Tag: inverse of a matrix and linear equations

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Questions Related to inverse of a matrix and linear equations

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$=4+4=8$

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

If $A$ is a nonsingular matrix satisfying $AB=BA+A$ then

  1. $\left|B\right|=\left|I+B\right|$
  2. $\left|B\right|=\left|2I+B\right|$
  3. $\left|B\right|=\left|B-I\right|$
  4. $\left|B\right|=\left|B-2I\right|$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given AB = BA + A, we can rewrite this as AB - BA = A. This is a commutation relation. Using properties of determinants and the given condition, it can be shown that |B| = |I + B|.

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

If $A$ and $B$ and square matrix of the same order such that $AB=A$ and $BA=B$, then $A$ and $B$ are both:

  1. Singular

  2. Non-singular

  3. Idempotent

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

If AB = A and BA = B, then A^2 = A(BA) = (AB)A = AA = A^2. Also, A = AB implies |A| = |A||B|. If |A| is not 0, then |B| = 1. However, if A and B are singular, |A| = 0 and |B| = 0, which satisfies the equations.

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

The number of $3\times 3$ non-singular matrices, with four entries as $1$ and all other entries as $0$ is 

  1. Less than $4$
  2. $5$
  3. $6$
  4. At least $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A 3x3 matrix with four 1s and five 0s has a determinant that can be calculated by expanding along rows or columns. By testing various configurations, one finds that many such matrices are singular, but there are indeed more than 7 non-singular configurations possible.

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-singular square matrices and AB=I, then which of the following is true ?

  1. $BA = I$
  2. ${ A }^{ -1 }=B$
  3. ${ B }^{ -1 }=A$
  4. ${ A }^{ 2 }=B$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For square matrices A and B, if AB = I, then A and B are inverses of each other. By the properties of inverse matrices, if AB = I, then BA = I must also hold.

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

If $A$ and $B$ are non-singular matrices, then _____

  1. $(AB)^{-1} = A^{-1}B^{-1}$
  2. $AB = BA$
  3. $(AB)^T = A^T. B^T$
  4. $(AB)^{-1} = B^{-1} A^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For any two invertible matrices A and B of the same order, the inverse of their product is the product of their inverses in reverse order, which is given by the formula (AB)^-1 = B^-1 A^-1.