Chapter 9 Quiz — The Inverse Matrix
Twelve conceptual checks. Try each before opening the answer. These test understanding, not arithmetic speed — if you can answer them, you have the chapter.
Q1. What is the single best one-sentence answer to "what does $A^{-1}$ do?"
Answer
$A^{-1}$ is the transformation that *exactly undoes* $A$ — it reverses the stretch, rotation, shear, or reflection, so that applying $A$ and then $A^{-1}$ returns every point to where it started ($A^{-1}A = AA^{-1} = I$). The inverse is the rewind button.Q2. A transformation collapses the plane onto a line. Is it invertible? Answer and explain why in terms of information.
Answer
No — it is **singular**. Collapsing onto a line means infinitely many input points get crushed onto each surviving point, so the information about *which* input you started from is destroyed. No transformation can re-inflate the line back into the plane, so there is no inverse. Singular = collapse = irreversible.Q3. State at least four conditions, each equivalent to "$A$ is invertible," for a square matrix.
Answer
Any four of: the map $\mathbf{x}\mapsto A\mathbf{x}$ is bijective; $\det(A)\neq 0$; $\operatorname{rank}(A) = n$ (full rank); a pivot in every row and column; the columns are linearly independent; the null space is trivial ($A\mathbf{x}=\mathbf{0}\Rightarrow\mathbf{x}=\mathbf{0}$); $A\mathbf{x}=\mathbf{b}$ has a unique solution for every $\mathbf{b}$. They are all the same geometric fact: the transformation loses no dimension.Q4. Why is $(AB)^{-1} = B^{-1}A^{-1}$ and not $A^{-1}B^{-1}$?
Answer
Because $AB$ means "do $B$, then $A$" (composition), and to undo a sequence you reverse the most recent action first — undo $A$, then undo $B$ — giving $B^{-1}A^{-1}$. Socks then shoes; to undo, remove shoes then socks. The order-reversal is forced because matrix multiplication doesn't commute. Algebraically: $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I$.Q5. How does Gauss-Jordan compute $A^{-1}$, and what happens to the algorithm if $A$ is singular?
Answer
Form the augmented matrix $[A\,|\,I]$ and row-reduce until the left block becomes $I$; the right block is then $A^{-1}$. (The row operations that turn $A$ into $I$ collectively *are* $A^{-1}$, and they turn $I$ into $A^{-1}$.) If $A$ is singular, the left block can never become $I$ — a row of zeros appears, signaling a missing pivot — and the algorithm reports "no inverse exists."Q6. Your matrix has $\det(A) = 0.000001$. Does the small determinant mean the matrix is nearly singular and the inverse is untrustworthy?
Answer
Not necessarily. The determinant scales with the entries (e.g. halving a $10\times 10$ matrix divides $\det$ by $2^{10}$ without changing invertibility), so its raw magnitude is a poor health check. A tiny nonzero $\det$ can come from a perfectly well-conditioned matrix with small entries. The right measure of "near-singular" is the **condition number** (Chapter 38), not the size of the determinant. Only $\det = 0$ *exactly* means singular.Q7. Why should you almost never solve $A\mathbf{x} = \mathbf{b}$ by computing np.linalg.inv(A) @ b?
Answer
Because solving directly (`np.linalg.solve`, via LU factorization and substitution) is both *faster* — inverting does about three times the arithmetic and then still needs a matrix-vector multiply — and *more accurate* — forming $A^{-1}$ and multiplying chains extra floating-point error, which can corrupt the answer for ill-conditioned matrices. The maxim: "never invert a matrix to solve a system." Use `inv` only when you need the inverse *as an object*.Q8. The inverse of a rotation $R(\theta)$ turns out to equal its transpose, $R^{-1} = R^{\mathsf{T}}$. Why does that make sense geometrically?
Answer
To undo a counterclockwise rotation by $\theta$, you rotate clockwise by $\theta$, i.e. by $-\theta$. Since $\cos(-\theta)=\cos\theta$ and $\sin(-\theta)=-\sin\theta$, the matrix $R(-\theta)$ is $R(\theta)$ with the off-diagonal signs swapped — which is exactly the transpose. (This is special to rotations and reflections — the *orthogonal* matrices of Chapter 21, where $A^{-1} = A^{\mathsf{T}}$ always.)Q9. Is the transpose $A^{\mathsf{T}}$ the same as the inverse $A^{-1}$?
Answer
No, in general they are completely different transformations. The transpose flips rows and columns; the inverse undoes the motion. For a horizontal shear, the transpose is a *vertical* shear (not the inverse, which is a horizontal shear the other way). They coincide only for orthogonal matrices (rotations/reflections). Confusing the two is a common error.Q10. Can a non-square matrix have a two-sided inverse? What about a one-sided inverse?
Answer
A non-square matrix cannot have a *two-sided* inverse (that would force the input and output dimensions to match). But it can have a *one-sided* inverse: a tall matrix with independent columns has a left inverse $B$ with $BA = I$ (recovering its input), and a wide matrix with independent rows has a right inverse. The canonical one for least squares is the pseudoinverse $(A^{\mathsf{T}}A)^{-1}A^{\mathsf{T}}$ (Chapter 19). "Inverse" unqualified means the two-sided inverse of a square matrix.Q11. For a square matrix, if you find a $B$ with $BA = I$, do you still need to check $AB = I$ separately?
Answer
No. For *square* matrices, a one-sided inverse is automatically two-sided: $BA = I$ forces $AB = I$ (and $B = A^{-1}$). The reason is that $BA = I$ makes $A$ injective, hence (being square) bijective, hence reversible on both sides. So one equation suffices — which is why Gauss-Jordan, which builds a right inverse, delivers the full inverse for free. This fails for non-square matrices.Q12. If $\det(A) = 5$, what is $\det(A^{-1})$, and why?