Linear Algebra

Inverses

10/13 in Linear Algebra. See all.

Inverse of a Matrix
The "opposite" of a matrix or transformation, A1A^{-1} (inverse of AA) has the property that applying AA, then A1A^{-1} gives the identity transformation. So, A1A=InA^{-1}\cdot A=I_n. Notes:

  • AB=BA=InAB=BA=I_n. BB is the inverse of AA, and AA is the inverse of BB.
  • Non-invertible matrices are called singular. Invertible ones are called non-singular.
  • Inverses are unique.
  • The following properties are true for invertible matrices AA and BB:
    • (A1)1=A(A^{-1})^{-1}=A
    • (AB)1=B1A1(AB)^{-1}=B^{-1}A^{-1}
    • (AT)1=(A1)T(A^T)^{-1}=(A^{-1})^T
    • (λA)1=1λA1(\lambda A)^{-1}=\frac{1}{\lambda}A^{-1}

Computing the Inverse of a Matrix
Lemma 1. A square matrix is invertible if and only if it can be transformed by elementary row operations to row-reduced form with all diagonal entries nonzero.

Lemma 2. Any row-reduced square matrix with all diagonal elements nonzero can be transformed to the identity matrix with elementary row operations.

By these lemmas, we can write EkEk1E2E1A=In,E_{k}E_{k-1}\dots E_2E_1A=I_n, so A1=EkEk1E2E1A^{-1}=E_{k}E_{k-1}\dots E_2E_1. So, we can compute the inverse of matrix AA by applying the elementary row operations Ek,,E1E_k,\dots, E_1 to the identity matrix! So, to compute A1A^{-1}, apply the same row operations to the identity matrix as you applied to AA in order to transform it to the identity. One way of finding the inverse of an n×nn\times n matrix AA is as follows:

  1. Form the n×2nn\times 2n matrix [AIn][A\,|\,I_n].
  2. Row-reduce the matrix to [InA1][I_n\,|\,A^{-1}].
  3. The matrix A1A^{-1} is the n×nn\times n matrix on the right side of the augmented matrix.

Ex. Find the inverse of [1234].[12103401][12100231][1210013/21/2][1021013/21/2]So, the inverse is [213/21/2].\begin{align*}\text{Find the inverse of }&\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}.\\\left[\begin{array}{cc|cc}1 & 2 & 1 & 0\\3 & 4 & 0 & 1\end{array}\right]&\rightarrow\left[\begin{array}{cc|cc}1 & 2 & 1 & 0\\0 & -2 & -3 & 1\end{array}\right]\rightarrow\\\left[\begin{array}{cc|cc}1 & 2 & 1 & 0\\0 & 1 & 3/2 & -1/2\end{array}\right]&\rightarrow\left[\begin{array}{cc|cc}1 & 0 & -2 & 1\\0 & 1 & 3/2 & -1/2\end{array}\right]\rightarrow\\\text{So, the inverse is }&\boxed{\begin{bmatrix}-2 & 1\\3/2 & -1/2\end{bmatrix}}.\end{align*} It's good practice to verify results, so you should multiply the inverse by the original matrix to check that the result is the identity matrix.

As a side note, for 2×22\times 2 matrices, we can find the determinant directly with the formula: [abcd]1=1adbc[dbca].\begin{bmatrix}a & b \\ c & d\end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}.

Invertible Matrix Theorem
For any n×nn\times n matrix AA, the following are equivalent:

  1. AA is invertible.
  2. AA is row equivalent to InI_n.
  3. AA has nn pivot positions.
  4. The equation Ax=0Ax=0 has only the trivial solution.
  5. The columns of AA are linearly independent.
  6. The equation Ax=bAx=b has at least one solution for each bb in Rn\mathbb{R}^n.
  7. The columns of AA span Rn\mathbb{R}^n.
  8. The equation Ax=bAx=b has exactly one solution for each bb in Rn\mathbb{R}^n.
  9. The determinant of AA is nonzero.
  10. The transpose of AA is invertible.

You don't need to memorize each one of these, but do take the time to understand why they are true — this will help you remember them.