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

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.

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

If $A = \begin{bmatrix}1 & k & 3\ 3 & k & -2 \ 2 & 3 & -4\end{bmatrix}$ is singular then $k = ?$

  1. $\dfrac {16}{3}$
  2. $\dfrac {34}{5}$
  3. $\dfrac {33}{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A matrix is singular if its determinant is 0. Calculating the determinant: 1(k*-4 - 3*-2) - k(3*-4 - 2*-2) + 3(3*3 - 2*k) = 0. This simplifies to 1(-4k + 6) - k(-12 + 4) + 3(9 - 2k) = 0, which is -4k + 6 + 8k + 27 - 6k = 0, or -2k + 33 = 0, so k = 33/2.

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 $\lambda $ is a characteristic root of $A$ and $n\in N$, then $\lambda ^{n}$ is a characteristic root of

  1. $A^n$
  2. $A^{n-1}$
  3. $A^{-n}$
  4. $A-A^n+A^{-n}$
Reveal answer Fill a bubble to check yourself
A 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.
We have            $A^2 :X=A(AX) =A(\lambda X) = \lambda(AX)$
                                   $=\lambda^2X$,
                          $A^3 :X=A(A^2X) =A(\lambda^2 X)$
                                   $=\lambda^2(AX)=\lambda^2(\lambda X)=\lambda^3 X$
      Continuing in this way, we obtain
                        ${A}^{n} X=\lambda^{n}: X: \forall : n\in N$

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

If $A =\begin{bmatrix}4 &x+2 \2x-3 &x+1 \end{bmatrix}$ is an invertible matrix, then $x$ cannot take value

  1. -1

  2. 2

  3. 3

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

A matrix is invertible if its determinant is non-zero. Determinant = 4(x+1) - (x+2)(2x-3) = 4x + 4 - (2x^2 + x - 6) = -2x^2 + 3x + 10. Setting this to 0: 2x^2 - 3x - 10 = 0. Using the quadratic formula, x = (3 +/- sqrt(9 - 4*2*-10)) / 4 = (3 +/- sqrt(89)) / 4. Since these values are not in the options, 'none of these' is the correct choice.

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 $n\times n$ and let $P$ be a non-singular matrix, then which of the following matrices have the same characteristic roots.

  1. $A$ and $PA$
  2. $A$ and $AP$
  3. $A$ and $P^{-1}AP$

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

Let the characteristic root of $A$ be $\lambda $.


$\displaystyle \Rightarrow \left| A-\lambda I \right| =0$

For $\displaystyle { P }^{ -1 }AP;\left| { P }^{ -1 }AP-\lambda I \right| =\left| { P }^{ -1 }AP-\lambda { P }^{ -1 }PI \right| =\left| { P }^{ -1 } \right| \left| A-\lambda I \right| \left| P \right| =\left| A-\lambda 1 \right| .$

$\Rightarrow \lambda $ is also characteristic roots of matrix $\displaystyle { P }^{ -1 }AP$

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

Let $A$ be an $n\times n$ matrix such that $A^n=\alpha A,$ where $\alpha$ is a real number different from $1$ and $-1$. Then, the matrix $A+I _n$ is

  1. singular

  2. non-singular, i.e., invertible

  3. scalar

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $B=A+{ I } _{ n }$.


Since $A=B-{ I } _{ n }$, the condition ${ A }^{ n }=\alpha A$ can be written in the form $\displaystyle { \left( { B-I } _{ n } \right)  }^{ n }\alpha \left( B-{ I } _{ n } \right) $

$\displaystyle \Rightarrow { B }^{ n }-^{ n }{ { C } _{ 1 } }{ B }^{ n-1 }+^{ n }{ { C } _{ 2 } }{ B }^{ n-2 }+...+{ \left( -1 \right)  }^{ n }{ I } _{ n }=\alpha B-\alpha { I } _{ n }$

$\displaystyle \Rightarrow { B }^{ n }-^{ n }{ { C } _{ 1 } }{ B }^{ n-1 }+^{ n }{ C } _{ 2 }{ B }^{ n-2 }+...+{ \left( -1 \right)  }^{ n-1 }B-\alpha B=-\alpha { I } _{ n }-{ \left( -1 \right)  }^{ n }{ I } _{ n },$

$\displaystyle \Rightarrow B\left( { B }^{ n-1 }-^{ n }{ { C } _{ 1 }{ B }^{ n-2 }+^{ n }{ { C } _{ 2 } }{ B }^{ n-3 }+...+{ \left( -1 \right)  }^{ n-1 }{ I } _{ n }-\alpha { I } _{ n } } \right) =\left[ { \left( -1 \right)  }^{ n+1 }-\alpha  \right] { I } _{ n }.$

Since $\displaystyle { \left( -1 \right)  }^{ n+1 }-\alpha \neq 0,\left[ \because \alpha \neq \pm 1 \right] $

$\therefore$ $B$ is invertible 

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

Matrix $\begin{bmatrix}a & b &(a\alpha -b) \b  & c & (b\alpha -c)\2 & 1 & 0\end{bmatrix}$ is non invertible if 

  1. $\alpha = 1/2$

  2. a, b, c are in A.P.

  3. a, b, c are in G.P.

  4. a, b, c are in H.P.
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$\Delta =\begin{vmatrix} a & b & (a\alpha -b) \ b & c & (b\alpha -c) \ 2 & 1 & 0 \end{vmatrix}\ =2\left( b(b\alpha -c)-c(a\alpha -b) \right) -1\left( a(b\alpha -c)-b(a\alpha -b) \right)$

 
Expanding along $R _3$

$ =2{ b }^{ 2 }\alpha -2bc-2ac\alpha +2bc-ab\alpha +ca+ab\alpha -{ b }^{ 2 }\\ =\left( { b }^{ 2 }-ac \right) \left( 2\alpha -1 \right) $

Matrix is non invertible when $\Delta =0$
i.e $\displaystyle\alpha=\frac{1}{2}$ and $a,b,c$ are in $G.P$

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

If $A$ and $B$ are any two matrices such that $AB = 0$ and $A$ is non-singular, then

  1. $B = 0$
  2. $B$ is singular
  3. $B$ is non-singular
  4. $B = A$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given that $AB = 0$, taking determinants on both sides, we have
$|A\cdot B| = 0$
$\therefore |A| \times |B| = 0$
Product of two numbers is zero if one of them is zero.
Since $A$ is non-singular matrix, $B$ has to be singular, meaning that its determinant has to be zero.
The matrix need not be zero.