Worked examples — Determinant of 3×3 matrix — cofactor expansion

This page is a drill hall. The parent note taught you the method; here we hunt down every kind of matrix a problem can throw at you and work each one to the end. Before we start, one reminder of the vocabulary, so no symbol is used before it is earned.

The checkerboard of signs you will glance at constantly: $\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right]$

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The scenario matrix

Every 3×3 determinant problem is one (or a mix) of the cases below. Each worked example names the cell(s) it covers, so by the end every cell is hit.

#	Case class	What makes it special	Example that hits it
A	All-nonzero, mixed signs	No shortcuts — brute-force one row carefully	Ex 1
B	Row/column full of zeros	Whole term vanishes → less work	Ex 2
C	Two expansions agree	Prove invariance (row vs column)	Ex 3
D	Triangular matrix	Answer = product of the diagonal	Ex 4
E	Degenerate (det = 0)	Rows linearly dependent → flat, zero volume	Ex 5
F	Sign / orientation flip	Negative determinant = mirrored orientation	Ex 6
G	Geometry / word problem	Volume of a real parallelepiped	Ex 7
H	Exam twist (unknown letter)	Solve for $k$ so that $\det =0$	Ex 8

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Example 1 — Case A: all-nonzero, mixed signs

Forecast: Row 2 is $(-1,0,4)$. The middle entry is $0$, so one of the three terms will die on its own. Guess the sign of the final answer before reading on — is it positive or negative?

1. Write the row-2 expansion with its checkerboard signs. Row 2 signs from the board are $-,+,-$. So $\det (A)=-(-1)M_{21}+(0)M_{22}-(4)M_{23}.$ Why this step? Row 2 gives cofactors $C_{21},C_{22},C_{23}$ whose signs are $(-1{)}^{2+1},(-1{)}^{2+2},(-1{)}^{2+3}=-,+,-$. Pinning the signs first stops the most common error (dropping a minus).

2. The middle term vanishes. Element $a_{22}=0$, so $0\cdot M_{22}=0$. We never even compute $M_{22}$. Why? A zero element kills its whole term — that is the entire reason we like zeros.

3. Compute $M_{21}$ (delete row 2, column 1): $M_{21}=\left|\begin{array}{cc}3& 2\\ 5& 1\end{array}\right|=3\cdot 1-2\cdot 5=3-10=-7.$ Why this step? The surviving 2×2 is exactly the four entries left when we shade out row 2 and column 1.

4. Compute $M_{23}$ (delete row 2, column 3): $M_{23}=\left|\begin{array}{cc}1& 3\\ 2& 5\end{array}\right|=1\cdot 5-3\cdot 2=5-6=-1.$

5. Assemble. $\det (A)=-(-1)(-7)+0-(4)(-1)=-(7)+4=-3.$ Why this step? Carefully: $-(-1)\times (-7)=(+1)\times (-7)=-7$, then $-(4)(-1)=+4$. Two sign flips per term — this is where beginners go wrong.

Verify: Expand instead along row 1 as an independent path. $1\left|\begin{array}{cc}0& 4\\ 5& 1\end{array}\right|-3\left|\begin{array}{cc}-1& 4\\ 2& 1\end{array}\right|+2\left|\begin{array}{cc}-1& 0\\ 2& 5\end{array}\right|=1(-20)-3(-9)+2(-5)=-20+27-10=-3.✓$ Same answer, different route — exactly what the parent note promised.

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Example 2 — Case B: a column full of zeros

Forecast: An entire column is zero. What does zero volume look like? Predict the answer in your head before any arithmetic.

1. Choose the zero column. Expand along column 2: entries $(0,0,0)$. Why this step? Every element there is $0$, so every term is $0\times (\text{something})=0$.
2. Add them up. $\det (B)=0\cdot C_{12}+0\cdot C_{22}+0\cdot C_{32}=0.$ Why this step? No minor even needs computing — the fastest possible determinant.

Geometric read: the three column-vectors of $B$ all have $y$-component $0$; they are trapped in the $xz$-plane. Three vectors squashed into one plane span a flat parallelepiped — zero volume.

Verify: Expanding along row 1 instead: $3\left|\begin{array}{cc}0& 5\\ 0& 7\end{array}\right|-0+2\left|\begin{array}{cc}1& 0\\ 4& 0\end{array}\right|=3(0)+2(0)=0.✓$

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Example 3 — Case C: two expansions must agree

Forecast: Column 1 is $(1,0,0)$ — two zeros. Row 1 is $(1,2,3)$ — no zeros. One path is short, one is long; the theorem says they land in the same place.

1. Expand along column 1 (the easy path). Column-1 signs are $+,-,+$. $\det (D)=1\cdot \left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|-0+0=1(4\cdot 6-5\cdot 0)=24.$ Why this step? Two zeros kill two of three terms, leaving one clean 2×2.

2. Expand along row 1 (the long path). Row-1 signs $+,-,+$. $\det (D)=1\left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|-2\left|\begin{array}{cc}0& 5\\ 0& 6\end{array}\right|+3\left|\begin{array}{cc}0& 4\\ 0& 0\end{array}\right|.$ Why this step? No zeros in row 1, so we honestly compute all three minors.

3. Evaluate each 2×2. $\left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|=24$, $\left|\begin{array}{cc}0& 5\\ 0& 6\end{array}\right|=0\cdot 6-5\cdot 0=0$, $\left|\begin{array}{cc}0& 4\\ 0& 0\end{array}\right|=0$. Why this step? Two of them have a zero column, so they vanish — the matrix's structure protects us.

4. Sum: $1(24)-2(0)+3(0)=24.$

Verify: Both paths give $24$ — invariance confirmed. This also demonstrates Case D below: $24=1\cdot 4\cdot 6$, the product of the diagonal. See properties of determinants for why invariance always holds.

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Example 4 — Case D: triangular matrix

Forecast: Guess the answer using only the diagonal $2,1,6$.

1. Expand along row 1 — it is $(2,0,0)$, two free zeros. $\det (T)=2\left|\begin{array}{cc}1& 0\\ 4& 6\end{array}\right|-0+0.$ Why this step? Zeros above the diagonal in row 1 kill two terms.
2. The surviving 2×2 is itself triangular. $\left|\begin{array}{cc}1& 0\\ 4& 6\end{array}\right|=1\cdot 6-0\cdot 4=6.$ Why this step? The anti-diagonal has a $0$, so only the main diagonal survives — the same trick one level down.
3. Multiply: $\det (T)=2\cdot 6=12=2\cdot 1\cdot 6.$

Verify: $2\cdot 1\cdot 6=12.✓$

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Example 5 — Case E: degenerate matrix (det = 0)

Forecast: The rows look suspiciously evenly spaced. Do you smell a hidden dependency?

1. Expand along row 1. $\det (C)=1\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|-2\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|+3\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|.$
2. Three 2×2s. $\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|=45-48=-3$, $\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|=36-42=-6$, $\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|=32-35=-3$.
3. Sum: $1(-3)-2(-6)+3(-3)=-3+12-9=0.$ Why this step? The alternating signs make the pieces cancel exactly — the signature of a collapsed shape.

Geometric read (see figure): row 3 = $2\times$row 2 $-$ row 1. The three row-vectors lie in a single tilted plane, so the parallelepiped they build is a flat sheet — zero thickness, zero volume.

Verify: $\det =0.✓$

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Example 6 — Case F: sign / orientation flip

Forecast: Swapping two axes is like flipping a glove inside-out. Positive or negative volume?

1. Expand along row 3 $=(0,0,1)$ — two zeros. $\det (P)=0-0+1\cdot (-1{)}^{3+3}\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|.$ Why this step? Row 3 is cheapest.
2. The minor. $\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|=0\cdot 0-1\cdot 1=-1.$
3. Assemble: $\det (P)=1\cdot (+1)\cdot (-1)=-1.$

Verify: Swapping two rows of the identity always flips the sign: $\det (I)=1\Rightarrow \det (P)=-1.✓$

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Example 7 — Case G: real-world volume (word problem)

Forecast: Volume is the magnitude of the determinant whose rows are $\mathbf{u},\mathbf{v},\mathbf{w}$. Estimate: bigger than $2\cdot 3\cdot 4=24$? Smaller?

1. Set up the determinant. Stack the vectors as rows: $V_{\text{signed}}=\left|\begin{array}{ccc}2& 1& 0\\ 0& 3& 1\\ 1& 0& 4\end{array}\right|.$ Why this step? The determinant of the edge-vectors is the signed volume — this is the whole geometric meaning from what is a determinant. (It is also the scalar triple product; see cross product and determinants.)
2. Expand along column 1 $(2,0,1)$ — one zero helps. $=2\left|\begin{array}{cc}3& 1\\ 0& 4\end{array}\right|-0+1\cdot (-1{)}^{3+1}\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|.$ Why this step? $a_{21}=0$ removes the middle term; the checkerboard gives $+,-,+$ down column 1.
3. Two minors. $\left|\begin{array}{cc}3& 1\\ 0& 4\end{array}\right|=12-0=12$, $\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|=1-0=1.$
4. Assemble: $2(12)+1(1)=24+1=25.$
5. Take the magnitude for physical volume: $|25|=25.$

Verify (units and sanity): each vector is in nm, so the volume is in ${\text{nm}}^3$: $25{\text{nm}}^3$. It exceeds the naive $2\cdot 3\cdot 4=24$ because the vectors are close to perpendicular — reasonable. Cross-check by expanding along row 1: $2(12)-1(0\cdot 4-1\cdot 1)+0=24-1(-1)=25.✓$

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Example 8 — Case H: exam twist with an unknown

Forecast: We'll get a cubic in $k$. How many real roots — one, two, or three?

1. Expand along column 1 $(1,0,k)$ — the middle zero saves one minor. $\det (K)=1\left|\begin{array}{cc}1& k\\ 0& 1\end{array}\right|-0+k\cdot (-1{)}^{3+1}\left|\begin{array}{cc}k& 2\\ 1& k\end{array}\right|.$ Why this step? Column 1 has a zero; the sign pattern down it is $+,-,+$.
2. Two minors. $\left|\begin{array}{cc}1& k\\ 0& 1\end{array}\right|=1-0=1$, $\left|\begin{array}{cc}k& 2\\ 1& k\end{array}\right|=k^2-2.$
3. Assemble as a polynomial in $k$. $\det (K)=1(1)+k(k^2-2)=k^3-2k+1.$ Why this step? Now "singular" becomes "solve $k^3-2k+1=0$."
4. Factor. Test $k=1$: $1-2+1=0$ ✓, so $(k-1)$ is a factor: $k^3-2k+1=(k-1)(k^2+k-1).$ The quadratic gives $k=\frac{-1\pm \sqrt{5}}{2}$. Why this step? Dividing out the found root leaves a quadratic we solve with the standard formula.
5. All three roots: $k=1,k=\frac{-1+\sqrt{5}}{2}\approx 0.618,k=\frac{-1-\sqrt{5}}{2}\approx -\mathrm{1.618.}$

Verify: Plug $k=1$ back: $\det =1^3-2(1)+1=0$ ✓. And $(k-1)(k^2+k-1)$ expanded $=k^3+k^2-k-k^2-k+1=k^3-2k+1$ ✓. (These $k$-values are exactly where the inverse fails and where a linear system in $K$ has no unique Cramer's-rule solution.)

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Recall Self-test: name the case, then solve

Which scenario cell does each belong to, and what's the answer? A row of zeros — determinant? ::: $0$ (Case B), space collapses. An upper-triangular matrix with diagonal $3,-2,5$ — determinant? ::: $3\cdot (-2)\cdot 5=-30$ (Case D). Swapping two rows of a matrix with $\det =7$ — new determinant? ::: $-7$ (Case F), orientation flips. $\det =0$ tells you about the inverse? ::: It doesn't exist; the matrix is singular (Case E).