Tag: business maths

Questions Related to business maths

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 :

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

  1. $A$ is non-singular

  2. $A$ is singular

  3. $A = 0$

  4. $A = I _n$

Show answer
Correct Option: B
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.

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

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

  1. $\dfrac {16}{3}$

  2. $\dfrac {34}{5}$

  3. $\dfrac {33}{2}$

  4. None of these

Show answer
Correct Option: C

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

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

  1. $A^n$

  2. $A^{n-1}$

  3. $A^{-n}$

  4. $A-A^n+A^{-n}$

Show answer
Correct Option: A
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
                        ${A}^{n} X=\lambda^{n}: X: \forall : n\in N$

If $A$ is an invertible matrix. then which of the followings are true:

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

  1. $A\neq 0$

  2. Adj. $A\neq 0$

  3. $|A|\neq 0$

  4. $A^{-1}=|A|:Adj. A.$

Show answer
Correct Option: A,B,C
Explanation:
A is invertible matrix
$\Rightarrow { A }^{ -1 }$ exists
$\Rightarrow \left| A \right| \neq 0$
$AdjA\neq 0$
$A\neq 0$
Option A, B, C are correct

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

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

  1. -1

  2. 2

  3. 3

  4. none of these

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Correct Option: D

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.

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

  1. $A$ and $PA$

  2. $A$ and $AP$

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

  4. none of these

Show answer
Correct Option: C
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 $\displaystyle { P }^{ -1 }AP$

If $A, : B : and : C$ are three square matrices of the same order, then $AB = AC\Rightarrow  B = C$ if

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

  1. $|A|\neq 0$

  2. $A$ is invertible

  3. $A$ is orthogonal

  4. $A$ is symmetric

Show answer
Correct Option: A,B,C
Explanation:

$AB=AC\Rightarrow B=C$

When $\left| A \right| \neq 0$

Hence $A$ in invertible and orthogonal

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

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

  1. singular

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

  3. scalar

  4. None of these

Show answer
Correct Option: B
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 

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

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

  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.

Show answer
Correct Option: A,C
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 $G.P$

If $\left |\begin{matrix}1 & -1 &x \ 1 & x & 1\ x & -1 & 1\end{matrix} \right|$ has no inverse, then the real value of $x$ can be is

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

  1. 2

  2. 3

  3. 0

  4. 1

Show answer
Correct Option: D
Explanation:
$\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}$
No inverse
$\Rightarrow \left| A \right| =0$
$x+1+1(1-x)+x(-1-{ x }^{ 2 })=0$
$x+1+1-x-x-{ x }^{ 3 }=0$
${ x }^{ 3 }-x-2=0$
$\Rightarrow { x }^{ 3 }+2x-x-2=0$
$\left( { x }^{ 2 }-1 \right) \left( x+2 \right) =0$
$\Rightarrow x=\pm 1,x=-2$
Real value of $x=1$
Option D