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 mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

$(x _1 - x _2)^2 + (y _1 - y _2)^2 = a^2$;
$(x _2 - x _3)^2 + (y _2 - y _3)^2 = b^2$;
$(x _3 - x _1)^2 + (y _3 - y _1)^2 = c^2$;
then find $4 \begin{vmatrix}x _1 & y _1 & 1\ x _2 & y _2 & 1\ x _3 & y _3 & 1\end{vmatrix}^2 = $

  1. $(a+b+c) (b+c - a) (c + a - b) (a + b - c)$
  2. $-(a+b+c) (b+c - a) (c + a - b) (a + b - c)$
  3. $-(a+b+c) (b+c - a) (c + a - b) (a + b - c)/2$
  4. $(a+b+c) (b+c - a) (c + a - b) (a + b - c)/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area of the triangle PQR with vertices $(x _1, y _1) , (x _2, y _2) $ and $(x _3, y _3)$ is

$\therefore

\Delta = \dfrac{1}{2} \begin{vmatrix}x _1 & y _1 & 1\ x _2 &

y _2 & 1\ x _3 & y _3 & 1\end{vmatrix}$          ....(1)

Now the area of $\Delta$ $PQR$ with sides $a, b, c$ is

$\Delta = \sqrt{s (s - a) (s - b) (s - c)}$

$= \displaystyle \sqrt{\frac{1}{16}(2s)(2s - 2a)(2s - 2b)(2s - 2c)}$

$\Delta = \displaystyle \sqrt{\left [ \frac{(a + b + c)(b + c - a)(c + a - b)(a + b - c)}{16}\right ]}$          ....(2)

From (1) and (2), we get

$\displaystyle

\frac{1}{2} \begin{vmatrix} x _1&  y _1& 1\ x _2 & y _2 &

1\ x _3 & y _3 & 1\end{vmatrix} = \sqrt{\left [

\frac{(a+b+c)(b+c- a)(c + a - b)(a + b - c)}{16} \right ]}$

Squaring both sides, we have

$\Rightarrow

4 \begin{vmatrix}x _1 & y _1 & 1\ x _2 & y _2 & 1\ x _3

& y _3 & 1\end{vmatrix}^2 = (a+b+c)(b+c-a)(c+a- b)(a+b-c)$

Multiple choice statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

Let $A$ be the set of all $3 \times  3$ symmetric matrices all of whose entries are either $0$ or $1$. Five of these entries are $1$ and four of them are $0$.
The number of matrices in $A$ is

  1. $12$
  2. $6$
  3. $9$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If two zeros are the entries in the diagonal, then
$^{3}\mathrm{C} _{2}\times^{3}\mathrm{C} _{1}$
If all the entries in the principle diagonal is 1, then
$^{3}\mathrm{C} _{1}$
$\Rightarrow$  Total matrix $= 12.$

Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

In a square matrix $A$ of order $3$ the elements, $a _{i\ i}s^{'}$ are the sum of the roots of the equation $x^{2}-(a+b)x+ab=0;\ a _{1,\ i+1}s^{'}$ are the product of the roots, $a _{1,\ i-i}s^{'}$ are all unity and the rest of the elements all zero. The value of the det. $(A)$ is equal to

  1. $0$
  2. $(a+b)^{3}$
  3. $a^{3}-b^{3}$
  4. $(a^{2}+b^{2})$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, square matrix $A$ of order $3$ with $a _{ii}= (a+b) $ (sum of roots)

$a _{1 i+1}=a _{12}=ab$  (product of roots)

$a _{1i-1}= a _{10}=1$

$|A|= \begin{vmatrix} (a+b) & 0 & 0 \\ 1 & (a+b) & ab \\ 0 & 0 & (a+b) \end{vmatrix}= (a+b)[(a+b)(a+b)-0]$

$=(a+b)^{3}$

$\therefore $ Option $B$ is correct.


Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\Delta =\begin{vmatrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{vmatrix}$ and ${ A } _{ ij }$ is cofactors of ${ a } _{ ij }$, then the value of $\Delta $ is given by

  1. ${ a } _{ 11 }{ A } _{ 31 }+{ a } _{ 12 }{ A } _{ 32 }+{ a } _{ 13 }{ A } _{ 33 }$
  2. ${ a } _{ 11 }{ A } _{ 11 }+{ a } _{ 12 }{ A } _{ 21 }+{ a } _{ 13 }{ A } _{ 31 }$
  3. ${ a } _{ 21 }{ A } _{ 11 }+{ a } _{ 22 }{ A } _{ 12 }+{ a } _{ 23 }{ A } _{ 13 }$
  4. ${ a } _{ 11 }{ A } _{ 11 }+{ a } _{ 21 }{ A } _{ 21 }+{ a } _{ 31 }{ A } _{ 31 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Delta =\begin{vmatrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{vmatrix}$


Also given ${ A } _{ ij }$ is cofactor of ${ a } _{ ij }$
$\Delta =a _{11} A _{11}+a _{21} A _{21}+a _{31} A _{31}$

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\triangle =\begin{bmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{bmatrix}$ and ${A} _{2},{B} _{2},{C} _{2}$ are respectively cofactors of ${a} _{2},{b} _{2},{c} _{2}$ then ${a} _{1}{A} _{2}+{b} _{1}{B} _{2}+{c} _{1}{C} _{2}$ is equal to ?

  1. $-\triangle$
  2. $0$
  3. $\triangle$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Co-factor of $ \displaystyle a _2 = (-1)^{i+j}\left|\begin{matrix} b _1 & c _1 \ b _3 & c _3 \end{matrix}\right| = - [b _1 \, c _3 - c _1 \, b _3] = A _2 $

i=2  j=1

of $ \displaystyle b _2 = (-1)^{2+2} \left|\begin{matrix} a _1 & c _1 \\ a _3 & c _3 \end{matrix}\right| = [a _1 \, c _3 - c _1 \, a _3] = B _2 $ 

of $ \displaystyle c _2 = (-1)^{2+3} \left|\begin{matrix} a _1 & b _1 \\ a _3 & b _3 \end{matrix}\right| = -[a _1 \, b _3 \, - b _1 \, a _3 ] = C _2 $

$ a _1 \, A _2 + b _1 \, B _2 +c _1 \, C _2 $

$ \displaystyle -a _1 \, b _1 \, c _3 + a _1 \, c _1 \, b _3 + b _1 \, a _1 \, c _3 - b _1 c _1 \, a _3 -c _1 \, a _1 \, b _3 + c _1 \, b _1 \, a _3 $

$ \displaystyle = 0 $

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\Delta = \begin{vmatrix}a _1 & b _1 & c _1 \ a _2 & b _2 & c _2\ a _3 & b _3 & c _3\end{vmatrix}$ and $A _1, B _1, C _1$ denote the co-factors of $a _1, b _1, c _1$ respectively, then teh value os the determinant $\begin{vmatrix}A _1 & B _1 & C _1\ A _2 & B _2 & C _2\ A _3 & B _3 & C _3\end{vmatrix}$ is-

  1. $\Delta$
  2. $\Delta^2$
  3. $\Delta^3$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

$\Delta = \begin{vmatrix}a _1 & b _1 & c _1 \ a _2 & b _2 & c _2\ a _3 & b _3 & c _3\end{vmatrix}$ and $A _1, B _1, C _1$ denote the co-factors of $a _1, b _1, c _1$ respectively.
Now,
 $\begin{vmatrix}A _1 & B _1 & C _1\ A _2 & B _2 & C _2\ A _3 & B _3 & C _3\end{vmatrix}$
$=|adj \Delta|$
$=\Delta^{3-1}$
$=\Delta^2$.

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\Delta  = \left| {\begin{array}{*{20}{c}}  {{a _1}}&{{b _1}}&{{c _1}} \   {{a _2}}&{{b _2}}&{{c _2}} \   {{a _3}}&{{b _3}}&{{c _3}} \end{array}} \right|$ and $A _2$, $B _2$, $C _2$ are respectively cofactors of $a _2,b _2,c _2$ then 


$a _1A _2+b _1B _2+c _1C _2$ is 

  1. $ - \Delta $
  2. $0$
  3. $\Delta $

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\Delta =\begin{vmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \\ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \\ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{vmatrix}$

$A _2=c _3\ b _1+b _1\ c _1$

$B _2=a _1\ c _3+a _3\ b _1$

$C _2=-a _1\ b _3+a _3\ b _1$

$\to \ a _1 A _2+b _1 B _2+c _1 C _2$

$a, b, c _3+a, c, b _3+b, a, c _3-b, a, c _3-b, c, a _3-c, a, b _3+b, c, a _3$

$=0$

$B$ is correct.
Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

The value of a third order determinant is $11$, then the value of the square of the determinant formed by the cofactors will be?

  1. $11$
  2. $121$
  3. $1331$
  4. $14641$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
third order determinant = determinant of $3\times 3$ matrix $A$
given $|A|=11$
det (cofactor matrix of $A$) =set (transpare of cofactor amtrix of $A$) (transpare does not change the det)
=det(adjacent of $A$)
$\left\{det\ (cofactor\ matrix\ of\ A) \right\}^2=\left\{det\ (adjacent\ of\ A)\right\}^2$
(Using for an $n\times n\ det\ (cofactor\ matrix\ of\ A)=det\ (A)^{n-1})$
we get
$det\ (cofactor\ matrix\ of\ A)^2=\left\{det (A)^{3-1}\right\}^2$
$=(11)^{2\times 2}=11^4$
$=146.41$
Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

Consider the determinant, $\Delta=\begin{vmatrix} p & q & r \ x & y & z \ l & m & n \end{vmatrix}$ ${M} _{0}$ denotes the minor of an element in $i$th row and $j$th column and ${C} _{ij}$ denotes the cofactor of an element in $i$th row and $j$th column.
The value of $p.{C} _{21}+q.{C} _{22}+r.{C} _{23}$ is equal to

  1. $0$
  2. $-\Delta$
  3. $\Delta$
  4. ${\Delta}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the property of determinants, if any element is multiplied with cofactor of corresponding element of another row, and summed up for each element of original row, the sum comes out zero $\Rightarrow (A)$

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $A=\left[ \begin{matrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{matrix} \right] $ and $C _{ij}$ is cofactor of $a _{ij}$ in $A$, then value of $|A|$ is given by

  1. $a _{11}C _{31}+a _{12}C _{32}+a _{13}C _{33}$
  2. $a _{11}C _{11}+a _{12}C _{21}+a _{13}C _{31}$
  3. $a _{21}C _{11}+a _{22}C _{21}+a _{23}C _{31}$
  4. $a _{11}C _{11}+a _{21}C _{21}+a _{31}C _{31}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\begin{vmatrix} { a }^{ 2 }+{ \lambda  }^{ 2 } & ab+c\lambda  & ca-b\lambda  \ ab-c\lambda  & { b }^{ 2 }+{ \lambda  }^{ 2 } & bc+a\lambda  \ ca+b\lambda  & bc-a\lambda  & { c }^{ 2 }+{ \lambda  }^{ 2 } \end{vmatrix}\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}={ \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$, then the value of $\lambda$ is

  1. 8

  2. 27

  3. 1

  4. -1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $D=\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}$


determinant of cofactors is

$D^{c}=\begin{vmatrix} { a }^{ 2 }+{ \lambda  }^{ 2 } & ab+c\lambda  & ca-b\lambda  \ ab-c\lambda  & { b }^{ 2 }+{ \lambda  }^{ 2 } & bc+a\lambda  \ ca+b\lambda  & bc-a\lambda  & { c }^{ 2 }+{ \lambda  }^{ 2 } \end{vmatrix}=D^2$

$\begin{vmatrix} { a }^{ 2 }+{ \lambda  }^{ 2 } & ab+c\lambda  & ca-b\lambda  \ ab-c\lambda  & { b }^{ 2 }+{ \lambda  }^{ 2 } & bc+a\lambda  \ ca+b\lambda  & bc-a\lambda  & { c }^{ 2 }+{ \lambda  }^{ 2 } \end{vmatrix}\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}={ \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$

$\Rightarrow D^3= { \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$ -------(1)
Now,
$D=\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}$

 $=\lambda(\lambda^2+a^2)-c(-\lambda c-ab)-b(ac-b\lambda)$
 $=\lambda(\lambda^2+a^2+b^2+c^2)$
from (1)
$\left(\lambda(\lambda^2+a^2+b^2+c^2)\right)^3={ \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$
comparing on both sides gives
$\lambda^3=1$ and $\lambda^2=1$
$\therefore \lambda=1$
Hence, option C.

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

$\begin{vmatrix}a^2 + x^2 & ab - cx & ac + bx\ ab+ cx & b^2 + x^2 & bc - ax\ ac - bx & bc + ax & c^2 + x^2\end{vmatrix} =$

  1. $ \begin{vmatrix}x & b & -c\\ -a & x & c\\ a & -b & x\end{vmatrix}^2$
  2. $ \begin{vmatrix}x & -b & c\\ a & x & -c\\ -a & b & x\end{vmatrix}^2$
  3. $ \begin{vmatrix}x & c & -b\\ -c & x & a\\ b & -a & x\end{vmatrix}^2$
  4. $ \begin{vmatrix}x & -c & b\\ c & x & -a\\ -b & a & x\end{vmatrix}^2$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Let $D = \begin{vmatrix} x & c & -b\ -c & x & a\ b & -a & x\end{vmatrix}$
Cofactors of 1st Row of D are
$x^2 + a^2 ,  ab + cx, ac - bx$
Cofactor of 2nd Row of D are
$ab - cx,x^2 + b^2, ax + bc$
and cofactors of 3rd row of D are
$ax + bx, bc - ax, x^2 + c^2$
$\therefore $ Determinant of cofactors of D is
$D^c

= \begin{vmatrix}x^2 +a^2 & ab + cx & ac - bx\ ab - cx &

x^2 + b^2 & ax + bc\ ac + bx & bc - ax & x^2 +

c^2\end{vmatrix}$
$= \begin{vmatrix}a^2 + x^2 & ab- cx & ac

+ bx\ ab + cx & b^2 + x^2 & bc - ax\ ac - bx & ax +

bc & c^2 + x^2\end{vmatrix}$              (Rows interchanging into columns)
$= D^2$
$= \begin{vmatrix} x & c &-b^2 \ -c

& x & a\ b & -a & x\end{vmatrix}^2$            

($\because D^c = D^2$, D is third order determinant)
Hence,
$\begin{vmatrix}a^2

+ x^2 & ab - cx & ac + bx\ ab+cx & b^2 + x^2 & bc -

ax\ ac - bx & ax + bc &c^2 + x^2 \end{vmatrix}

=  \begin{vmatrix}x & c & -b\ -c & x & a\ b & -a

& x\end{vmatrix}^2$

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\Delta = \begin{vmatrix}a _1 & b _1 & c _1\a _2 & b _2 & c _2\a _3 & b _3 & c _3\end{vmatrix}$ and $A _1, B _1, C _1$ denote the co-factors of $a _1, b _1, c _1$ respectively, then the value of the determinant $\begin{vmatrix}A _1 & B _1 & C _1\A _2 & B _2 & C _2\ A _3 & B _3 & C _3\end{vmatrix}$ is

  1. $\Delta$
  2. $\Delta^2$
  3. $\Delta^3$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $A=\begin{bmatrix} a & c & b\ b & a & c\ c & b & a\end{bmatrix}$ then the cofactor of $a _{32}$ in $A+A^T$ is?

  1. $-(2a(b+c)-(b+c)^2)$
  2. $ac-b^2$
  3. $a^2-bc$
  4. $2a(a+c)-(a+c)^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A + A^T is a symmetric matrix. The element a32 of A+A^T is the sum of the (3,2) element of A and A^T, which is b + c. The cofactor of this element in the resulting matrix involves calculating the minor of the (3,2) position.