Invertible Matrices

  • If A and B are two square matrices such that AB = BA = I, then B is the inverse matrix of A and is denoted by A-1 and A is the inverse of B
  • Inverse of a matrix if it exists is unique.
  • AA−1 = A−1A = In
  • (A−1)−1 = A
  • (kA)−1 = k−1A−1 if k ≠ 0
  • (AB)−1 = B−1 A−1

View the Topic in this video From 24:58 To 48:27

Disclaimer: may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. makes no representations whatsoever concerning the content of these sites and the fact that has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by with respect to such sites, its services, the products displayed, its owners, or its providers.

  •  For a matrix A = \tt \begin{bmatrix}a & b \\c & d \end{bmatrix},\ adj\ A=\begin{bmatrix}d & -b \\-c & a \end{bmatrix}\ and\ A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b \\-c & a \end{bmatrix} if ad − bc ≠ 0
  • If A is a triangular matrix, then A−1, if it exists is a triangular matrix of the same kind.
    In fact if \tt A=\begin{bmatrix}a_{11} & 0 & 0 \\a_{21} & a_{22} & 0 \\a_{31} & a_{32} & a_{33} \end{bmatrix} and a11 a22 a33 ≠ 0, then
    \tt A^{-1}=\frac{1}{a_{11}a_{22}a_{33}}\begin{bmatrix}a_{22}a_{33} & 0 & 0 \\-a_{21}a_{33} & a_{33}a_{11} & 0 \\A_{13} & -a_{32}a_{11} & a_{11}a_{22} \end{bmatrix} where A13 = cofactor of (1,3)th element in A i.e.
    \tt A_{13}=\begin{vmatrix}a_{21} & a_{22} \\a_{31} & a_{32} \end{vmatrix}
  • If A = diag (λ1, λ,...,λn) then A−1 exists if and only if λi ≠ 0 ∀ i and
    A−1 = diag (λ1−12−1,....,λn−1)
    Also Am = diag (λ1m2m,....,λnm) if m ∈ N
  • If a square matrix A satisfies the equation a0 + a1x + a2x+ ... + arx=0, then A is invertible if a0 ≠ 0 and its inverse is given by A^{-1}= − \frac{1}{a_0}[a_1I+a_2A + ...+a_r\ A^{r-1}]
  • If AB = I, then BA = I and B = A−1 
  • If If AB = CA = I, then B = C