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

If $A = \begin{bmatrix}1 & k & 3\ 3 & k & -2 \ 2 & 3 & -4\end{bmatrix}$ is singular then $k = ?$

  1. $\dfrac {16}{3}$
  2. $\dfrac {34}{5}$
  3. $\dfrac {33}{2}$

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

A matrix is singular if its determinant is 0. Calculating the determinant: 1(k*-4 - 3*-2) - k(3*-4 - 2*-2) + 3(3*3 - 2*k) = 0. This simplifies to 1(-4k + 6) - k(-12 + 4) + 3(9 - 2k) = 0, which is -4k + 6 + 8k + 27 - 6k = 0, or -2k + 33 = 0, so k = 33/2.

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  $AX=\lambda X$

If $\lambda $ is a characteristic root of $A$ and $n\in N$, then $\lambda ^{n}$ is a characteristic root of

  1. $A^n$
  2. $A^{n-1}$
  3. $A^{-n}$
  4. $A-A^n+A^{-n}$
Reveal answer Fill a bubble to check yourself
A 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
                        ${A}^{n} X=\lambda^{n}: X: \forall : n\in N$

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

If $A =\begin{bmatrix}4 &x+2 \2x-3 &x+1 \end{bmatrix}$ is an invertible matrix, then $x$ cannot take value

  1. -1

  2. 2

  3. 3

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

A matrix is invertible if its determinant is non-zero. Determinant = 4(x+1) - (x+2)(2x-3) = 4x + 4 - (2x^2 + x - 6) = -2x^2 + 3x + 10. Setting this to 0: 2x^2 - 3x - 10 = 0. Using the quadratic formula, x = (3 +/- sqrt(9 - 4*2*-10)) / 4 = (3 +/- sqrt(89)) / 4. Since these values are not in the options, 'none of these' is the correct choice.

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 $n\times n$ and let $P$ be a non-singular matrix, then which of the following matrices have the same characteristic roots.

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

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

Let the characteristic root of $A$ be $\lambda $.


$\displaystyle \Rightarrow \left| A-\lambda I \right| =0$

For $\displaystyle { P }^{ -1 }AP;\left| { P }^{ -1 }AP-\lambda I \right| =\left| { P }^{ -1 }AP-\lambda { P }^{ -1 }PI \right| =\left| { P }^{ -1 } \right| \left| A-\lambda I \right| \left| P \right| =\left| A-\lambda 1 \right| .$

$\Rightarrow \lambda $ is also characteristic roots of matrix $\displaystyle { P }^{ -1 }AP$

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

Let $A$ be an $n\times n$ matrix such that $A^n=\alpha A,$ where $\alpha$ is a real number different from $1$ and $-1$. Then, the matrix $A+I _n$ is

  1. singular

  2. non-singular, i.e., invertible

  3. scalar

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

Let $B=A+{ I } _{ n }$.


Since $A=B-{ I } _{ n }$, the condition ${ A }^{ n }=\alpha A$ can be written in the form $\displaystyle { \left( { B-I } _{ n } \right)  }^{ n }\alpha \left( B-{ I } _{ n } \right) $

$\displaystyle \Rightarrow { B }^{ n }-^{ n }{ { C } _{ 1 } }{ B }^{ n-1 }+^{ n }{ { C } _{ 2 } }{ B }^{ n-2 }+...+{ \left( -1 \right)  }^{ n }{ I } _{ n }=\alpha B-\alpha { I } _{ n }$

$\displaystyle \Rightarrow { B }^{ n }-^{ n }{ { C } _{ 1 } }{ B }^{ n-1 }+^{ n }{ C } _{ 2 }{ B }^{ n-2 }+...+{ \left( -1 \right)  }^{ n-1 }B-\alpha B=-\alpha { I } _{ n }-{ \left( -1 \right)  }^{ n }{ I } _{ n },$

$\displaystyle \Rightarrow B\left( { B }^{ n-1 }-^{ n }{ { C } _{ 1 }{ B }^{ n-2 }+^{ n }{ { C } _{ 2 } }{ B }^{ n-3 }+...+{ \left( -1 \right)  }^{ n-1 }{ I } _{ n }-\alpha { I } _{ n } } \right) =\left[ { \left( -1 \right)  }^{ n+1 }-\alpha  \right] { I } _{ n }.$

Since $\displaystyle { \left( -1 \right)  }^{ n+1 }-\alpha \neq 0,\left[ \because \alpha \neq \pm 1 \right] $

$\therefore$ $B$ is invertible 

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

Matrix $\begin{bmatrix}a & b &(a\alpha -b) \b  & c & (b\alpha -c)\2 & 1 & 0\end{bmatrix}$ is non invertible if 

  1. $\alpha = 1/2$

  2. a, b, c are in A.P.

  3. a, b, c are in G.P.

  4. a, b, c are in H.P.
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$\Delta =\begin{vmatrix} a & b & (a\alpha -b) \ b & c & (b\alpha -c) \ 2 & 1 & 0 \end{vmatrix}\ =2\left( b(b\alpha -c)-c(a\alpha -b) \right) -1\left( a(b\alpha -c)-b(a\alpha -b) \right)$

 
Expanding along $R _3$

$ =2{ b }^{ 2 }\alpha -2bc-2ac\alpha +2bc-ab\alpha +ca+ab\alpha -{ b }^{ 2 }\\ =\left( { b }^{ 2 }-ac \right) \left( 2\alpha -1 \right) $

Matrix is non invertible when $\Delta =0$
i.e $\displaystyle\alpha=\frac{1}{2}$ and $a,b,c$ are in $G.P$

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

If $A$ and $B$ are any two matrices such that $AB = 0$ and $A$ is non-singular, then

  1. $B = 0$
  2. $B$ is singular
  3. $B$ is non-singular
  4. $B = A$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given that $AB = 0$, taking determinants on both sides, we have
$|A\cdot B| = 0$
$\therefore |A| \times |B| = 0$
Product of two numbers is zero if one of them is zero.
Since $A$ is non-singular matrix, $B$ has to be singular, meaning that its determinant has to be zero.
The matrix need not be zero.
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 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

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

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

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

$\therefore\,{u} _{1}+{u} _{2}=\left[ \begin{matrix} 1  \\  -1 \\ -1 \end{matrix}\right]$