Worked examples — Determinants — cofactor expansion along any row - column

Before anything, the two base facts everything rests on. First the $2\times 2$ determinant:

Second, the cofactor symbol $C_{ij}$ we will use in every formula below:

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The scenario matrix

Here is every class of case cofactor expansion can present. Each row is a distinct trap or twist; the last column says which worked example kills it.

#	Case class	What makes it different	Killed by
A	Plain $3\times 3$, no zeros	Full 3-cofactor grind	Ex 1
B	A row/column with zeros	80/20: pick the sparse line	Ex 2
C	Sign-trap position $(i,j)$ with $i+j$ odd	The $(-1{)}^{i+j}=-1$ that people forget	Ex 3
D	"Same answer any line" proof-in-practice	Expand two ways, compare	Ex 4
E	Make-zeros-first (row ops)	Row-add is free; row-swap costs $-1$	Ex 5
F	Degenerate / singular ($\det =0$)	Detect collapse, meaning	Ex 6
G	$4\times 4$ recursion	Cofactor of a cofactor	Ex 7
H	Real-world word problem (area)	Determinant = signed area	Ex 8
I	Exam twist: symbolic / unknown entry	Solve for $x$ making $\det =0$	Ex 9

Work through all nine and you have touched every cell.

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Example 1 — Case A: plain $3\times 3$, no free zeros

1. Write the row-1 formula. Why this step? Cofactor expansion along row $i=1$ says walk the top row, each entry times its signed minor: $\det A=a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}.$
2. Read the signs off the checkerboard. $(-1{)}^{1+1}=+$, $(-1{)}^{1+2}=-$, $(-1{)}^{1+3}=+$. Why this step? The sign depends only on the position $(i,j)$, never on the entry's value.
3. Compute each minor by covering that entry's row and column. Why this step? Deleting the entry's row and column is the definition of its minor — it strips the problem down to a $2\times 2$ we already know how to solve. $M_{11}=\det \left(\begin{array}{cc}4& 1\\ 2& -2\end{array}\right)=4(-2)-1(2)=-10,$ $M_{12}=\det \left(\begin{array}{cc}0& 1\\ 5& -2\end{array}\right)=0(-2)-1(5)=-5,$ $M_{13}=\det \left(\begin{array}{cc}0& 4\\ 5& 2\end{array}\right)=0(2)-4(5)=-20.$
4. Combine with entries and signs. Why this step? Now plug $a_{1j}$, sign, minor together: $\det A=2(+)(-10)+(-1)(-)(-5)+3(+)(-20).$ $=-20-5-60=-85.$

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Example 2 — Case B: chase the zeros

1. Pick column 2 and write the formula. Why this step? 80/20: two of the three terms are multiplied by $0$, so they vanish before you lift a pencil. $\det A=a_{12}{C}_{12}+a_{22}{C}_{22}+a_{32}{C}_{32}=0+4C_{22}+0.$
2. Sign of $C_{22}$: $(-1{)}^{2+2}=+$. Why this step? Position $(2,2)$ sits on a $+$ square.
3. One minor: delete row 2, column 2. Why this step? Removing the surviving entry's row and column leaves the $2\times 2$ block whose determinant is the minor. $M_{22}=\det \left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)=4-1=3.$
4. Combine: Why this step? Entry $\times$ sign $\times$ minor is the single surviving term. $\det A=4\cdot (+)\cdot 3=12.$

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Example 3 — Case C: the odd-position sign trap

1. List the position signs. Why this step? Get every sign settled before touching numbers so a stray minus can't sneak in. $(-1{)}^{1+3}=+,(-1{)}^{2+3}=-,(-1{)}^{3+3}=+.$
2. Three minors (delete each entry's row and column 3). Why this step? Each minor is what survives after covering the entry's own row and column 3 — the same shading picture as the intro figure. $M_{13}=\det \left(\begin{array}{cc}0& 4\\ 5& 2\end{array}\right)=-20,M_{23}=\det \left(\begin{array}{cc}2& -1\\ 5& 2\end{array}\right)=4-(-5)=9,$ $M_{33}=\det \left(\begin{array}{cc}2& -1\\ 0& 4\end{array}\right)=8-0=8.$
3. Combine entry $\times$ sign $\times$ minor. Why this step? The middle term is where people crash — entry $1$, sign $-$, minor $9$: $\det A=3(+)(-20)+1(-)(9)+(-2)(+)(8)=-60-9-16=-85.$

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Example 4 — Case D: prove "any line" on a fresh matrix

1. Expand along row 1. Why this step? Standard start. $\det C=1(+)\det \left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)+2(-)\det \left(\begin{array}{cc}2& 2\\ 2& 1\end{array}\right)+2(+)\det \left(\begin{array}{cc}2& 1\\ 2& 2\end{array}\right).$
2. Minors. Why this step? Each $2\times 2$ is the block left after covering a row-1 entry's row and column. $\det \left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)=-3,\det \left(\begin{array}{cc}2& 2\\ 2& 1\end{array}\right)=-2,\det \left(\begin{array}{cc}2& 1\\ 2& 2\end{array}\right)=2.$
3. Combine. Why this step? Entries $\times$ signs $\times$ minors, summed: $1(-3)+2(-)(-2)+2(2)=-3+4+4=5.$

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Example 5 — Case E: make zeros first (row ops), mind the swap

1. Row-add operations. Why this step? Adding a multiple of one row to another leaves $\det$ unchanged — pure profit. Do $R_2\to R_2-2R_1$ and $R_3\to R_3-R_1$: $\to \left(\begin{array}{ccc}1& 2& 3\\ 0& 1& -3\\ 0& -2& 5\end{array}\right).$
2. Expand down column 1 — only the top entry is nonzero. Why this step? The two engineered zeros kill two of the three cofactors, leaving one. $\det B=1(+)\det \left(\begin{array}{cc}1& -3\\ -2& 5\end{array}\right)=1(5-6)=-1.$
3. A cautionary variant. Why this step? Suppose instead you had swapped $R_1\leftrightarrow R_3$ somewhere. A swap multiplies $\det$ by $-1$; scaling a row by $k$ multiplies by $k$. Row-adds are the only free operation.

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Example 6 — Case F: the degenerate (singular) matrix

1. Expand along row 1. Why this step? No zeros, so grind it honestly to prove the collapse. $\det S=1(+)\det \left(\begin{array}{cc}4& 6\\ 8& 9\end{array}\right)+2(-)\det \left(\begin{array}{cc}2& 6\\ 7& 9\end{array}\right)+3(+)\det \left(\begin{array}{cc}2& 4\\ 7& 8\end{array}\right).$
2. Minors. Why this step? Cover each row-1 entry's row and column to read off the surviving $2\times 2$. $\det \left(\begin{array}{cc}4& 6\\ 8& 9\end{array}\right)=36-48=-12,\det \left(\begin{array}{cc}2& 6\\ 7& 9\end{array}\right)=18-42=-24,\det \left(\begin{array}{cc}2& 4\\ 7& 8\end{array}\right)=16-28=-12.$
3. Combine. Why this step? Entries $\times$ signs $\times$ minors, summed: $1(-12)+2(-)(-24)+3(-12)=-12+48-36=0.$

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Example 7 — Case G: a $4\times 4$ (cofactor of a cofactor)

1. Expand along row 2. Why this step? One nonzero entry $\Rightarrow$ one cofactor. $\det D=3\cdot (-1{)}^{2+2}{M}_{22}=3M_{22},$ where $M_{22}$ deletes row 2 and column 2: $M_{22}=\det \left(\begin{array}{ccc}1& 2& -1\\ 2& 1& 0\\ 0& 0& 4\end{array}\right).$
2. Now expand this $3\times 3$ along its bottom row $(0,0,4)$ — again sparse. Why this step? Recursion: a $4\times 4$ cofactor is itself solved by cofactor expansion. $M_{22}=4\cdot (-1{)}^{3+3}\det \left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)=4(1-4)=4(-3)=-12.$
3. Combine. Why this step? Feed the inner minor back into the single row-2 term. $\det D=3\cdot (-12)=-36.$

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Example 8 — Case H: real-world word problem (signed area)

1. Build the edge vectors from $P$. Why this step? A determinant measures the area of the parallelogram spanned by two vectors; the triangle is half of it. $\overrightarrow{PQ}=(4,1),\overrightarrow{PR}=(1,3).$
2. Form and expand the $2\times 2$ determinant (base case, no cofactors needed). Why this step? At $2\times 2$ we are already at the base case $ad-bc$ — nothing left to expand. $\det \left(\begin{array}{cc}4& 1\\ 1& 3\end{array}\right)=4\cdot 3-1\cdot 1=11.$
3. Halve it. Why this step? The determinant gives the parallelogram area; a triangle is exactly half. $\text{Area}=\frac{1}{2}(11)=5.5{\text{m}}^2.$ Why positive? The sign of the determinant is $+$, meaning $P\to Q\to R$ runs counter-clockwise. A clockwise ordering would give $-5.5$; the magnitude is still the area.

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Example 9 — Case I: exam twist with an unknown

1. Expand along row 1 (has a zero at position $(1,3)$). Why this step? One term vanishes. $\det M=x(+)\det \left(\begin{array}{cc}x& 1\\ 1& x\end{array}\right)+1(-)\det \left(\begin{array}{cc}1& 1\\ 0& x\end{array}\right)+0.$
2. Minors. Why this step? Cover each row-1 entry's row and column to read the surviving $2\times 2$. $\det \left(\begin{array}{cc}x& 1\\ 1& x\end{array}\right)=x^2-1,\det \left(\begin{array}{cc}1& 1\\ 0& x\end{array}\right)=x.$
3. Combine. Why this step? Entries $\times$ signs $\times$ minors gives the determinant as a polynomial in $x$: $\det M=x(x^2-1)-1\cdot x=x^3-x-x=x^3-2x=x(x^2-2).$
4. Set to zero and solve. Why this step? Singular $\iff \det =0$. $x(x^2-2)=0\Rightarrow x=0,x=\sqrt{2},x=-\sqrt{2}.$

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Recall Self-test across all cells

Cover the answers and reproduce each number. Ex 1 answer ::: $-85$ Ex 3 answer ::: $-85$ Ex 4 answer ::: $5$ Ex 5 answer ::: $-1$ Ex 6 answer (singular) ::: $0$ Ex 7 answer ::: $-36$ Ex 8 area ::: $5.5{\text{m}}^2$ Ex 9 singular values ::: $x=0,\pm \sqrt{2}$

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Connections

• Parent: Cofactor expansion
• Leibniz formula for determinants — where the signs ultimately come from.
• Determinant via row reduction — Example 5's trick pushed all the way.
• Properties of determinants (multilinearity, alternation) — Examples 6 & 8 use these.
• Invertibility and singular matrices — Examples 6 & 9.
• Adjugate matrix and inverse and Cramer's rule — both built on cofactors.