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

If the matrix $A = \begin{bmatrix}8 & -6 & 2 \ -6 & 7 & -4 \ 2 & -4 & \lambda\end{bmatrix}$ is singular, then $\lambda = $

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

Given, the matrix $A$ is singular.

$\Rightarrow |A|=0$

$\begin{vmatrix} 8 & -6 & 2 \ -6 & 7 & -4 \ 2 & -4 & \lambda  \end{vmatrix}=0$

$8(7\lambda -16)+6(-6\lambda +8)+2(10)=0$

$\Rightarrow 20\lambda -60=0$

$\Rightarrow \lambda=3$

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

Suppose $A$ is any $3\times3$ non-singular matrix and $(A-3I)(A-5I)=O$,where $I=I _{3}$ and $O=O _{3}$.If $\alpha A+\beta A^{-1}=8I$ ,then $\alpha+\beta$ is equal to:

  1. $8$
  2. $7$
  3. $16$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,  $(A-3I)(A-5I)=O$ for a $3\times 3$ non-singular matrix.

or, $A^2-8A+15I=O$
or, $A-8I+15A^{-1}=O$ [ Since $A$ is non-singular ($A^{-1}$ exists) then multiplying both sides with $A^{-1}$]
or, $A+15A^{-1}=8I$.
Comparing this with the given equation we get, $\alpha=1, \beta=15$.
So $\alpha+\beta=1+15=16$.

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


Let $A$ and $B$ are two matrices of same order $\displaystyle 3\times 3$ given by $\displaystyle A=\begin{bmatrix}1 &3  &\lambda+2 \2  &4  &6 \3  &5  &8 \end{bmatrix}$ $\displaystyle B= \begin{bmatrix}3 &2  &4 \3  &2  &5 \2
 &1  &4 \end{bmatrix}$If $2A + B$ is singular, then $\displaystyle 2\lambda$ equals

  1. $3$
  2. $5$
  3. $7$
  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here, $2A+B=\begin{bmatrix} 2 & 6 & 2\lambda +4 \ 4 & 8 & 12 \ 6 & 10 & 16 \end{bmatrix}+\begin{bmatrix} 3 & 2 & 4 \ 3 & 2 & 5 \ 2 & 1 & 4 \end{bmatrix}$

$\Rightarrow 2A+B=\begin{bmatrix} 5 & 8 & 2\lambda +8 \ 7 & 10 & 17 \ 8 & 11 & 20 \end{bmatrix}$

Given , $|2A+B|=0$


$\begin{vmatrix} 5 & 8 & 2\lambda +8 \ 7 & 10 & 17 \ 8 & 11 & 20 \end{vmatrix}=0$

$\Rightarrow 9-6\lambda=0$

$\Rightarrow 2\lambda=3$

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 of the following is true?

  1. If $\displaystyle \left | A\right | \neq \pm 1 $, then $\displaystyle A^{-1} $ exist & all its entries are non-integer
  2. If $\displaystyle \left | A\right | = \pm 1 $, then $\displaystyle A^{-1} $ exist & all its entries are integer
  3. If $\displaystyle \left | A\right | = \pm 1 $,then $\displaystyle A^{-1} $ need not exist
  4. If $\displaystyle \left | A\right | = \pm 1 $, then $\displaystyle A^{-1} $ exist but all its entries are not necessarily integers.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given A is a square matrix with all entries as integer
(a) Now $\displaystyle \left|A\right|\neq 1,-1$ mean it may be zero also $\displaystyle \left|A\right| =0,$then $\displaystyle A^{-1}$ does'not exist $\displaystyle \therefore $ choice (a) is false.

(b) If $\displaystyle \left|A\right| =1,-1$ then $\displaystyle A^{-1}$ certainly exist but A is a square matrix with all integral entries so all cofactors are integers .So adj. A matrix has all integral entries .
$\displaystyle A^{-1}=\frac{1}{\left|A\right| }(adj A)=\pm(adj A) $
$\therefore $ choice (b) is correct.

(c) $\displaystyle \left|A\right| =1,-1 $ $\displaystyle \therefore $ $\displaystyle A^{-1}$ must exist but given $\displaystyle A^{-1}$ does not exist which is false result $\displaystyle \therefore $  choice (c) is incorrect

(d) $\displaystyle \left|A\right| =1,-1$ it is true that $\displaystyle A^{-1}$ exist but we have to follow that A has all integral entries but choice (d) says that it is not necessarily that entries are integer which mean adj. A, may or may not be with integral entries. Hence choice (d) is false.

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 of order $n \times  n$. A constant $\lambda $ is said to be characteristic root of $A$ if there exists a $n \times  1$ matrix $X$ such that  $AX=\lambda X$

If $0$ is a characteristic root of $A$, then :

  1. $A$ is non-singular
  2. $A$ is singular
  3. $A = 0$
  4. $A = I _n$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $X\neq 0$ is such that $(A-\lambda I)X=0,:|A-\lambda I|=0\Leftrightarrow A-\lambda I$ is singular. If $A-\lambda I$ is non-singular the then equation $(A-\lambda I)X=0\Rightarrow X=0$
If $\lambda= 0$, we get $|A|=0\Rightarrow A$ is singular.