Tag: inverse of a matrix and linear equations

Questions Related to inverse of a matrix and linear equations

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:

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

  1. $8$

  2. $7$

  3. $16$

  4. $12$

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Correct Option: C
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 $\alpha+\beta=1+15=16$.


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

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

  1. $3$

  2. $5$

  3. $7$

  4. $9$

Show answer
Correct Option: A
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 , $|2A+B|=0$


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

$\Rightarrow 9-6\lambda=0$

$\Rightarrow 2\lambda=3$

Let $A$ be a square matrix all of whose entries are integers, then which of the following is true?

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

  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.

Show answer
Correct Option: B
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.

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