Inverse matrix

Definition . Let A ∈ M_n,n(𝕂) a square matrix, we call inverse matrix of A (if exists) the matrix A⁻¹, such that:

A ⋅ A⁻¹ = A⁻¹ ⋅ A = I_n

being I_n the identity matrix with order n. The set of invertible matrices of order n is denoted by GL_n(𝕂), where GL stands for "linear group". (GL_n(𝕂), ⋅) is a non commutative group with respect to multiplication.

Proposition I.1 Given A ∈ M_n,n(𝕂), if the inverse of A exists is unique.

Proof. Let A⁻¹ its inverse. If B is a matrix such that:

BA = I

then:

(BA) A^-1 = I A⁻¹

(BAA^-1) = B I = B I A^-1 = A^-1I

si ha quindi:

B = A^-1

Theorem I.0 Sufficient condition for a matrix A ∈ M_n,n(𝕂) to possess an inverse A⁻¹ is that det A ≠ 0. In this case the inverse is given by the following relation

A^{- 1} = \frac{1}{det A} \cdot {(\begin{array}{cccc} A_{11} & A_{12} & \dots & A_{1 n} \\ A_{21} & A_{22} & \dots & A_{2 n} \\ ⋮ & ⋮ & ⋮ \\ A_{n 1} & A_{n 2} & \dots & A_{n n} \end{array})}^{T}

where T indicates the "transposition", A_ij are the matrices obtained by removing the i-th row and j-th column from A.

Proof. - Proviamo prima che se esiste la matrice inversa di A, allora necessariamente det A ≠ 0. Infatti poichè A⋅A^-1 = I_n, det(A⋅A⁻¹) = det I_n, owing to Binet's Theorem we have

1 = det A ⋅ det (A⁻¹)

and this implies det A ≠ 0.

Supposing that det A ≠ 0 we prove that the inverse is calculated as stated above. L'elemento di posto ki nella matrice identità A⋅A^-1 è in base alla (4.10)

b_{k i} = \sum_{j = 1}^{n} a_{k j} \cdot \frac{1}{det A} A_{i j} = \frac{1}{det A} \cdot \sum_{j = 1}^{n} a_{k j} A_{i j}

ora se k=i, la sommatoria scritta vale det A, per la definizione di determinante (1.3), e quindi b_ki =1; se invece k≠ i, mostriamo che tale somma vale 0. Infatti sia A' la matrice ottenuta da A rimpiazzando la riga i con una coppia della riga k. Alterare la riga i non altera i numeri A_i1 ...,A_in. Si vede allora che l'espressione

a_k1A_i1+ a_k2A_i2 +....+ a_knA_in

non è altro che det A', per la definizione di determinante. Ma A' ha due righe uguali, quindi il suo determinante è nullo. Abbiamo quindi provato che l'elemento di posto ki nella matrice A ⋅ A^-1 vale 1 se k=i, 0 se k≠i.

Example. Calculate the inverse of the matrix

A = (\begin{array}{ccc} 1 & 2 & 3 \\ 0 & - 1 & 3 \\ 0 & 0 & 2 \end{array})

Essendo triangolare det A = 1 ⋅ (−1) ⋅ 2 = −2 ≠0, then A⁻¹ exists. We have

\begin{aligned} A_{11} & = | \begin{array}{cc} 1 & 3 \\ 0 & 2 \end{array} | = - 2 & A_{12} & = - | \begin{array}{cc} 0 & 3 \\ 0 & 2 \end{array} | = 0 & A_{13} & = | \begin{array}{cc} 0 & - 1 \\ 0 & 0 \end{array} | = 0, \\ A_{21} & = - | \begin{array}{cc} 2 & 3 \\ 0 & 2 \end{array} | = - 4 & A_{22} & = | \begin{array}{cc} 1 & 3 \\ 0 & 2 \end{array} | = 2 & A_{23} & = - | \begin{array}{cc} 1 & 2 \\ 0 & 0 \end{array} | = 0, \\ A_{31} & = | \begin{array}{cc} 2 & 3 \\ - 1 & 3 \end{array} | = 9 & A_{32} & = - | \begin{array}{cc} 1 & 3 \\ 0 & 3 \end{array} | = 0 & A_{33} & = | \begin{array}{cc} 1 & 2 \\ 0 & - 1 \end{array} | = - 1 \end{aligned}

A^{- 1} = \frac{1}{2} {(\begin{array}{ccc} - 2 & 0 & 0 \\ - 4 & 2 & 0 \\ 9 & 3 & - 1 \end{array})}^{T} = (\begin{array}{ccc} 1 & 2 & - 9 / 2 \\ 0 & - 1 & - 3 / 2 \\ 0 & 0 & 1 / 2 \end{array})

In the case of order 2 the calculation is particularly simple

A = (\begin{array}{cc} a & b \\ c & d \end{array})

In the hyphothesys det A = ad − bc ≠ 0, we have

A = (\begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}) = (\begin{array}{cc} d & - c \\ - b & a \end{array})

thus

A^{- 1} = \frac{1}{a d - b c} (\begin{array}{cc} d & - b \\ - c & a \end{array})

Theorem 4.11 The set of invertible n × n matrices over R forms a group with respect to the matrix multiplication. We denote this group by GL_n(R) (“GL” abbreviates “general linear (group)”).

Proof The associativity of the multiplication in GL_n(ℝ) is clear. As shown in (2) in Lemma 4.10, the product of two invertible matrices is an invertible matrix. The neutral element in GL_n(ℝ) is the identity matrix I_n , and since every A ∈ GL_n(ℝ) is assumed to be invertible, A⁻¹ exists with (A⁻¹)⁻¹ = A ∈ GL_n(ℝ). □

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