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

The matrix form of the linear system $\frac{dx}{dt}=3x-5y\hspace{0.2cm} and\frac {dy}{dt}=4x+8y$ is
  1. $\frac{d}{dt}\bigg(\begin{matrix} \ x \\ \ y \\ \end{matrix}\bigg) = \begin{bmatrix} \ 3 & -5 \\ \ 4 & 8 \\ \end{bmatrix}\bigg(\begin{matrix} \ x \\ \ y \\ \end{matrix}\bigg)$
  2. $\frac{d}{dt}\bigg(\begin{matrix} \ x \\ \ y \\ \end{matrix}\bigg) = \begin{bmatrix} \ 3 & 8 \\ \ 4 & -5 \\ \end{bmatrix}\bigg(\begin{matrix} \ x \\ \ y \\ \end{matrix}\bigg)$
  3. $\frac{d}{dt}\bigg(\begin{matrix} \ x \\ \ y \\ \end{matrix}\bigg) = \begin{bmatrix} \ 4 & -5 \\ \ 3 & 8 \\ \end{bmatrix}\bigg(\begin{matrix} \ x \\ \ y \\ \end{matrix}\bigg)$
  4. $\frac{d}{dt}\bigg(\begin{matrix} \ x \\ \ y \\ \end{matrix}\bigg) = \begin{bmatrix} \ 4 & 8 \\ \ 3 & -5 \\ \end{bmatrix}\bigg(\begin{matrix} \ x \\ \ y \\ \end{matrix}\bigg)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Multiple choice

Multiplication of matrices E and F is G. Matrices E and G are

$E=\begin{bmatrix} \ cos\theta & -sin\theta & 0 \ \ sin\theta & cos\theta & 0 \ \ 0 & 0 & 1 \ \end{bmatrix}and\hspace{0.1cm} G=\begin{bmatrix} \ 1 & 0 & 0 \ \ 0 & 1 & 0 \ \ 0 & 0 & 1 \ \end{bmatrix}.\text{What is the matrix F}?$
  1. $\begin{bmatrix} \ cos\theta & -sin\theta & 0 \\ \ sin\theta & cos\theta & 0 \\ \ 0 & 0 & 1 \\ \end{bmatrix}$
  2. $\begin{bmatrix} \ sin\theta & cos\theta & 0 \\ \ -cos\theta & sin\theta & 0 \\ \ 0 & 0 & 1 \\ \end{bmatrix}$
  3. $\begin{bmatrix} \ cos\theta & sin\theta & 0 \\ \ -sin\theta & cos\theta & 0 \\ \ 0 & 0 & 1 \\ \end{bmatrix}$
  4. $\begin{bmatrix} \ sin\theta & -cos\theta & 0 \\ \ cos\theta & sin\theta & 0 \\ \ 0 & 0 & 1 \\ \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

If $\Delta =\begin{vmatrix}
x+1 & x+2 & x+a\
x+2 & x+3 & x+b\
x+3 & x+4 & x+c
\end{vmatrix}=0$, then
the family of lines $ax+by+c=0$ passes through

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

$\Delta =\begin{vmatrix} x+1 & x+2 & x+a \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c \end{vmatrix}=0$


Applying ${ C } _{ 2 }\rightarrow { C } _{ 2 }-{ C } _{ 1 },{ C } _{ 3 }\rightarrow { C } _{ 3 }-{ C } _{ 1 }$

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

Applying ${ R } _{ 2 }\rightarrow { R } _{ 2 }-{ R } _{ 1 },{ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 1 }$

$\Delta =\begin{vmatrix} x+1 & 1 & a-1 \ 1 & 0 & b-a-1 \ 2 & 0 & c-a-2 \end{vmatrix}=0$

Expanding along ${ C } _{ 2 }$
$1\left( c-a-2 \right) -2\left( b-a-1 \right) =0\ \Rightarrow a-2b+c=0$
From options for point $(1,-2)$ lies on the line $ax+by+c=0$

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

If $\begin{vmatrix} x _1 & y _1 & 1 \ x _2 & y _2 & 1 \ x _3 & y _3 & 1\end{vmatrix}=\begin{vmatrix} a _1 & b _1 & 1\ a _2 & b _2 & 1 \ a _3 & b _3 & 1\end{vmatrix}$, then the two triangles with vertices $(x _1, y _1), (x _2, y _2), (x _3, y _3)$ and $(a _1,b _1)$, $(a _2, b _2)$, $(a _3, b _3)$ must be congruent.

  1. True

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

The two determinants denote twice the area of $\Delta^s$ whose vertices are $(x _1,y _1), (x _2, y _2)$, $(x _3, y _3)$ and $(a _1, b _1)$, $(a _2, b _2), (a _3, b _3)$. This the equality of two determinants implies that their areas are equal. But equality of the areas of two triangles does not imply that they are congruent.

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

If ${ \left( { x } _{ 1 }-{ { x } _{ 2 } } \right)  }^{ 2 }+{ \left( { y } _{ 1 }-{ y } _{ 2 } \right)  }^{ 2 }={ a }^{ 2 }$, ${ \left( x _{ 2 }-{ x } _{ 3 } \right)  }^{ 2 }+{ \left( { y } _{ 2 }-{ y } _{ 3 } \right)  }^{ 2 }={ b }^{ 2 }$, ${ \left( { x } _{ 3 }-{ x } _{ 1 } \right)  }^{ 2 }+{ \left( { y } _{ 3 }-{ y } _{ 1 } \right)  }^{ 2 }={ c }^{ 2 }$ and $k\begin{vmatrix} { x } _{ 1 } & { y } _{ 1 } & 1 \ { x } _{ 2 } & { y } _{ 2 } & 1 \ { x } _{ 3 } & { y } _{ 3 } & 1 \end{vmatrix}=(a+b+c)(b+c-a)(c+a-b)\times (a+b-c)$, then the value of $k$ is

  1. $1$
  2. $2$
  3. $4$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given : $k\begin{vmatrix} { x } _{ 1 } & { y } _{ 1 } & 1 \\ { x } _{ 2 } & { y } _{ 2 } & 1 \\ { x } _{ 3 } & { y } _{ 3 } & 1 \end{vmatrix}^{2}=(a+b+c)(b+c-a)(c+a-b)\times (a+b-c)$ ...... $(i)$

Area of any triangle $PQR$ with vertices $(x _1, y _1) , (x _2, y _2) $ and $(x _3, y _3)$ is given by
$\Delta = \dfrac{1}{2} \begin{vmatrix}x _1 & y _1 & 1\\ x _2 & y _2 & 1\\ x _3 & y _3 & 1\end{vmatrix}$ ......... $(ii)$

${ \left( { x } _{ 1 }-{ { x } _{ 2 } } \right)  }^{ 2 }+{ \left( { y } _{ 1 }-{ y } _{ 2 } \right)  }^{ 2 }={ a }^{ 2 }$, ${ \left( x _{ 2 }-{ x } _{ 3 } \right)  }^{ 2 }+{ \left( { y } _{ 2 }-{ y } _{ 3 } \right)  }^{ 2 }={ b }^{ 2 }$, ${ \left( { x } _{ 3 }-{ x } _{ 1 } \right)  }^{ 2 }+{ \left( { y } _{ 3 }-{ y } _{ 1 } \right)  }^{ 2 }={ c }^{ 2 }$

$\implies a,b,c$ are the length of sides of a $\triangle PQR$

Also, area of $\triangle PQR$ with sides $a, b, c$ is

$\Delta = \sqrt{s (s - a) (s - b) (s - c)}$ ......... $\left[\text{Where } s=\dfrac{a+b+c}{2}\right]$

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

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

From $(ii)$ and $(iii)$, 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

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

Comparing above equation with $(i)$, we get $k=4$