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How to quickly verify if a matrix is the inverse of another using row-column pairing?

To verify if a matrix is the inverse of another, check the inner product of each row of the first matrix with each column of the second matrix. The product of the ii-th row and jj-th column must equal 1 if i=ji=j (diagonal entries) and 0 if i≠ji \neq j (off-diagonal entries). This ensures the resulting product matrix is the identity matrix. In multiple-choice problems, testing just one or two specific row-column pairs can often eliminate incorrect options quickly.

Conditions

  • You are given two square matrices of the same dimension.
  • You need to verify if their product is the identity matrix.

Reasoning, step by step

  1. Identify the rows of the first matrix and the columns of the second matrix.
  2. Select a row ii and a column jj.
  3. Compute the dot product of row ii and column jj.
  4. Check if the result is 1 when i=ji=j, and 0 when i≠ji \neq j.
  5. Repeat for all combinations to fully verify, or use specific combinations to eliminate options in a multiple-choice setting.

Example

To verify if C=[c′a′b′f′d′e′i′g′h′]C = \begin{bmatrix} c' & a' & b' \\ f' & d' & e' \\ i' & g' & h' \end{bmatrix} is the inverse of N=[ghiabcdef]N = \begin{bmatrix} g & h & i \\ a & b & c \\ d & e & f \end{bmatrix}, check the (2,2)(2,2) entry: the second row of NN is [a b c][a\ b\ c] and the second column of CC is [a′d′g′]\begin{bmatrix} a' \\ d' \\ g' \end{bmatrix}. Their dot product is aa′+bb′+cc′aa' + bb' + cc', which equals 1 because it was the (1,1)(1,1) entry of AA−1=IAA^{-1}=I. Checking the (1,2)(1,2) entry: the first row of NN is [g h i][g\ h\ i] and the second column of CC is [a′d′g′]\begin{bmatrix} a' \\ d' \\ g' \end{bmatrix}. Their dot product is ga′+hd′+ig′ga' + hd' + ig', which equals 0 because it was the (3,1)(3,1) entry of AA−1=IAA^{-1}=I.

Common misconceptions

  • Assuming that checking only the diagonal entries is sufficient to prove a matrix is the inverse, without verifying the off-diagonal entries are zero.
  • Believing that having the same set of elements in the candidate matrix automatically makes it the inverse.

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