Exercises — Determinants — cofactor expansion along any row - column

WHAT $M_{23}$ means: cover row 2 and column 3. Row 2 is $(0,3,4)$; column 3 is $(1,4,8)$. Delete both. $M_{23}=\det \left(\begin{array}{cc}2& 5\\ 7& 6\end{array}\right)=2\cdot 6-5\cdot 7=12-35=-23.$ So $M_{23}=-23$. (Notice: $M$ carries no sign yet — that is the minor's job, not the cofactor's.)

The cofactor is the signed minor: $C_{23}=(-1{)}^{2+3}{M}_{23}$. Since $2+3=5$ is odd, $(-1{)}^5=-1$. $C_{23}=(-1)\cdot (-23)=+23.$ The sign flipped the minor's sign — this is exactly the checkerboard doing its job.

$\left(\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right)$

• $(1,2)$: $(-1{)}^{1+2}=(-1{)}^3=-$
• $(2,2)$: $(-1{)}^{2+2}=(-1{)}^4=+$
• $(3,1)$: $(-1{)}^{3+1}=(-1{)}^4=+$ The sign depends only on $i+j$ — the actual position, never on whether you call it a row or column expansion.

Expand along row 1: $\det A=1\cdot C_{11}+2\cdot C_{12}+3\cdot C_{13}$. $C_{11}=(+)\det \left(\begin{array}{cc}5& 6\\ 8& 10\end{array}\right)=50-48=2.$ $C_{12}=(-)\det \left(\begin{array}{cc}4& 6\\ 7& 10\end{array}\right)=-(40-42)=-(-2)=2.$ $C_{13}=(+)\det \left(\begin{array}{cc}4& 5\\ 7& 8\end{array}\right)=32-35=-3.$ $\det A=1(2)+2(2)+3(-3)=2+4-9=-3.$

WHY choose row 1: it has two zeros, so two of the three cofactor terms die instantly. Only $a_{13}=4$ survives. $\det =4\cdot C_{13}=4\cdot (-1{)}^{1+3}\det \left(\begin{array}{cc}1& 2\\ 5& 6\end{array}\right)=4\cdot (+)(6-10)=4\cdot (-4)=-16.$

Column 1 is $(0,1,5{)}^T$; the top zero drops out. $\det =1\cdot C_{21}+5\cdot C_{31}.$ $C_{21}=(-1{)}^{2+1}\det \left(\begin{array}{cc}0& 4\\ 6& 7\end{array}\right)=-(0-24)=24.$ $C_{31}=(-1{)}^{3+1}\det \left(\begin{array}{cc}0& 4\\ 2& 3\end{array}\right)=+(0-8)=-8.$ $\det =1(24)+5(-8)=24-40=-16.✓$ Same value — the "expand along any row/column" theorem in action.

WHY row-add is free: adding a multiple of one row to another leaves $\det$ unchanged. Kill the first column below the top entry. Do $R_2\to R_2-2R_1$ and $R_3\to R_3-3R_1$: $\to \left(\begin{array}{ccc}2& 1& 3\\ 0& 1& 2\\ 0& 2& 4\end{array}\right).$ Expand down column 1 (only the top entry $2$ survives): $\det B=2\cdot (-1{)}^{1+1}\det \left(\begin{array}{cc}1& 2\\ 2& 4\end{array}\right)=2\cdot (4-4)=2\cdot 0=0.$ $\det B=0$ — the matrix is singular (its rows are dependent: $R_3=R_1+R_2$).

Expand along column 1 (a zero in the bottom slot): $\det =k\cdot (-1{)}^{1+1}\det \left(\begin{array}{cc}k& 1\\ 1& k\end{array}\right)+1\cdot (-1{)}^{2+1}\det \left(\begin{array}{cc}2& 0\\ 1& k\end{array}\right).$ $=k(k^2-1)-(2k-0)=k^3-k-2k=k^3-3k.$ Set to zero: $k^3-3k=k(k^2-3)=0$, so $k=0,k=\sqrt{3},k=-\sqrt{3}.$ At these $k$ the matrix collapses space → non-invertible.

WHY subtract rows: $R_2\to R_2-R_1$, $R_3\to R_3-R_1$ zeros out column 1 below the top and leaves clean factors: $\to \left(\begin{array}{ccc}1& x& x^2\\ 0& y-x& y^2-x^2\\ 0& z-x& z^2-x^2\end{array}\right).$ Expand down column 1 (only top entry $1$ survives): $\det A=1\cdot \det \left(\begin{array}{cc}y-x& y^2-x^2\\ z-x& z^2-x^2\end{array}\right).$ Factor each entry: $y^2-x^2=(y-x)(y+x)$, $z^2-x^2=(z-x)(z+x)$. Pull $(y-x)$ from row 1, $(z-x)$ from row 2: $=(y-x)(z-x)\det \left(\begin{array}{cc}1& y+x\\ 1& z+x\end{array}\right)=(y-x)(z-x)[(z+x)-(y+x)].$ $=(y-x)(z-x)(z-y).$

$\det (2A)$: scaling the whole $3\times 3$ matrix by $2$ scales each of the 3 rows by $2$, and each row-scaling multiplies $\det$ by $2$. So $\det (2A)=2^3\det A=8\cdot 7=56$. $\det (A^T)$: the Leibniz sum is invariant under transpose, so $\det (A^T)=\det A=7$. (General rule: $\det (kA)=k^n\det A$ for an $n\times n$ matrix.)

For a $2\times 2$, each minor is a single leftover entry. $C_{11}=+(4)=4,C_{12}=-(3)=-3,C_{21}=-(2)=-2,C_{22}=+(1)=1.$ The adjugate is the transpose of the cofactor matrix: $\mathrm{adj}(A)=\left(\begin{array}{cc}{C}_{11}& C_{21}\\ C_{12}& C_{22}\end{array}\right)=\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right).$ With $\det A=1\cdot 4-2\cdot 3=-2$: $A^{-1}=\frac{1}{\det A}\mathrm{adj}(A)=\frac{1}{-2}\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right)=\left(\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right).$ This links to Adjugate matrix and inverse.

Coefficient matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$, $\det A=6-1=5$. Replace column 1 (the $x$-column) with the right-hand side $(5,10{)}^T$: $A_x=\left(\begin{array}{cc}5& 1\\ 10& 3\end{array}\right),\det A_x=15-10=5.$ $x=\frac{\det A_x}{\det A}=\frac{5}{5}=1.$ (Then $y=(10-1)/3=3$ from the second equation; check $2(1)+3=5$.✓) See Cramer's rule.

Say rows $2$ and $3$ are equal. Swapping those two identical rows visibly leaves the matrix unchanged, so its determinant is unchanged: $\det A=\det A$. But a single row-swap multiplies the determinant by $-1$ (an alternation property): $\det A=-\det A$. Adding $\det A$ to both sides: $2\det A=0\Rightarrow \det A=0$. Cofactor cross-check: subtract $R_3-R_2$ (free) to make row 3 all zeros; expanding along that zero row gives ${\sum }_{j}0\cdot C_{3j}=0$. Both routes agree.

First the row-2 cofactors: $C_{21}=(-1{)}^3\det \left(\begin{array}{cc}2& 3\\ 8& 10\end{array}\right)=-(20-24)=4,$ $C_{22}=(-1{)}^4\det \left(\begin{array}{cc}1& 3\\ 7& 10\end{array}\right)=(10-21)=-11,$ $C_{23}=(-1{)}^5\det \left(\begin{array}{cc}1& 2\\ 7& 8\end{array}\right)=-(8-14)=6.$ Now dot them against row 1 $=(1,2,3)$: $1(4)+2(-11)+3(6)=4-22+18=0.✓$ WHY zero: this expression is the determinant of the matrix whose rows 1 and 2 are both row 1 of $A$ — a repeated row, so $\det =0$ by L5.1. This "alien cofactor = 0" fact is exactly what makes $A\cdot \mathrm{adj}(A)=(\det A)I$ work.

Column 1 is $(a_{11},0,0{)}^T$ — expand there. Only the top term survives: $\det =a_{11}\cdot (-1{)}^{1+1}\det \left(\begin{array}{cc}{a}_{22}& a_{23}\\ 0& a_{33}\end{array}\right).$ The inner $2\times 2$ is $a_{22}{a}_{33}-a_{23}\cdot 0=a_{22}{a}_{33}$. So $\det =a_{11}{a}_{22}{a}_{33}$ — the diagonal product. Numerically: $3\cdot 5\cdot 4=60$. Every step "chased the zeros" that triangularity handed us for free.