Mathematics

Linear Algebra

510 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 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

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

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.

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

If 3, -2 are the Exigent values of non-singular matrix A and |A|=4. Then Exigent values of Adj(A) are

  1. 3/4, -1/2

  2. 4/3, -2

  3. 12, -8

  4. -12, 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} { \lambda _{ 1 } }=3,\, \, { \lambda _{ 2 } }=-2 \ \left| A \right| =4 \ adj\left( A \right) =\left( A \right) \cdot { A^{ -1 } } \ Ax=\lambda x \ \frac { 1 }{ \lambda  } x={ A^{ -1 } }x \ \left( { \lambda I\cdot { A^{ -1 } } } \right) =\frac { { \left( A \right) \cdot \left( { \lambda I-{ A^{ -1 } } } \right)  } }{ { \left( A \right)  } }  \ exigent\, value\, of\, adj\left( A \right) \, is\, \frac { { \left( A \right)  } }{ { exigent\, value\, of\, A } }  \ =\frac { 4 }{ 3 } ,\frac { 4 }{ { -8 } }  \ =\left( { \frac { 4 }{ 3 } ,-2 } \right)  \ Hence,\, option\, B\, is\, correct\, answer. \end{array}$

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

The values of K for which matrix $A = \begin{bmatrix} 1& 0 & - K\ 2 & 1 & 3\ K & 0 & 1\end{bmatrix}$ is invertible are

  1. $\displaystyle \{-1,1 \}$
  2. $\displaystyle R$
  3. $\displaystyle R\backslash \{-1,1\}$
  4. $\displaystyle no\space real\space values$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Matrix A is invertible if $|A| \neq 0$, i.e.,
$\begin{bmatrix}1 & 0 & -K\ 2 & 1 & 3\ K & 0 & 1\end{bmatrix} \neq 0$
or $1(1) - K (-K) \neq 0$
Expanding along second column
$|A| =-0+1(1-(-K)(K))=1+K^2 \neq 0$ which is true for all real K.
Hence, A is invertible for all real values of K.

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

$\displaystyle \begin{bmatrix} 1 & -2 & 3 \ 2 & -1 & 4 \ 3 & 4 & 1 \end{bmatrix}$ is a

  1. rectangular matrix

  2. singular matrix

  3. square matrix

  4. nonsingular matrix
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

It is a $3 \times 3$ so it is a square matrix,


$\displaystyle \begin{bmatrix} 1 & -2 & 3 \ 2 & -1 & 4 \ 3 & 4 & 1 \end{bmatrix}$


$=1(-17)+2(-10)+3(11)$

$=-17-20+33$

$=33-37=-4$

so, it is not singular