Example 1, row reduce a 3×3 matrix
Suppose you want to reduce the matrix below to reduced row echelon form.
[ 1 2 1 ]
[ 2 4 0 ]
[ 3 6 3 ]
Result
[ 1 2 0 ]
[ 0 0 1 ]
[ 0 0 0 ]
- Subtract 2 times row 1 from row 2.
- Subtract 3 times row 1 from row 3.
- Swap the second and third rows.
- Scale the new second row to make the pivot 1.
Example 2, determinant check before finding an inverse
Before trying to invert a matrix, check whether its determinant is zero.
[ 4 7 ]
[ 2 6 ]
Determinant: 10
Because the determinant is not zero, the matrix is invertible. That makes it worth moving to the inverse calculator instead of wasting time on a singular matrix.
Example 3, inverse of a 2×2 matrix
For a small invertible matrix, the inverse gives you the matrix that returns the identity when multiplied back.
A = [ 4 7 ]
[ 2 6 ]
A⁻¹ = [ 0.6 -0.7 ]
[ -0.2 0.4 ]
This gives you a fast answer and a simple path to double-check it with the matching calculator.
Example 4, add two 2×2 matrices
Matrix addition works entry by entry, so each value in the result comes from the matching positions in both matrices.
A = [ 1 4 ]
[ -2 3 ]
B = [ 5 -1 ]
[ 6 2 ]
A + B:
[ 6 3 ]
[ 4 5 ]
This is a quick check when you want to verify the result of a simple matrix operation before moving on to something larger.
Example 5, trace of a 3×3 matrix
The trace is the sum of the main diagonal. It gives you one quick summary value without needing the full determinant or inverse.
[ 3 1 0 ]
[ 2 5 4 ]
[ 7 6 -2 ]
Trace: 3 + 5 + (-2) = 6
That makes trace a fast check when you want a compact value from the matrix before moving on to a bigger calculation.
Example 6, rank of a 3×3 matrix
The rank tells you how many independent rows or columns the matrix really has after row reduction.
[ 1 2 3 ]
[ 2 4 6 ]
[ 1 1 1 ]
Rank: 2
One row depends on another here, so the matrix is not full rank. That makes rank a fast dependency check before solving bigger linear algebra problems.
Example 7, identity matrix for an inverse check
The identity matrix is the target result when a matrix is multiplied by its inverse.
I₃ = [ 1 0 0 ]
[ 0 1 0 ]
[ 0 0 1 ]
If your inverse check works, the final multiplication should match this pattern.