Tag: inverse of a matrix and linear equations

Questions Related to inverse of a matrix and linear equations

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$

Matrices A and B satisfy $AB = B^{-1}$, where $ B\quad =\quad \begin{bmatrix} 2 & -1 \ 2 & 0 \end{bmatrix}$, then find without finding $A^{-1}$, the matrix X satisfying $A^{-1}XA = ?$

  1. $B$

  2. $B^2$

  3. $A$

  4. None of these


Correct Option: A
Explanation:

Given, $A^{-1} XA = B$


$AA^{-1} XA = AB$

$IXA =AB$

$XAB =AB^2$ 

$XAB =I$ since $\left[AB=B^{-1}\Rightarrow AB^2=I\right]$

$XAB^2 = B$

$XI = B$

$\therefore X = B$

Hence option $'A'$ is the answer.

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

  1. $k\neq -1$

  2. $k\neq 0$

  3. $k\neq 1$

  4. none of these


Correct Option: B
Explanation:

Since A satisfies the given equation, therefore ${ A }^{ 3 }-5{ A }^{ 2 }+4A+kI=0$

${ A }^{ -1 }$ exits if $k\neq 0$ since if $k=0$ then the above equation gives $A=0$ and in that case ${ A }^{ -1 }$ wont exist.

If $A^3 = O$, then $I + A + A^2$ equals

  1. $I - A$

  2. $(I + A^1)^{-1}$

  3. $(I - A)^{-1}$

  4. none of these


Correct Option: C
Explanation:

Given, $A^{3} = 0$
$\Rightarrow I-A^{3} = I$

Using the identity, we get
$\Rightarrow (I-A)(I+A+A^{2}) = I$
$\therefore I+A+A^{2} = (I-A)^{-1}$

If $A$ and $B$ are symmetric matrices and $AB=BA$, then ${ A }^{ -1 }B$ is a

  1. Symmetric matrix

  2. Skew-symmetric matrix

  3. Identity matrix

  4. None of these


Correct Option: A
Explanation:

We have $AB=BA=B'A'=\left( AB \right) '$

$\Rightarrow AB$ is symmetric.
Also, $AB{ A }^{ -1 }=BA{ A }^{ -1 }=B\quad \quad \left( \because AB=BA \right) $
$\Rightarrow { A }^{ -1 }AB{ A }^{ -1 }={ A }^{ -1 }B\Rightarrow B{ A }^{ -1 }={ A }^{ -1 }B$
Therefore, $\left( { A }^{ -1 }B \right) '=\left( B{ A }^{ -1 } \right) =\left( { A }^{ -1 } \right) 'B'=A'B$   $[\because { A }^{ -1 }$ and $B$ are symmetric $]$
Thus, the matrix ${ A }^{ -1 }B$ is symmetric.

If $A^2 + A - I = 0$, then $A^{-1}$ =

  1. $I + A$

  2. $I - A$

  3. $-I + A$

  4. $-I - A$


Correct Option: A
Explanation:

$A^2 + A - I = 0$


multiplying both sides with $A^{-1}$ gives

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

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

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

Hence, option A.

IF $A,B,C$ are non-singular $n\times n$ matrices, then $(ABC)^{-1}$ = ____________.

  1. $A^{-1}C^{-1}B^{-1}$

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

  3. $C^{-1}A^{-1}B^{-1}$

  4. $B^{-1}C^{-1}A^{-1}$


Correct Option: B
Explanation:

using the property $(AB)^{-1}=B^{-1}A^{-1}$


$(ABC)^{-1}=(BC)^{-1}A^{-1}=C^{-1}B^{-1}A^{-1}$

Hence, option B.