Your matrix has det(A) = −15, so an inverse exists. This page walks every row operation on the augmented matrix [A | I], shows the reasoning behind each one, then verifies the answer by hand and by a second independent method. Work through it once and the method becomes mechanical.
Find A⁻¹ by row reducing the augmented matrix [A | I]. The question gives two choices: give the inverse, or declare that no inverse exists.
Augment A with the 3×3 identity. The identity on the right is not decoration — it is a ledger. Every row operation you perform gets applied to both halves simultaneously, so the right half quietly records the sequence of operations that undoes A.
Before you touch a single row, spend ten seconds on the determinant. For a square matrix, invertible ⇔ det(A) ≠ 0. Computing det(A) with the 3×3 formula is faster than the row reduction, and it tells you which multiple-choice option to pick.
A⁻¹ is the matrix B that undoes A. Formally, A⁻¹ satisfies both A·A⁻¹ = I and A⁻¹·A = I — the identity matrix, which is the multiplicative identity: I·M = M·I = M for every matrix M.
Read it as an equation system: multiplying by A⁻¹ is the same as dividing by A. That is why every linear system Ax = b can be solved as x = A⁻¹b.
Every elementary row operation is itself a matrix multiplication. Swapping rows, scaling a row, adding a multiple of one row to another — each one can be written as left-multiplying the whole augmented matrix by some matrix E.
Now suppose the reduction reaches [I | E]. Then E·A = I, so E is a left inverse of A; and because A is square, that makes E = A⁻¹. Whatever lands on the right when the left side becomes I is the inverse. That is the whole method in one sentence.
The determinant of the triangular matrix you end up with (before scaling the last row) is the product of its diagonal: 1 · 1 · (−15) = −15. That matches det(A) — and it is not a coincidence. Row addition does not change the determinant, so the determinant of the reduced form equals the determinant of the original.
One caveat worth knowing: scaling a row multiplies the determinant. Row additions preserve it; row swaps negate it; scaling by k multiplies it by k. So if you scale the last row by −1/15, the reduced matrix has determinant 1 while A has determinant −15.
Think of A as a linear map on R³ — a stretch, a shear, a rotation, maybe a flip. A⁻¹ is the map that undoes all of it: it stretches by the reciprocal amount along each direction and undoes every shear.
This is why invertibility is really about information loss. If A squashes the cube flat along one axis (det = 0), no map can un-squash it — the data along that axis is gone, and no inverse exists. A det of −15 means A is a volume-expanding, orientation-flipping map: it multiplies volumes by 15 and reverses handedness. A⁻¹ multiplies volumes by 1/15 and flips handedness back.
In the classroom you compute A⁻¹ because the exercise asks for it. On the job, you almost never do — and knowing why is part of understanding the math:
Row reduction can fail silently if you slip on an arithmetic sign. Multiplying A by your candidate inverse and confirming you get the identity takes 60 seconds and catches every sign error. Nine dot products, each must equal 1, 0, or 0.
Every entry lands on the identity — no rounding, no decimals. That is the signature of a correct inverse: the products cancel exactly.
A⁻¹ = adj(A) / det(A), where adj(A) is the transpose of the cofactor matrix. Slower by hand than row reduction — but it is the fastest way to catch an arithmetic slip, because it uses completely different arithmetic. If both routes give the same matrix, the answer is right.
Identical to the row-reduction result. Two independent methods, same matrix — that is how you know.
Enter the nine entries of A⁻¹. Fractions are fine (2/15), decimals are fine
(0.1333), and a leading minus works either way. It checks each cell the moment you
leave the field.
Select option A and enter:
If the homework box wants plain ASCII, type:
-1/5 -2/5 1/5 · 2/5 -1/5
3/5 · 2/5 2/15 -1/15