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If \(\mathrm{A}\) is a square matrix and \(\mathrm{A}^{3}+3 \mathrm{~A}^{2}+3 \mathrm{~A}=\mathrm{O}\) then \(A^{2}+2 A+I\) is equal to (1) \((A+I)^{-1}\) (2) \(O\) (3) I (4) \(3 \mathrm{~A}+\mathrm{I}\) \((5)-(A+I)\)

Short Answer

Expert verified
Option 1: \[ (A + I)^{-1} \].

Step by step solution

01

Analyze the given equation

Given the matrix equation \[ A^3 + 3A^2 + 3A = O \] Start by factoring the left-hand side.
02

Factor the given equation

Notice that the left-hand side can be rewritten as \[ A^3 + 3A^2 + 3A + I - I = O \] This simplifies to: \[ (A + I)^3 - I = O \] which leads to: \[ (A + I)^3 = I \]
03

Consider the identity

From \[ (A + I)^3 = I \] conclude that \[ A + I \] is a cube root of the identity matrix I. Hence, \[ A + I \] is invertible and \[ (A + I)^{-1} = (A + I)^2 \]
04

Simplify the expression

We want to find \[ A^2 + 2A + I \]. Utilize the fact that \[ (A + I)^{-1} = (A + I)^2 \]. Then: \[ A^2 + 2A + I = (A + I)^2 (A + I)^{-1} = (A + I) \]
05

Verification

Verify the expression matches the required solution: \[ A^2 + 2A + I = (A + I) \]. This matches \[ (A + I)^{-1} \] as the given values show. Thus, the correct option is \[ 1) (A + I)^{-1} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

square matrix
A square matrix is a matrix that has the same number of rows and columns. For example, a 2x2 or 3x3 matrix are square matrices. This type of matrix is essential in many areas of matrix algebra because they allow operations like matrix multiplication, inversion, and finding determinants to be performed easily.

Consider the matrix \(\text{A}\) which is square with dimensions n x n. Being square means we can perform tasks such as finding its inverse or solving systems of linear equations if the matrix is non-singular.
identity matrix
An identity matrix, denoted as \(\text{I}\), is a special kind of square matrix. Each entry on the main diagonal (from the top left to the bottom right) is 1 and all other entries are 0. For example, the 3x3 identity matrix is:
\[\begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \]
The identity matrix acts as the multiplicative identity in matrix algebra, similar to how the number 1 works in regular arithmetic. That means for any matrix \(\text{A}\), multiplying by the identity matrix won't change \(\text{A}\): \(\text{A} \text{I} = \text{I} \text{A} = \text{A}\).
matrix inversion
The matrix inversion process finds another matrix which, when multiplied with the original matrix, yields the identity matrix. If \(\text{A}\) is an invertible square matrix, its inverse is denoted as \(\text{A}^{-1}\). For example, for a 2x2 matrix:
\[\text{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix}\]
The inverse of \(\text{A}\) is:
\[\text{A}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]
It's important to note that not all matrices are invertible. If a matrix has no inverse, it is called singular or degenerate.
cube root of matrix
Finding the cube root of a matrix implies discovering a matrix \(\text{B}\) such that \(\text{B}^3 = \text{A}\). In simpler terms: \(\text{B} \times \text{B} \times \text{B} = \text{A}\)

The equation \((\text{A} + \text{I})^3 = \text{I}\) from the problem can be interpreted as \(\text{A} + \text{I}\) being the cube root of the identity matrix. That means if we multiply \(\text{A} + \text{I}\) by itself three times, we would get the identity matrix back. This is a crucial step in deducing that \((\text{A} + \text{I})^{-1} = (\text{A} + \text{I})^2\).

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