Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

For two suitable order matrices $A, B$; correct statement is-

  1. ${(AB)}^{-1}={A}^{-1}{B}^{-1}$

  2. ${(AB)}^{-1}={B}^{-1}{A}^{-1}$

  3. ${(AB)}^{-1}={(BA)}^{-1}$

  4. none of these


Correct Option: B
Explanation:

It is fundamental concept that, $(AB)^{-1}=B^{-1}A^{-1}$

If A is a $3 \times 3$ matrix such that $\left| A \right| = 4\ than\ \left| {{{\left( {adjA} \right)}^{ - 1}}} \right| = $

  1. $16$

  2. $64$

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

  4. None


Correct Option: A
Explanation:

We know, $A^{-1}=\dfrac{adjA}{|A|}$

Multiplying above equation with A both sides,
$AA^{-1}=\dfrac{A\times adjA}{|A|}\Rightarrow|A|=A\times adjA$
Multiplying with $(adjA)^{-1}$ both sides ,
$|A|\times(adjA)^{-1}=A\times adjA \times(adjA)^{-1}\Rightarrow|A|\times(adjA)^{-1}=A$
Taking determinant both sides,$||A|\times(adjA)^{-1}|=|A|\Rightarrow ||A||\times |(adjA)^{-1}|=|A|\Rightarrow |A|^n\times |(adjA)^{-1}|=|A|$
Where n is the order of matrix A, i.e. $n=3$ and $|A|=4$
Thus, $|(adjA)^{-1}|=\dfrac{4}{4^3}=\dfrac{1}{16}$

If the matrices $A, B, (A+B)$ are non singular then ${[A{(A+B)}^{-1}B]}^{-1}$ is equal to-

  1. $A+B$

  2. ${A}^{-1}+{B}^{-1}$

  3. $A{(A+B)}^{-1}$

  4. None


Correct Option: B
Explanation:
$ (A(A+B)^{-1}B)^{-1}$

$ = [(A(A^{-1}+B^{-1}))B]^{-1}$

$ = [(AA^{-1}+AB^{-1})B]^{-1}$

$ = ((I+AB^{-1})B)^{-1} = (B+AB^{-1}B)^{-1}$

$ = (B+A)^{-1} = A^{-1}+B^{-1}$

If $A$ is an invertible matrix of order $2$, then $det({A}^{-1})$ is equal to

  1. $det(A)$

  2. $\cfrac{1}{det(A)}$

  3. $1$

  4. $0$


Correct Option: B
Explanation:

We know that 

$AA^{-1}=I$
Taking determinant both sides
$|AA^{-1}|=|I|$
$|A||A^{-1}|=|I|$       $[\because |AB|=|A||B|]$
$|A||A^{-1}|=1$         $[\because |I|=1]$
$|A^{-1}|=\dfrac{1}{|A|}$
Since $|A|\neq0$
Hence, $|A^{-1}|=\dfrac{1}{|A|}$

Let $A,B$ and $C$ be square matrices of order $3\ \times 3$. If $A$ invertible $(A-B)C=BA^{-1}$, then

  1. $C\ (A-B)=A^{-1}B$

  2. $C\ (A-B)=BA^{-1}$

  3. $(A-B)C=A^{-1}B$

  4. $All\ the\ above$


Correct Option: A

A square non-singular matrix A satisfies $\displaystyle A^{2}-A+2I=0$, then $\displaystyle A^{-1}=$

  1. $\displaystyle I-A$

  2. $\displaystyle \frac{1}{2}\left ( I-A \right )$

  3. $\displaystyle I+A$

  4. $\displaystyle \frac{1}{2}\left ( I+A \right )$


Correct Option: B
Explanation:

Given, $\displaystyle A^{2}-A+2I=0$


$\Rightarrow A^{2}A^{-1}-AA^{-1}+2IA^{-1}=0$

$\Rightarrow A-I+2A^{-1}=0$

$\Rightarrow 2A^{-1}=I-A$

$\Rightarrow A^{-1}=\displaystyle \frac{1}{2}(I-A)$

If $A$ satisfies the equation $\displaystyle x^{3}-5x^{2}+4x+\lambda =0$, then $\displaystyle A^{-1}$ exists if

  1. $\displaystyle \lambda \neq 1$

  2. $\displaystyle \lambda \neq 2$

  3. $\displaystyle \lambda \neq -1$

  4. $\displaystyle \lambda \neq 0$


Correct Option: D
Explanation:

Since, A satisfies the equation
$\displaystyle x^{3}-5x^{2}+4x+\lambda =0$
$\Rightarrow A^{3}-5A^{2}+4A+\lambda=O$
$\Rightarrow A^{3}A^{-1}-5A^{2}A^{-1}+4AA^{-1}+\lambda A^{-1}=O$
$\Rightarrow A^{2}-5A+4I+\lambda A^{-1}=O$
So, $A^{-1}$ exists if $\lambda\ne 0$

If $A$ is an invertiable idempotent matrix and $B=7A^{7}+6A^{6}+5A^{5}+......+A$ then $|B|$ is equal to 

  1. $7$

  2. $14$

  3. $28$

  4. $35$


Correct Option: A

If $\begin{bmatrix} 1 & -1 & x \ 1 & x & 1 \ x & -1 & 1 \end{bmatrix}$ has no inverse, then the real value of $x$ is 

  1. $2$

  2. $3$

  3. $0$

  4. $1$


Correct Option: C

Let p be a nonsingular matrix, and $I + p + p^2 + ..... + p^n = 0$, then find $p^{-1}$.

  1. $I$

  2. $p^{n+1}$

  3. $p^n$

  4. $\left( p^{n+1} - I\right) \left( p-I\right)$


Correct Option: C
Explanation:

We have $I + p + p^2 + ..... + p^n = O$  ----------$(1)$
Since p is a nonsingular matrix, p is invertible.
Multiplying both sides of (1) by $p^{-1}$, we get
$p^{-1} + I + Ip + ..... + p^{n - 1} I = O. p^{-1}$
or $ p^{-1} + I (1 + p + ...... + p^{n-1}) = O$
or $ p^{-1} = - I (I + p + p^2 + ..... + p^{n - 1}) = - I(-p^n) = p^n$