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Topos Uranos
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The inverse matrix

Summary

This lesson studies the inverse matrix, which does for square matrices what the inverse does for nonzero numbers. It proves that the inverse, when it exists, is unique; that a matrix is invertible if and only if its determinant is nonzero, and that in that case the inverse is the adjugate matrix, formed by the cofactors, divided by the determinant; from this it obtains the formula for the inverse of order two. It proves the properties of the inverse of the product, of the transpose and of the powers, translates each elementary row operation into multiplication by an elementary matrix and deduces from this the Gauss-Jordan method, which obtains the inverse by transforming the matrix (A,I)\left( A, I \right) into (I,A−1)\left( I, A^{-1} \right). Finally, it solves matrix equations and gathers in a single theorem the conditions equivalent to invertibility, which relate the inverse to the determinant, to the rank and to systems of linear equations; from them it deduces that it suffices to check one of the two products in the definition, and with elementary matrices it proves that the determinant of a product is the product of the determinants.

Learning objectives

  1. Define the inverse matrix, prove its uniqueness and decide whether a matrix is invertible by means of its determinant.
  2. Compute the inverse by means of the adjugate matrix, in particular with the formula of order two, and by means of the Gauss-Jordan method, justifying both procedures.
  3. Apply the properties of the inverse of the product, of the transpose and of the powers, and solve matrix equations respecting the order of the factors.
  4. Relate the invertibility of a matrix to the rank, to elementary matrices and to the existence and uniqueness of the solutions of linear systems, and prove with elementary matrices that the determinant of a product is the product of the determinants.

The inverse matrix

In the lesson of the course Algebra and Geometry I on number systems we saw that every nonzero number aa has an inverse a−1a^{-1}, which makes it possible to solve the equation ax=bax = b by multiplying both its sides by a−1a^{-1}. In the lesson on matrices and matrix operations, a system of linear equations was written as a single equation AX=BAX = B, and the identity matrix II turned out to be the identity element of the product. If this is so, the natural question is whether there exists a matrix that plays, with respect to AA, the role of the inverse, so that the system is solved with a single multiplication. Now, the matrix product is not commutative, so the definition must require both equalities; and for ABAB and BABA to have the same size as II, the matrix AA must be square.

A square matrix AA of order nn is invertible when there exists a matrix BB of order nn such that AB=BA=IAB = BA = I, where II is the identity of order nn. A matrix BB with that property is called an inverse of AA.

For example, the two matrices below are inverses of each other, as is checked by multiplying them in both orders:

(2153)(3−1−52)=(3−1−52)(2153)=(1001)\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Not every nonzero matrix is invertible, and in this matrices differ from numbers. A matrix with a zero row is not, because the product of that row by any column is 00, so that ABAB has a zero row for every BB and cannot be the identity. Nor is a matrix AA for which there exists a nonzero CC with AC=0AC = 0, one of the zero divisors of the lesson on matrices: if AA had an inverse BB, we would have C=(BA)C=B⋅(AC)=0C = (BA)C = B \cdot (AC) = 0; and neither is CC, because if it had an inverse BB we would have A=A⋅(CB)=(AC)B=0A = A \cdot (CB) = (AC)B = 0. For example, the matrix with rows 1,11, 1 and 1,11, 1 multiplied by the one with rows 1,−11, -1 and −1,1-1, 1 gives the zero matrix, and neither of the two is invertible.

LemmaUniqueness of the inverse

A square matrix has at most one inverse: if AB=BA=IAB = BA = I and AD=DA=IAD = DA = I, then B=DB = D.

The proof is that of the uniqueness of the opposite and of the inverse in the Course I lesson on number systems, except that commutativity cannot be used; instead one equality on each side is used, together with the associativity of the matrix product and the fact that the identity is the identity element, proved in the lesson on matrices. The figure writes each product as a row of factors and marks the two that are grouped.

Proof

  1. AB=BA=I,AD=DA=I\dato{b}{AB = BA = I}, \qquad \dato{d}{AD = DA = I}

    We start from the two hypotheses: BB and DD each satisfy the two equalities of the definition.

  2. B=BI=B⋅(AD)B = B \resaltar{I} = B \cdot \resaltar{(AD)}

    Since II is the identity element of the product, B=BIB = BI; we replace II by ADAD, which is the equality of DD on the right.

  3. B⋅(AD)=(BA)D=ID=DB \cdot (AD) = \resaltar{(BA)} D = \resaltar{I} D = D

    By associativity we group differently, and we use the equality of BB on the left, BA=IBA = I. The two factors that are grouped change, but the order of the three matrices does not.

The unique inverse of AA, when it exists, is denoted by A−1A^{-1}. It must not be written as a fraction 1A\frac{1}{A}, nor the product BA−1BA^{-1} as a quotient BA\frac{B}{A}: since BA−1BA^{-1} and A−1BA^{-1}B are, in general, different, that notation would not indicate on which side we multiply. Arthur Cayley, who in 1858 founded the calculus of matrices, already defined the inverse by means of the adjugate matrix of the section “Determinant and invertibility”.