beta  \right)  }^{ -1 }$ is equal to 

  1. $A{ \left( -\alpha ,-\beta \right) }$
  2. $A{ \left( -\alpha ,\beta \right) }$
  3. $A{ \left(\alpha ,-\beta \right) }$
  4. $A{ \left(\alpha ,\beta \right) }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The matrix A(alpha, beta) is a rotation matrix combined with a scaling factor e^beta. The inverse of a rotation matrix is obtained by replacing alpha with -alpha, and the inverse of the scaling factor e^beta is e^-beta.

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 $a, b, c$ are non real number satisfying equation $x^{5}=1$ and $S$ be the set of all non-invertible matrices of the from $\begin{bmatrix} 1 & a & b \ w & 1 & c \ { w }^{ 2 } & w & 1 \end{bmatrix}$ where $w={ e }^{ \dfrac { 12\pi  }{ 5 }  }$. The number of distinct matrices in set $S$ is 

  1. $1$
  2. $28$
  3. $32$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The matrix is non-invertible if its determinant is zero. Given the structure and the roots of unity, the determinant evaluates to zero only under specific conditions related to the values of 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 is an invertible matrix, then det $\displaystyle :\left ( A^{-1} \right )$ is equal to

  1. $\displaystyle \:det\left ( A \right )$
  2. $\displaystyle \:\frac{1}{det\left ( A \right )}$
  3. $1$

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

We know that $|A^{n}|=|A|^{n}$  n be any integer
$\Rightarrow |A^{-1}|=|A|^{-1}=\displaystyle \frac{1}{|A|}$

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 $3 \times 3$ matrix such that $\left| A \right| = 4\ than\ \left| {{{\left( {adjA} \right)}^{ - 1}}} \right| = $

  1. $16$
  2. $64$
  3. $\dfrac{1}{{16}}$

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

We know, $A^{-1}=\dfrac{adjA}{|A|}$

Multiplying above equation with A both sides,
$AA^{-1}=\dfrac{A\times adjA}{|A|}\Rightarrow|A|=A\times adjA$
Multiplying with $(adjA)^{-1}$ both sides ,
$|A|\times(adjA)^{-1}=A\times adjA \times(adjA)^{-1}\Rightarrow|A|\times(adjA)^{-1}=A$
Taking determinant both sides,$||A|\times(adjA)^{-1}|=|A|\Rightarrow ||A||\times |(adjA)^{-1}|=|A|\Rightarrow |A|^n\times |(adjA)^{-1}|=|A|$
Where n is the order of matrix A, i.e. $n=3$ and $|A|=4$
Thus, $|(adjA)^{-1}|=\dfrac{4}{4^3}=\dfrac{1}{16}$