Tag: determinants

Quizzes (5) Questions (79)

Questions Related to determinants

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

The number of value of $x$ in the closed interval $[-4,-1]$, the matrix $\begin{bmatrix} 3 & -1+x & 2 \ 3 & -1 & x+2 \ x+3 & -1 & 2 \end{bmatrix}$ is singular is 

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For a singular matrix the value of the determinant = 0 
$ [3(-1)(2)+(-1+x)(x+2)(x+3)+(2)(3)(-1)] $
$\times  [-(2)(-1)(x+3)-(-1+x)(3)(2)-(3)(x+2)(x+3)] = 0 $
$-6+x^3+4x^2+x-6-6+2x+6+6-6x-3x^2-15x-18 = 0 $
$ x^3+x^2-18x-24 = 0 $
$ x = -4 $
$x =(3-\sqrt{33})/2=-1.372 $
$x =(3+\sqrt{33})/2= 4.372 $

Value of x in the closed interval $[-4,-1]$ are $-4,   -1.372$

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

If $\left[ {\begin{array}{*{20}{c}}1&{ - 1}&x\1&x&1\x&{ - 1}&1\end{array}} \right]$ has no inverse, then the real value of $x$ is 

  1. $2$
  2. $3$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A=\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}has\quad no\quad inverse\quad x$

$if\,\left| A \right| = 0$

$\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}=0$

$1\left( {x + 1} \right) - 1\left( { - 1 + x} \right) + x\left( { - 1 - {x^2}} \right) = 0$

$x + 1 + 1 - x - x - {x^3} = 0$

$ - {x^3} - x + 2 = 0$

${x^3} + x - 2 = 0$

$\left( {x - 1} \right)\left( {{x^2} + x + 2} \right) = 0$

$x = 1\,is\,real\,value\,$

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

The matrix $\begin{bmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \ 3 & 1 & 1 \end{bmatrix}$ is:

  1. nonsingular

  2. singular

  3. skew symmetric

  4. symmetric
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{vmatrix} 1 & 0 & 1 \ 2 & 1 & 0 \end{vmatrix}=1\begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix}-0\begin{vmatrix} 2 & 0 \ 3 & 1 \end{vmatrix}+1\begin{vmatrix} 2 & 1 \ 3 & 1 \end{vmatrix}$

$=1(1-0)-0+1(2-3)$
$=1-1$
$=0$
Therefore, it is a singular matrix.

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

If $\begin{bmatrix} 1 & 2 & x \  4 & -1 & 7  \  2 & 4 & 6  \end{bmatrix}$ is a singular matrix, then $x=$

  1. $0$
  2. $1$
  3. $-3$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{vmatrix} 1 & 2 & x \ 4 & -1 & 7 \ 2 & 4 & -6  \end{vmatrix}=0$

$\Rightarrow 1\begin{vmatrix} -1 & 7 \ 4 & -6 \end{vmatrix}-2\begin{vmatrix} 4 & 7 \ 2 & -6 \end{vmatrix}+x\begin{vmatrix} 4 & -1 \ 2 & 4 \end{vmatrix}=0$

$\Rightarrow (6-28)-2(-24-14)+x(16+2)=0$
$\Rightarrow -22+76+18x=0$
$\Rightarrow 18x=-54$

$\Rightarrow x=-3$

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

If the matrix $\begin{bmatrix} \alpha  & 2 & 2 \ -3 & 0 & 4 \ 1 & -1 & 1 \end{bmatrix}$ is not invertible, then:

  1. $\alpha =-5$
  2. $\alpha =5$
  3. $\alpha =-0$
  4. $\alpha =-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a matrix to be not invertible, determinant of the matrix should be $0$
So, $\left| \begin{matrix} \alpha  & 2 & 2 \ -3 & 0 & 4 \ 1 & -1 & 1 \end{matrix} \right| =0$

$\Rightarrow -(-3)(2-(-2))-4(-\alpha -2)=0\ \Rightarrow 12+4\alpha +8=0\ \Rightarrow 4\alpha =-20\ \Rightarrow \alpha =-5$
Hence, option A is correct

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

Consider the following statements:
1. The matrix
               $\begin{pmatrix} 1 & 2 & 1 \ a & 2a & 1 \ b & 2b & 1 \end{pmatrix}$ is singular.
2. The matrix
              $\begin{pmatrix} c & 2c & 1 \ a & 2a & 1 \ b & 2b & 1 \end{pmatrix}$ is non-singular.
Which of the above statements is/are correct?

  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

For matrix

  $\begin{vmatrix} 1 & 2 & 1 \ a & 2a & 1 \ b & 2b & 1 \end{vmatrix}$


$C _2\rightarrow C _2-2C _1 $

$\begin{vmatrix} 1 & 0 & 1 \\ a & 0 & 1 \\ b & 0 & 1 \end{vmatrix}$ The determinant is $0$. So the matrix is singular.

For matrix,
$ \begin{vmatrix} c & 2c & 1 \\ a & 2a & 1 \\ b & 2b & 1 \end{vmatrix}$

$C _2\rightarrow C _2-2C _1 $

$\begin{vmatrix} c & 0 & 1 \\ a & 0 & 1 \\ b & 0 & 1 \end{vmatrix}$ The determinant is $0$. So the matrix is singular.

Thus, only $1$ is true.
Hence, option A is correct.

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 all of whose entries are integers. Then which one of the following is true?

  1. If $det(A)=\pm 1$, then ${A}^{-1}$ exists but all its entries are not necessarily integers.
  2. If $det(A)=\pm 1$, then ${A}^{-1}$ exists and all its entries are non integers
  3. If $det(A)=\pm 1$, then ${A}^{-1}$ exists and all its entries are integers
  4. If $det(A)=\pm 1$, then ${A}^{-1}$ need not exist
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a matrix A with integer entries, the inverse is given by (1/det(A)) * adj(A). If det(A) = +/- 1, then 1/det(A) is +/- 1. Since the adjugate matrix consists of determinants of integer submatrices, it also contains only integers. Thus, A^-1 has only integer entries.

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

If $A$ and $B$ are two non-zero square matrices of the same order such that the product $AB=0$, then

  1. both $A$ and $B$ must be singular

  2. exactly one of them must be singular

  3. both of them are non singular

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

$AB=O$
taking determinant on both sides
$|AB|=|A||B|=0$
$\Rightarrow$ Either $A$ or $B$ should be a singular matrix.
Hence, option D.

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

Let $A=\begin{bmatrix} a & b\ c & d\end{bmatrix}$ be a $2\times 2$ matrix, where a, b, c and d take the values $0$ or $1$ only. The number of such matrices which have inverses is?

  1. $8$
  2. $7$
  3. $6$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ A }^{ -1 }=\frac { adj\quad A }{ \left| A \right|  } $

so $\left| A \right| \neq 0\ ad-bc\neq 0$
for  $ ad-bc\neq 0$ 
either $ad=1 ,bc=0$ or $ad=0 ,bc=1$
$ad=1 ,bc=0$ can be chosen in 4 ways 
similarly $ad=0 ,bc=1$ can be chosen in 4 ways
so number of matrices  formed$=4+4=8$