Exercises — Properties — row operations, multiplicativity

An elementary matrix is what you get by doing ONE row operation to the identity matrix $I$ (the matrix with $1$s on the diagonal, $0$s elsewhere). Because doing the operation to $A$ equals multiplying $A$ on the left by $E$, the factor $\det E$ is the effect.

• (a) Swap $\Rightarrow$ $\det$ multiplied by $-1$; $\det E=-1$.
• (b) Scale by $5$ $\Rightarrow$ $\det$ multiplied by $5$; $\det E=5$.
• (c) Replacement $\Rightarrow$ $\det$ unchanged; $\det E=1$.

One swap flips the sign once. So $\det A^{\prime }=-7$.

Step (why): clear column 1 below the pivot using replacements — these never change $\det$. $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$: $\left(\begin{array}{ccc}1& 2& 0\\ 0& 1& 3\\ 0& 1& 5\end{array}\right).$ Step: clear the last entry in column 2: $R_3\to R_3-R_2$ (replacement, no change): $\left(\begin{array}{ccc}1& 2& 0\\ 0& 1& 3\\ 0& 0& 2\end{array}\right).$ Step: now triangular. No swaps, no scalings, so $\det A=1\cdot 1\cdot 2=2$.

Multiplying the whole matrix by $2$ scales each of the $4$ rows by $2$. Each scaling pulls out one factor of $2$: $\det (2A)=2^4\det A=16\cdot 3=48.$

$\det A=1\cdot 4-2\cdot 3=-2$. $\det B=0\cdot 0-1\cdot 1=-1$. $AB=\left(\begin{array}{cc}2& 1\\ 4& 3\end{array}\right)$, so $\det (AB)=2\cdot 3-1\cdot 4=2$. Check: $\det A\det B=(-2)(-1)=2$. ✓

The top-left entry is $0$, so we cannot use it as a pivot directly. Step (why): swap $R_1\leftrightarrow R_2$ to bring a nonzero pivot up. This is $1$ swap, so $s=1$. $\left(\begin{array}{ccc}1& 1& 3\\ 0& 2& 1\\ 2& 0& 1\end{array}\right).$ Step: clear column 1: $R_3\to R_3-2R_1$ (replacement, no change): $\left(\begin{array}{ccc}1& 1& 3\\ 0& 2& 1\\ 0& -2& -5\end{array}\right).$ Step: clear column 2: $R_3\to R_3+R_2$ (replacement): $\left(\begin{array}{ccc}1& 1& 3\\ 0& 2& 1\\ 0& 0& -4\end{array}\right).$ Step: pivots product $=1\cdot 2\cdot (-4)=-8$. Apply the swap sign: $\det B=(-1{)}^s\cdot (-8)=(-1{)}^1\cdot (-8)=8.$

Reduce to $I$ means the final triangular product is $\det I=1$. Undo the bookkeeping. Each operation multiplied the current determinant by its factor. Starting from $\det C$ and ending at $1$: $1=(-1{)}^1\cdot \frac{1}{2}\cdot 3\cdot \det C(\text{replacements contribute }\times 1).$ So $(-1)\cdot \frac{3}{2}\cdot \det C=1\Rightarrow \det C=\frac{1}{-\frac{3}{2}}=-\frac{2}{3}.$

• $\det (A^{-1})=\frac{1}{\det A}=\frac{1}{4}$, because $\det A\cdot \det A^{-1}=\det I=1$ (see Invertibility & Singular Matrices).
• $\det (A^3)=(\det A{)}^3=4^3=64$ (multiplicativity applied twice).
• $\det (2A^{-1})=2^3\det (A^{-1})=8\cdot \frac{1}{4}=2$ (three rows scaled by $2$).

(a) Look at the figure. $S$ stretches the horizontal direction by $2$ (a $2\times 1$ rectangle) — a genuine area change. $H$ is a shear (a replacement operation in matrix form): it slides the top edge sideways but keeps the base and height, so the parallelogram has the same area as the square. This is the geometric face of "replacement doesn't change $\det$." (b) $\det S=2\cdot 1-0=2$. $\det H=1\cdot 1-1\cdot 0=1$. (c) Multiplicativity predicts the area factor of doing $S$ then $H$ (the map $HS$) is $\det H\cdot \det S=1\cdot 2=2$. Confirm: $HS=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$, and $\det (HS)=2\cdot 1-1\cdot 0=2$. ✓ The shear added no area.

Proof: suppose rows $i$ and $j$ are equal. Do the replacement $R_i\to R_i-R_j$. Replacement doesn't change $\det$. But now row $i$ is all zeros, and a matrix with a zero row has $\det =0$ (a zero row means one direction is crushed to nothing — zero volume). So $\det A=0$. Illustration: rows $1$ and $2$ are equal. $R_2\to R_2-R_1$ gives a zero row $2$; hence $\det A=0$.

A cyclic shift of $3$ rows can be built from $2$ swaps: start $I$ rows $(1,2,3)$. Swap $R_1\leftrightarrow R_2\to (2,1,3)$; swap $R_2\leftrightarrow R_3\to (2,3,1)$ — the target ordering. That's $2$ swaps, so $\det P=(-1{)}^2=1$.

Peel it apart with the tools: $\det (cA)=c^n\det A=c^{n}d,\det ((cA{)}^{-1})=\frac{1}{\det (cA)}=\frac{1}{c^{n}d}.$ For $n=3,c=2,d=5$: $\frac{1}{2^3\cdot 5}=\frac{1}{40}$.

Apply $\det$ to both sides of $A^2=A$. By multiplicativity $\det (A^2)=(\det A{)}^2$, and the right side is $\det A$: $(\det A{)}^2=\det A\Rightarrow (\det A{)}^2-\det A=0\Rightarrow \det A(\det A-1)=0.$ So $\det A\in \{0,1\}$. Connection: an idempotent's eigenvalues are all $0$ or $1$, and $\det A$ is the product of eigenvalues. If any eigenvalue is $0$ the product is $0$ (singular projection); if all are $1$ the product is $1$ (then $A=I$). Both cases match.

Why this is true: multiplicativity says $\det (E_1{E}_2\cdots E_m)=\det E_1\cdot \det E_2\cdots \det E_m$. Each factor is a known number:

• each swap contributes $\det E=-1$ — there are $s$ of them, giving $(-1{)}^s$;
• each scale-by-$k_r$ contributes $\det E=k_r$ — giving ${\prod }_r{k}_r$;
• each replacement contributes $\det E=1$ — leaving the product untouched. Multiplying all factors: $\det Q=(-1{)}^s({\prod }_r{k}_r)\cdot 1=(-1{)}^s{\prod }_r{k}_r$. $■$ This is the abstract engine behind the "REF bookkeeping" formula — see Gaussian Elimination & Row Echelon Form.

• L1: name the effect $\to$ swap $-1$, scale $k$, replacement $1$.
• L2: reduce & multiply pivots; watch $2^n$ for $\det (2A)$.
• L3: carry $(-1{)}^s$ and every scaling factor; pay back divisions.
• L4: shear = area-preserving; identical rows $\Rightarrow \det =0$.
• L5: $\det (cA{)}^{-1}=\frac{1}{c^{n}d}$; idempotent $\Rightarrow \det \in \{0,1\}$; $\det Q=(-1{)}^s\prod k_r$.