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 applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

One of the roots of $\begin{vmatrix} x+a & b & c\  a & x+b & c\  a & b & x+c \end{vmatrix}=0$ is :

  1. $abc$
  2. $a+b+c$
  3. $-(a+b+c)$
  4. $-abc$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For the determinant
$ column\ 1=column\ 1+column\ 2+column \ 3\\ \left| \begin{matrix} x+a+b+c & b & c \\ x+a+b+c & \quad x+b & c \\ x+a+b+c & b & x+c \end{matrix} \right| =0\\ \\ row2=row2-row1\\ row3=row3-row1\\ \begin{vmatrix} x+a+b+c & \quad b & \quad c \\ 0 & \quad x & \quad 0 \\ 0 & \quad 0 & \quad x \end{vmatrix}=0$
Expanding the determinant 
$\Rightarrow  (x+a+b+c)(x.x)-b.0+c.0=0$ 
$\Rightarrow { x }^{ 2 }(x+a+b+c)=0$ 
Then, the roots are $x=0$ or $x=-(a+b+c)$
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

If AX = B where A is $3 \times 3$ and X and B are $3\times 1$ matrices then which of the following is correct?

  1. If | A | = 0 then AX = B has infinite solutions

  2. If AX = B has infinite solutions then | A | = 0
  3. If (adj (A)) B = 0 and | A | $\neq$ 0 then AX = B has unique solution
  4. If (adj (A)) B $\neq$ 0 & |A| = 0 then AX = B has no solution
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

Let $S$ be the set of all column matrices $\begin{bmatrix}b _{1}\b _{2} \ b _{3}
\end{bmatrix}$ such that $b _{1}, b _{2}, b _{3} \  \epsilon \  \mathbb {R}$ and the system of equation (in real variables) 
$-x + 2y + 5z = b _{1}$
$2x - 4y + 3z = b _{2}$
$x - 2y + 2z = b _{3}$
has at least one solution. Then, which of the following system(s) (in real variables) has/have at least one solution of each $\begin{bmatrix}b _{1}\ b _{2}\ b _{3}
\end{bmatrix}\epsilon \  S$?

  1. $x + 2y + 3z = b _{1}, 4y + 5z = b _{2}$ and $x + 2y + 6z = b _{3}$
  2. $x + y + 3z = b _{1}, 5x + 2y + 6z = b _{2}$ and $-2x - y - 3z = b _{3}$
  3. $-x + 2y - 5z = b _{1}, 2x - 4y + 10z = b _{2}$ and $x - 2y + 5z = b _{3}$
  4. $x + 2y + 5z = b _{1}, 2x + 3z = b _{2}$ and $x + 4y - 5z = b _{3}$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

We find $D = 0$, where $D$ is the determinant formed by the coefficients of $x,\ y,\ z$ in the three equations and since no pair of planes are parallel, so there is an infinite number of solutions.
Let $\alpha P _{1} + \lambda P _{2} = P _{3}$
$\Rightarrow P _{1} + 7P _{2} = 13P _{3}$
$\Rightarrow b _{1} + 7b _{2} = 13b _{3}$
(A) $D\neq 0\Rightarrow$ unique solution for any $b _{1}, b _{2}, b _{3}$
(B) $D = 0$ but $P _{1} + 7P _{2} \neq 13P _{3}$
(C) $D = 0$ Also $b _{2} = -2b _{1}, b _{3} = -b _{1}$
Satisfied $b _{1} + 7b _{2} = 13b _{3}$ (Actually all three planes are co-incident)
(D) $D\neq 0$.

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

If $a{ e }^{ x }+b{ e }^{ y }=c;\quad p{ e }^{ x }+q{ e }^{ y }=d$ and $\quad { \Delta  } _{ 1 }=\begin{vmatrix} a & b \ p & q \end{vmatrix};{ \Delta  } _{ 2 }=\begin{vmatrix} c & b \ d & q \end{vmatrix};{ \Delta  } _{ 3 }=\begin{vmatrix} a & c \ p & d \end{vmatrix}$ then the value of $(x,y)$ is:

  1. $\left( \cfrac { { \Delta } _{ 2 } }{ { \Delta } _{ 1 } } ,\cfrac { { \Delta } _{ 3 } }{ { \Delta } _{ 1 } } \right) $
  2. $\left( \log { \cfrac { { \Delta } _{ 2 } }{ { \Delta } _{ 1 } } } ,\log { \cfrac { { \Delta } _{ 3 } }{ { \Delta } _{ 1 } } } \right) $
  3. $\left( \log { \cfrac { { \Delta } _{ 1 } }{ { \Delta } _{ 3 } } } ,\log { \cfrac { { \Delta } _{ 1 } }{ { \Delta } _{ 2 } } } \right) $
  4. $\left( \log { \cfrac { { \Delta } _{ 1 } }{ { \Delta } _{ 2 } } } ,\log { \cfrac { { \Delta } _{ 1 } }{ { \Delta } _{ 3 } } } \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let u=e^x and v=e^y. The system is au+bv=c and pu+qv=d. By Cramer's Rule, u = Delta2/Delta1 and v = Delta3/Delta1. Since u=e^x and v=e^y, x=log(Delta2/Delta1) and y=log(Delta3/Delta1).

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $A=\begin{bmatrix} a & b\ 0 & a\end{bmatrix}$ is nth root of $I _2$, then choose the correct statements.

  1. If n is odd, $a=1$, $b=0$
  2. If n is odd, $a=-1, b=0$
  3. If n is even, $a=1, b=0$
  4. If n is even, $a=-1, b=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a matrix A = [[a, b], [0, a]] to be an nth root of I, A^n = I. This implies a^n = 1. If n is odd, a=1, b=0 is a solution. If n is even, a=1 or a=-1 are possible.

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

If $A=\begin{bmatrix} \cos { x }  & \sin { x }  \ -\sin { x }  & \cos { x }  \end{bmatrix}$ and $A(AdjA)=k\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ then the value of $k$ is

  1. $\sin{x}\cos{x}$
  2. $1$
  3. $-1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$A(AdjA)=\begin{bmatrix}c^{2}+s^{2}&cs-cs\-cs+cs&c^{2}+s^{2}\end{bmatrix}=\begin{bmatrix}1&0\0&1\end{bmatrix}\implies k=1$

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

If $A=\left[ \begin{matrix} 2 & -3 \ -4 & 7 \end{matrix} \right] $, then ${2A}^{-1}=$

  1. $81-2A$
  2. $91-A$
  3. $31-2A$
  4. $A-91$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$A=\left[ { \begin{array} { *{ 20 }{ c } }2 & { -3 } \ { -4 } & 7 \end{array} } \right]  \ \left| { A-\lambda I } \right| =0 \ \left| { \begin{array} { *{ 20 }{ c } }{ 2-\lambda  } & { -3 } \ { -4 } & { 7-\lambda  } \end{array} } \right| =0 \ \left( { 2-\lambda  } \right) \left( { 7-\lambda  } \right) -12=0 \ \left( { \lambda -7 } \right) \left( { \lambda -2 } \right) -12=0 \ { \lambda ^{ 2 } }-9\lambda +2=0 \ { A^{ 2 } }-9A+2I=0 \ A-9I+2{ A^{ -1 } }=0 \ 2{ A^{ -1 } }=9I-0 \ 2{ A^{ -1 } }=9I-A$


$ \ Hence,\, option\, B\, is\, the\, correct\, answer.$

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

$A=\begin{bmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos\theta\end{bmatrix}$ and $AB=BA=I$, then B is equal to

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

Given, $A=\begin{bmatrix} \cos\theta & -\sin\theta \ \sin\theta & \cos\theta \end{bmatrix}$ and $AB=BA=I$
$\Rightarrow B=A^{-1}I=A^{-1}$
$=\displaystyle\frac{1}{\cos^2\theta +\sin^2\theta}\begin{bmatrix} \cos\theta & \sin\theta \ -\sin\theta & \cos\theta\end{bmatrix}$
$\Rightarrow B=\begin{bmatrix}\cos\theta & \sin\theta \ -\sin\theta & \cos\theta\end{bmatrix}$

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

$A=\begin{bmatrix} 2&2&1\0&1&4\0&2&6\end{bmatrix}$, $B=\begin{bmatrix} 2&2&1\0&1&4\0&0&1\end{bmatrix}$

To obtain B from the matrix A, order of operations would be   

  1. $R _3 \rightarrow R _3-3R _1$, $R _3\rightarrow R _2-R _1$
  2. $R _3 \rightarrow R _1-2R _2$, $R _3 \rightarrow (R _3 \times {-2})$
  3. $R _2 \rightarrow R _2-2R _2$, $R _3 \rightarrow (R _3 \div {2})$
  4. $R _3 \rightarrow R _3-2R _2$, $R _3 \rightarrow (R _3 \div {-2})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$A=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 2 & 6 \end{bmatrix}B=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$
From A to B
${ R } _{ 1 }$ & ${ R } _{ 2 }$ are same
changing ${ R } _{ 3 }$ w.r.t. ${ R } _{ 2 }$
Clearly the term $'2'$ in ${ R } _{ 3 }{ C } _{ 2 }$ in A is change to $'0'$ in comparing them we find
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 2 }$
We get the matrix
$\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & -2 \end{bmatrix}$
Now if we divide ${ R } _{ 3 }$ w.r.t. $'-2'$
We get the derived term $1$ in ${ R } _{ 3 }{ C } _{ 3 }$ of B
$\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}=B$
$\therefore $Operations are
${ R } _{ 3 }\rightarrow { R } _{ 3 }-2{ R } _{ 2 }$
 and ${ R } _{ 3 }\rightarrow { R } _{ 3 }-(-2)$
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

A= $\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$. 

B is matrix obtained by subtracting $4 \ times \ 1^{st}\ row\ from \ 2^{nd} \ row$ of A. Find matrix B

  1. $\begin{bmatrix} 1&2&3\\0&3&6\\0&6&12\end{bmatrix}$
  2. $\begin{bmatrix} 1&2&3\\7&0&0\\4&5&6\end{bmatrix}$
  3. $\begin{bmatrix} 1&2&3\\0&1&2\\3&4&5\end{bmatrix}$
  4. $\begin{bmatrix} 1&2&3\\0&-3&-6\\7&8&9\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given $A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
$\Rightarrow { R } _{ 2 }\rightarrow { R } _{ 2 }-4{ R } _{ 1 }$ (for matrix B)
$\Rightarrow B-\begin{bmatrix} 1 & 2 & 3 \\ 4-4(1) & 5-4(2) & 6-4(3) \\ 7 & 8 & 9 \end{bmatrix}=\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 7 & 8 & 9 \end{bmatrix}$
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

$A=\begin{bmatrix} 2&2&1\4&5&6\6&8&9\end{bmatrix}$, $B=\begin{bmatrix} 2&2&1\0&1&4\0&2&6\end{bmatrix}$

To convert matrix A into matrix B, the order of row operations are 

  1. $R _1\rightarrow R _2-2R _1$, $R _3 \rightarrow R _3-R _1$
  2. $R _2\rightarrow R _2-2R _1$, $R _3 \rightarrow R _3-3R _1$
  3. $R _1\rightarrow R _1-2R _2$, $R _3 \rightarrow R _1-R _3$
  4. $R _2\rightarrow R _3-2R _3$, $R _3 \rightarrow R _1-R _1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$A=\begin{bmatrix} 2 & 2 & 1 \\ 4 & 5 & 6 \\ 6 & 8 & 9 \end{bmatrix}\quad B=\begin{bmatrix} 2 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 2 & 6 \end{bmatrix}$
Clearly in order to convert matrix A to B. Both ${ R } _{ 2 }$ and ${ R } _{ 3 }$ are changed w.r.t. ${ R } _{ 1 }$
${ R } _{ 2 }^{ 1 }\leftrightarrow X{ R } _{ 2 }-Y{ R } _{ 1 }$
For ${ R } _{ 2 }=4$ and ${ R } _{ 1 }=2$ then ${ R } _{ 2 }^{ 1 }=0$
$\Rightarrow 0=X4-2$Y$
$\Rightarrow 2X=Y$
For ${ R }_{ 2 }=5$ and ${ R }_{ 1 }=1$ then ${ R }_{ 2 }^{ 1 }=1$
$1=X(5)-Y(2)$
$\Rightarrow X=1\quad Y=2$
Relation is ${ R }_{ 2 }\leftrightarrow { R }_{ 2 }-2{ R }_{ 21}$
Parallely ${ R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }$
Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Multiply the fourth row by $3$.
$\begin{bmatrix}3&4&2&11\9&1&0&0\0&1&0&2\0&0&6&1\end{bmatrix}$

  1. $0, 0, 18, 3$
  2. $0, 3, 0, 6$
  3. $0, 0, 24, 4$
  4. $9, 12, 6, 33$
  5. $0, 0, 12, 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Fourth row = $0,0,6,1$
multiplying by $3$
$(0\times 3 , 0\times 3 , 6\times 3 , 1\times 3)$
$(0,0,18,3)$

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

$\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$

The new matrix obtained after  adding $2^{nd} \  row \ to\  3\ times\  3^{rd} \ row $ is

  1. $\begin{bmatrix} 1&2&3\\4&5&6\\25&29&33\end{bmatrix}$
  2. $\begin{bmatrix} 1&2&3\\25&-29&-33\\4&5&6\end{bmatrix}$
  3. $\begin{bmatrix} 1&2&3\\7&8&9\\4&5&6\end{bmatrix}$
  4. $\begin{bmatrix} 1&2&3\\-25&-29&-33\\4&5&6\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
$\Rightarrow { R } _{ 3 }\rightarrow 3{ R } _{ 3 }+{ R } _{ 2 }$ for new matrix
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3(7)+4 & 3(8)+5 & 3(9)+6 \end{bmatrix}$
New matrix $=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 25 & 29 & 33 \end{bmatrix}$
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= $\begin{bmatrix} 1&2&3\4&5&6\7&8&9\end{bmatrix}$.

B is matrix obtained by subtracting $4\ times 1^{st}\ row from \ 2^{nd} \ row$ of A.
C is matrix obtained by subtracting $7 \ times \ 1^{st}\ row\ from\ 3^{rd} row$, then $C$ is 

  1. $\begin{bmatrix} 1&2&3\\0&3&6\\0&6&12\end{bmatrix}$
  2. $\begin{bmatrix} 1&2&3\\7&0&0\\4&5&6\end{bmatrix}$
  3. $\begin{bmatrix} 1&2&3\\0&1&2\\3&4&5\end{bmatrix}$
  4. $\begin{bmatrix} 1&2&3\\0&-3&-6\\0&-6&-12\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given $A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
B is obtained by substracting $4{ R } _{ 1 }$ from ${ R } _{ 2 }$
$\Rightarrow { R } _{ 2 }\leftrightarrow { R } _{ 2 }-4{ R } _{ 1 }$$\rightarrow B\sim \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 7 & 8 & 9 \end{bmatrix}$
C is obtained by subtracting $7{ R } _{ 1 }$ from ${ R } _{ 3 }$
$\Rightarrow { R } _{ 3 }\leftrightarrow { R } _{ 3 }-7{ R } _{ 1 }$
$C\sim \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{bmatrix}$
$\therefore $option D is correct