Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — the determinant of a 3 × 3 matrix
Find det214101328.
- 1Expand along the first row with signs + − +.The alternating sign is where most marks are lost.
- 220128=2(0−2)=−4Delete row 1 and column 1 for the first minor.
- 3−11428=−1(8−8)=0Note the minus in front.
- 4+31401=3(1−0)=3
- 5−4+0+3=−1Non-zero, so this matrix has an inverse.
Example 2 — solving a system by row reduction
Solve x+y+z=6, 2x−y+z=3, x+2y−z=2.
- 1Write the augmented matrix.Coefficients on the left, constants after the bar.
- 2R2−2R1 gives the row 0 −3 −1∣−9Clearing the first column.
- 3R3−R1 gives 0 1 −2∣−4
- 43R3+R2 gives 0 0 −7∣−21Now the matrix is triangular.
- 5−7z=−21, so z=3.Read the last row.
- 6Back-substitute: −3y−3=−9 gives y=2, then x=1.Check all three: 6,3,2 ✓
Example 3 — the inverse of a 3 × 3 matrix
Find the inverse of A=105216340.
- 1detA=1(0−24)−2(0−20)+3(0−5)=−24+40−15=1Non-zero, so an inverse exists. Check this before doing nine cofactors.
- 2Find the nine minors, then apply the sign pattern to get the cofactors.
- 3Cofactor matrix: −2418520−15−4−541
- 4Transpose to get the adjoint.A real step, not a formality.
- 5A−1=11−2420−518−1545−41
- 6Check: AA−1=I ✓Thirty seconds, and conclusive.