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 matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

lf $\mathrm{A}$ is $\left{\begin{array}{lll}
8 & -6 & 2\
-6 & 7 & -4\
2 & -4 & \lambda
\end{array}\right}$  is a singular matrix then  $\lambda =$ 

  1. 3

  2. 4

  3. 2

  4. 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $A=\begin{pmatrix}
8 & -6 & 2\
-6 & 7 & -4\
2 & -4 & \lambda
\end{pmatrix}$ is a singular matrix
So, det A=0
$\therefore $ BY operation of matrix (s),
$det A=8(7 \lambda-16)+6[-6 \lambda + 8]+2[24-14]$
$=56 \lambda - 128 -36 \lambda +48 +20$
$=20 \lambda - 60$
So, $det A = 0= 20 \lambda -60$
$\lambda =3$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

A= $\begin{bmatrix}
cos\alpha  & -sin\alpha \
sin\alpha  & cos\alpha
\end{bmatrix}$ ,then find which of the following are correct 
I) A is singular matrix
II) $A^{-1}$=$A^{T}$
III) A is symmetric matrix
IV) $A^{-1}= -A$

  1. only I and II

  2. only II and III

  3. only II

  4. only IV
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\left | A \right |= \cos ^{2}\alpha + \sin ^{2}\alpha = 1$
$A^{T}= \begin{bmatrix}\cos\alpha    & \sin \alpha \ -\sin \alpha  & \cos\alpha \end{bmatrix}\neq A$
$A^{-1}= \frac{1}{1}\begin{bmatrix}\cos\alpha    & \sin \alpha \ -\sin \alpha  & \cos\alpha \end{bmatrix}\neq -A$
$\begin{bmatrix}A^{T}= A^{-1}\end{bmatrix}$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If A=$\displaystyle \begin{vmatrix} 5 & -3   \ 4 & 2   \end{vmatrix}$ then find $\displaystyle AA^{-1}$

  1. $\displaystyle \begin{vmatrix} 0 & 0 \\ 0 & 0 \end{vmatrix}$
  2. $\displaystyle \begin{vmatrix} -1 & 0 \\ 0 & -1 \end{vmatrix}$
  3. $\displaystyle \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$

  4. Does not exist
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For any given square matrix , $ A{A}^{-1} $ is always equal to Identity Matrix $ I $

So, $ A{A}^{-1} = \begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} $

Multiple choice maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Which of the following matrices is not invertible?

  1. $\displaystyle \left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right] $
  2. $\displaystyle \left[ \begin{matrix} -1 & -1 \\ -1 & 2 \end{matrix} \right] $
  3. $\displaystyle \left[ \begin{matrix} 2 & 3 \\ 4 & 6 \end{matrix} \right] $
  4. $\displaystyle \left[ \begin{matrix} 2 & -2 \\ 1 & 1 \end{matrix} \right] $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A square matrix that is not invertible is called singular matrix in which its determinant is 0.
$\displaystyle \left[\begin{matrix} 2 & 3 \ 4 & 6 \end{matrix}  \right]$ is a non invertible matrix.
$\because$ its determinant is    $12-12 =0$
Option C is correct.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If the matrix $\displaystyle \left[ \begin{matrix} a \ c \end{matrix}\begin{matrix} b \ d \end{matrix} \right] $ is commutative with the matrix $\displaystyle \left[ \begin{matrix} 1 \ 0 \end{matrix}\begin{matrix} 1 \ 1 \end{matrix} \right] $, then

  1. $a=0, b=c$
  2. $b=0, c=d$
  3. $c=0, d=a$
  4. $d=0, a=b$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is commutative with the matrix $\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$


So,

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$\begin{bmatrix} a & a+b \\ c & c+d \end{bmatrix}=\begin{bmatrix} a+c & b+d \\ c & d \end{bmatrix}$

$a=a+c,\quad a+b=b+d$
$c=0,\quad a=d$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A$ is a square matrix of order $3$ and det $A = 5$, then what is det $[(2A)^{-1}]$ equal to?

  1. $\dfrac{1}{10}$
  2. $\dfrac{2}{5}$
  3. $\dfrac{8}{5}$
  4. $\dfrac{1}{40}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If $A$ is of order $3$ then,$A^-1$ is also of order $3$.

Now, $\text{det} (cA)=c^n(\text{ det} A)$ where $n$ is the order of the matrix.
And, $\text {det } A=\dfrac{1}{\text{det} A^{-1}}$
Thus $\text{det} [(2A)^{-1}]=2^3\text{det}[A]$
and $\text{det}[A^{-1}]=\dfrac{1}{8.5}$
$=\dfrac{1}{40}$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If A is a square matrix such that $A^2 = I $ where I is the identity matrix, then what is $A^{-1}$ equal to ?

  1. A + 1

  2. Null matrix

  3. A

  4. Transpose of A
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$A\rightarrow $square matrix
${ A }^{ 2 }=I$
To find ${ A }^{ -1 }=?$
Given ${ A }^{ 2 }=I$
$\Rightarrow A\times A=I$
$\Rightarrow A=\cfrac { I }{ A } $
$\Rightarrow A={ A }^{ -1 }$
$\therefore { A }^{ -1 }=A$
Option C is correct
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If A is an orthogonal matrix of order 3 and $B=\begin{bmatrix}1&2&3\-3&0&2\2&5&0\end{bmatrix}$, then which of the following is/are correct?
1. $|AB|= \pm 47$
2. $AB=BA$
Select the correct answer using the code given below :

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solution:

$A$ is an orthogonal matrix.
$A^2=1$
$|A|^2=1$
$|A|=\pm 1$
$|AB|=\pm|B|$
Now, $|B|=\begin{bmatrix}1&2&3\-3&0&2\2&5&0\end{bmatrix}=47$
$\therefore |AB|=\pm47$
And $AB=\begin{bmatrix}3&7&3\-3&0&2\2&5&0\end{bmatrix}$ and 
$BA=\begin{bmatrix}1&2&3\-3&0&2\2&5&0\end{bmatrix}$
Since, $AB\neq BA$
Hence, A is the correct option.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If A is a non singular matrix satisfying $A=AB-BA$, then which one of the following holds true

  1. $det. B=0$
  2. $B=0$
  3. $det. A=1$
  4. $det(B+I) =det(B-I)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A=AB-BA$
$AI+BA=AB$
$A(I+B)A=AB$
$\left| I+B \right| \left| A \right| =\left| A \right| \left| B \right| $
$\left| B \right| =\left| I+B \right| \longrightarrow 1$
$A=AB-BA$
$BA=AB-A$
$BA=A(B-I)$
$\left| B \right| \left| A \right| =\left| A \right| \left| B-I \right| $
$\left| B \right| =\left| B-I \right| \longrightarrow 2$
From equation 1 and 2
$\left| B-I \right| =\left| B+I \right| $

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $AB=0$ for the matrices
$A=\left[ \begin{matrix} \cos ^{ 2 }{ \theta  }  & \cos { \theta  } \sin { \theta  }  \ \cos { \theta  } \sin { \theta  }  & \sin ^{ 2 }{ \theta  }  \end{matrix} \right] $ and $B=\left[ \begin{matrix} \cos ^{ 2 }{ \phi  }  & \cos { \phi  } \sin { \phi  }  \ \cos { \phi  } \sin { \phi  }  & \sin ^{ 2 }{ \phi  }  \end{matrix} \right] $ then $\theta-\phi $ is

  1. an odd multiple of $\dfrac{\pi}{2}$
  2. an odd multiple of ${\pi}$
  3. an odd even of $\dfrac{\pi}{2}$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A=\begin{bmatrix} { \cos {  }  }^{ 2 }\theta  & \cos { \theta  } \sin { \theta  }  \\ \cos { \theta  } \sin { \theta  }  & { \sin {  }  }^{ 2 }\theta  \end{bmatrix}$      $B=\begin{bmatrix} { \cos {  }  }^{ 2 }  & \cos { \phi  } \sin { \phi  }  \\ \cos { \phi  } \sin { \phi  }  & { \sin {  }  }^{ 2 }\phi  \end{bmatrix}$

$AB = \begin{bmatrix} { \cos {  }  }^{ 2 }\theta  & \cos { \theta  } \sin { \theta  }  \\ \cos { \theta  } \sin { \theta  }  & { \sin {  }  }^{ 2 }\theta  \end{bmatrix}$  $ \begin{bmatrix} { \cos {  }  }^{ 2 }\phi  & \cos { \phi  } \sin { \phi  }  \\ \cos { \phi  } \sin { \phi  }  & { \sin {  }  }^{ 2 }\phi  \end{bmatrix}$

$=\begin{bmatrix} { \cos {  }  }^{ 2 }\theta \cos^{2}\phi + \cos { \theta  } \sin { \theta  } \cos\phi \sin\phi & {\cos^2\theta \cos\phi \sin\phi+\sin^2\phi \sin\theta \cos\theta} \\ \cos { \theta  } \sin { \theta  } \cos^{2}\phi  + { \sin {  }  }^{ 2 }\theta \cos\phi  \sin\phi  & \cos\theta \sin\theta \cos\phi  { \sin {  }  }^{ 2 }\phi  +\sin^{2}\theta\sin^{2}\phi   \end{bmatrix}$

$= \begin{bmatrix} 0  & 0  \\ 0  & 0  \end{bmatrix}$  

$\Rightarrow $ $\begin{bmatrix} { \cos {  }  }^{ 2 }\theta \cos \phi \cos (\theta -\phi)  & \cos { \theta  } \sin { \phi  } \cos (\theta-\phi) \\ \sin {\phi  } \cos {\phi   }  & { \sin {  }  }^{ 2 }\theta \sin \phi \cos (\theta -\phi)  \end{bmatrix}$ = $\begin{bmatrix} 0  & 0  \\ 0  & 0  \end{bmatrix}$

$\Rightarrow$ $\cos (\theta - \phi )$$\begin{bmatrix} { \cos {  }  }^{ 2 }\theta \cos \phi  & \cos { \theta  } \sin { \phi  }  \\ \sin {\phi  } \cos {\phi   }  & { \sin {  }  }^{ 2 }\theta \sin \phi   \end{bmatrix}$ = $\begin{bmatrix} 0  & 0  \\ 0  & 0  \end{bmatrix}$

$\Rightarrow \cos (\theta-\phi) (\cos \theta \cos \phi \sin \theta \sin \phi - \cos \theta \cos \phi \sin \theta \sin \phi ) = 0$

$\Rightarrow \cos (\theta - \phi) = 0$

$\theta - \phi = (2n+1)\dfrac{\pi}{2}$

i.e an odd multiple of $\dfrac{\pi}{2}$ 
Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices
Let $A$ be a matrix of order $2 \times 2$ such that $A^2 = 0$ then $A^2 - (a + d)A + (ad - bc) I$ is equal to
  1. $I$
  2. $0 _{2\times 2}$
  3. $-I$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$

$A^2=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}=\begin{pmatrix} a^2+bc & ab+bd \\ ca+cd & bc+d^2 \end{pmatrix}............................(1)$

$-(a+d)A=\begin{pmatrix} -a^2-ad & -ab-bd \\ -ac-cd & -ad-d^2 \end{pmatrix}..............(2)$

$(ad-bc)I=\begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix}.............(3)$
Adding 1,2,3 we get,
$A^2-(a+d)A+(ad-bc)I=\begin{pmatrix} a^2+bc-a^2-ad+ad-bc & ab+bd-ab-bd \\ ac+cd-ac-cd & bc+d^2-ad-d^2+ad-bc \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

Hence, the value of $A^2-(a+d)A+(ad-bc)I=0 _{2\times 2}$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Let $A$ and $B$ are two matrices such that $AB =BA$, then for every $n\in N$,

  1. $A^nB=BA^n$
  2. $(AB)^n = A^nB^n$
  3. $(A+B)^n=$ $^nC _0A^n+$ $^nC _1A^{n-1}B^1+$ $^nC _2A^{n-2}B^2+ ... + ^nC _n\:B^n$.
  4. $A^{2n}-B^{2n}=(A^n-B^n)(A^n+B^n)$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

$A^2B =A(AB) =A(BA) =(AB)A$

$= (BA)A =BA^2$

Similarly, $A^3B=BA^3$

In general $A^nB=BA^n: \forall : n\geq 1$

(b) and (c) hold as $AB =BA$.

Also, $(A^n -B^n) (A^n + B^n)$

$=A^nA^n-B^n:A^n+A^n:B^n-B^n:B^n$

$=A^{2n}-B^{2n}$

Hence, options A,B,C and D.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $D _1$ and $D _2$ are two $3\times 3$ diagonal matrices, then

  1. $D _1\:D _2$ is diagonal matrix
  2. $D _1\:D _2=D _2\:D _1$
  3. $D _1^2+D _2^2$ is a diagonal matrix

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

Let ${ D } _{ 1 }=\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix},{ D } _{ 2 }=\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}$

Then

${ D } _{ 1 }{ D } _{ 2 }=\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}=\begin{bmatrix} ax & 0 & 0 \ 0 & by & 0 \ 0 & 0 & cz \end{bmatrix}\$


$ { D } _{ 2 }{ D } _{ 1 }=\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}=\begin{bmatrix} xa & 0 & 0 \ 0 & yb & 0 \ 0 & 0 & zc \end{bmatrix}$

As $ax=xa,by=yb,cz=zc$

${ D } _{ 1 }{ D } _{ 2 }={ D } _{ 2 }{ D } _{ 1 }$

${ { D } _{ 1 } }^{ 2 }+{ { D } _{ 2 } }^{ 2 }=\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}\begin{bmatrix} a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{bmatrix}+\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}\$

$ =\begin{bmatrix} { a }^{ 2 } & 0 & 0 \ 0 & { b }^{ 2 } & 0 \ 0 & 0 & { c }^{ 2 } \end{bmatrix}+\begin{bmatrix} { x }^{ 2 } & 0 & 0 \ 0 & { y }^{ 2 } & 0 \ 0 & 0 & { z }^{ 2 } \end{bmatrix}=\begin{bmatrix} { { a }^{ 2 }+x }^{ 2 } & 0 & 0 \ 0 & { { b }^{ 2 }+y }^{ 2 } & 0 \ 0 & 0 & { { c }^{ 2 }+z }^{ 2 } \end{bmatrix}$