If A is a non-zero square matrix of order n w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $\det(I+A) 
eq 0$, which implies that $(I+A)$ is an invertible matrix.We are given $A^3 = O$.We know the algebraic identity $x^3 + y^3

Vedclass · Accepted Answer

$I-A+A^2$

Vedclass · Answer

(A) Given that $\det(I+A) 
eq 0$, which implies that $(I+A)$ is an invertible matrix.We are given $A^3 = O$.We know the algebraic identity $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$.Substituting $x = A$ and $y = I$, we get $A^3 + I^3 = (A+I)(A^2 - A + I)$.Since $A^3 = O$ and $I^3 = I$, the equation becomes $O + I = (A+I)(A^2 - A + I)$.This simplifies to $I = (A+I)(A^2 - A + I)$.Multiplying both sides by $(A+I)^{-1}$ from the left, we get $(A+I)^{-1} I = (A+I)^{-1} (A+I)(A^2 - A + I)$.Since $(A+I)^{-1}(A+I) = I$, we have $(A+I)^{-1} = I(A^2 - A + I) = A^2 - A + I$.Thus, $(I+A)^{-1} = I - A + A^2$.

If $A$ is a non-zero square matrix of order $n$ with $\det(I+A) \neq 0$ and $A^3=O$, where $I$ and $O$ are the identity and null matrices of order $n \times n$ respectively, then $(I+A)^{-1}$ is equal to

Similar Questions

Let the determinant of a square matrix $A$ of order $m$ be $m-n$,where $m$ and $n$ satisfy $4m + n = 22$ and $17m + 4n = 93$. If $\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(mA))) = 3^a 5^b 6^c$,then $a + b + c$ is equal to:

If $A = \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix}$ where $i = \sqrt{-1}$ and $B = A^{2029}$,then $B^{-1} =$

If $A = \begin{bmatrix} 1 & 2 \\ -4 & -1 \end{bmatrix}$, then $A^{-1}$ is

If $A$ and $B$ are square matrices of the same order and $AB = 3I$, then $A^{-1}$ is equal to

If $A = \begin{bmatrix} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{bmatrix}$, then $|(\operatorname{Adj} A)^{-1}| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module