Exercises — Determinant of 3×3 matrix — cofactor expansion

WHAT we do: add the two indices, then ask "even or odd?" Even $\to +$, odd $\to -$.

• $(1,3)$: $1+3=4$, even $\to +$
• $(2,3)$: $2+3=5$, odd $\to -$
• $(3,2)$: $3+2=5$, odd $\to -$
• $(2,2)$: $2+2=4$, even $\to +$

WHAT IT LOOKS LIKE: check them against the checkerboard picture — the four cells light up exactly $+,-,-,+$. The signs never depend on the numbers in the matrix, only on the position.

WHAT we look for: the line (row or column) with the most zeros. Every zero entry kills its whole cofactor term (zero times anything is zero), so a zero means "one fewer $2\times 2$ determinant to compute."

Scan: column 2 is $(0,0,6)$ — two zeros. No other row or column beats that.

Answer: expand along column 2. Only one term survives: $\det (A)=6\cdot C_{32}.$ WHY it matters: the lazy-but-correct choice cuts three minors down to one.

WHY row 2? It contains a $0$ at position $(2,2)$, so one term vanishes.

Row-2 entries are $a_{21}=-1,a_{22}=0,a_{23}=4$. Their checkerboard signs (read the middle row of the pattern) are $-,+,-$.

$\det (A)=-1\cdot C_{21}+0\cdot C_{22}+4\cdot C_{23}.$

Cofactor $C_{21}$ — sign $(-1{)}^{2+1}=-1$; cross out row 2, column 1: $M_{21}=\left|\begin{array}{cc}3& 2\\ 5& 1\end{array}\right|=3\cdot 1-2\cdot 5=3-10=-7,C_{21}=(-1)(-7)=7.$

Cofactor $C_{23}$ — sign $(-1{)}^{2+3}=-1$; cross out row 2, column 3: $M_{23}=\left|\begin{array}{cc}1& 3\\ 2& 5\end{array}\right|=1\cdot 5-3\cdot 2=5-6=-1,C_{23}=(-1)(-1)=1.$

Sum: $\det (A)=(-1)(7)+4(1)=-7+4=-3.$

Check with row 1 (WHY: to feel the invariance — different path, same answer). Entries $1,3,2$, signs $+,-,+$:

• $M_{11}=\left|\begin{array}{cc}0& 4\\ 5& 1\end{array}\right|=0-20=-20$
• $M_{12}=\left|\begin{array}{cc}-1& 4\\ 2& 1\end{array}\right|=-1-8=-9$
• $M_{13}=\left|\begin{array}{cc}-1& 0\\ 2& 5\end{array}\right|=-5-0=-5$ $\det =1(-20)-3(-9)+2(-5)=-20+27-10=-3.✓$

WHY column 1? It has a $0$ at $(2,1)$, so one of three terms disappears.

Column-1 entries $a_{11}=2,a_{21}=0,a_{31}=1$, signs (left column of checkerboard) $+,-,+$.

$C_{11}$ — sign $+$; delete row 1, column 1: $M_{11}=\left|\begin{array}{cc}3& 4\\ 0& 2\end{array}\right|=6-0=6,C_{11}=+6.$

$C_{31}$ — sign $(-1{)}^{3+1}=+$; delete row 3, column 1: $M_{31}=\left|\begin{array}{cc}1& -1\\ 3& 4\end{array}\right|=4-(-3)=7,C_{31}=+7.$

Sum: $\det (C)=2(6)+1(7)=12+7=19.$

Along column 1. Entries $1,0,0$ with signs $+,-,+$. Only the first survives: $\det (U)=1\cdot C_{11}=1\cdot \left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|=1\cdot (24-0)=24.$

Along row 1. Entries $1,2,3$ with signs $+,-,+$:

• $M_{11}=\left|\begin{array}{cc}4& 5\\ 0& 6\end{array}\right|=24$
• $M_{12}=\left|\begin{array}{cc}0& 5\\ 0& 6\end{array}\right|=0-0=0$
• $M_{13}=\left|\begin{array}{cc}0& 4\\ 0& 0\end{array}\right|=0-0=0$ $\det (U)=1(24)-2(0)+3(0)=24.$

Same answer, $24$ — confirming the parent's "different path, same destination."

General pattern: for any triangular matrix (all zeros on one side of the diagonal) the determinant is the product of the diagonal entries. Here $1\cdot 4\cdot 6=24$. WHY: expanding down the zero column always leaves exactly one nonzero term — the diagonal entry times the smaller triangular block — and that recursion multiplies the diagonal all the way down. See properties of determinants.

Structural observation first. Look at rows 1 and 2: $(2,4,6)=2\cdot (1,2,3)$. Row 2 is a scalar multiple of row 1 — the two rows point along the same line in space.

Expand along row 3 to confirm. Entries $5,1,0$, signs $+,-,+$:

• $M_{31}=\left|\begin{array}{cc}2& 3\\ 4& 6\end{array}\right|=12-12=0$
• $M_{32}=\left|\begin{array}{cc}1& 3\\ 2& 6\end{array}\right|=6-6=0$ $\det (D)=5(0)-1(0)+0\cdot C_{33}=0.$

Geometric meaning (WHAT IT LOOKS LIKE): three row-vectors normally span a solid box (a parallelepiped) with some volume. Here two of them lie on one line, so the "box" is squashed flat — zero thickness, hence zero volume, zero determinant. See the figure below.

WHY set $\det =0$? A matrix is singular exactly when its determinant vanishes — that is when it has no inverse; see determinant and inverse.

Expand along column 1 (it has a zero at $(3,1)$). Entries $k,1,0$.

$C_{11}$ — sign $(-1{)}^{1+1}=+1$; delete row 1, column 1: $M_{11}=\left|\begin{array}{cc}k& 1\\ 2& k\end{array}\right|=k^2-2,C_{11}=+(k^2-2).$

$C_{21}$ — sign $(-1{)}^{2+1}=-1$; delete row 2, column 1: $M_{21}=\left|\begin{array}{cc}2& 0\\ 2& k\end{array}\right|=2k-0=2k,C_{21}=(-1)(2k)=-2k.$

The third entry $a_{31}=0$ kills its term. Assemble, entry-times-cofactor: $\det =k\cdot C_{11}+1\cdot C_{21}+0=k(k^2-2)+1(-2k)=k^3-2k-2k=k^3-4k.$

Set to zero and factor: $k^3-4k=k(k^2-4)=k(k-2)(k+2)=0.$ $\boxed{k=0,k=2,k=-2.}$

Sanity check $k=2$: $\det =8-8=0.✓$ And $k=1$ (should be nonzero): $1-4=-3\ne 0$, so $M(1)$ is invertible.

WHY expand along row 1? Row 1 holds the basis vectors $\mathbf{i},\mathbf{j},\mathbf{k}$; expanding there peels off one component of the result per cofactor. Signs $+,-,+$.

$\mathbf{u}\times \mathbf{v}=\mathbf{i}\left|\begin{array}{cc}2& 3\\ 5& 6\end{array}\right|-\mathbf{j}\left|\begin{array}{cc}1& 3\\ 4& 6\end{array}\right|+\mathbf{k}\left|\begin{array}{cc}1& 2\\ 4& 5\end{array}\right|.$

Compute each $2\times 2$:

• $\left|\begin{array}{cc}2& 3\\ 5& 6\end{array}\right|=12-15=-3$
• $\left|\begin{array}{cc}1& 3\\ 4& 6\end{array}\right|=6-12=-6$
• $\left|\begin{array}{cc}1& 2\\ 4& 5\end{array}\right|=5-8=-3$

Assemble, minding the middle minus sign: $\mathbf{u}\times \mathbf{v}=(-3)\mathbf{i}-(-6)\mathbf{j}+(-3)\mathbf{k}=(-3,6,-3).$

Check (perpendicularity): a cross product is perpendicular to both inputs, so its dot with $\mathbf{u}$ should be $0$: $(-3)(1)+(6)(2)+(-3)(3)=-3+12-9=0.✓$

WHAT Cramer's rule says: for $A\mathbf{x}=\mathbf{b}$, the unknown $y=\frac{\det (A_y)}{\det (A)}$, where $A_y$ is $A$ with its second column replaced by the right-hand side $\mathbf{b}=(6,13,13)$. WHY it works: replacing a column by $\mathbf{b}$ and taking determinants isolates that one variable through the linearity of the determinant.

Coefficient matrix and its determinant. WHY expand along row 1? No row or column of $A$ has a zero, so no choice saves work; we default to row 1 for consistency (any row gives the same number, as L3.1 showed). Signs $+,-,+$: $A=\left[\begin{array}{ccc}1& 2& 1\\ 2& 1& 3\\ 1& 3& 2\end{array}\right]$

• $M_{11}=\left|\begin{array}{cc}1& 3\\ 3& 2\end{array}\right|=2-9=-7$
• $M_{12}=\left|\begin{array}{cc}2& 3\\ 1& 2\end{array}\right|=4-3=1$
• $M_{13}=\left|\begin{array}{cc}2& 1\\ 1& 3\end{array}\right|=6-1=5$ $\det (A)=1(-7)-2(1)+1(5)=-7-2+5=-4.$

Since $\det (A)=-4\ne 0$, the system has a unique solution — Cramer's rule is valid.

Build $A_y$ (replace column 2 with $(6,13,13)$). WHY row 1 again? Same reason — no zeros anywhere, so we keep the same row for a clean parallel comparison. Signs $+,-,+$: $A_y=\left[\begin{array}{ccc}1& 6& 1\\ 2& 13& 3\\ 1& 13& 2\end{array}\right]$

• $M_{11}=\left|\begin{array}{cc}13& 3\\ 13& 2\end{array}\right|=26-39=-13$
• $M_{12}=\left|\begin{array}{cc}2& 3\\ 1& 2\end{array}\right|=4-3=1$
• $M_{13}=\left|\begin{array}{cc}2& 13\\ 1& 13\end{array}\right|=26-13=13$ $\det (A_y)=1(-13)-6(1)+1(13)=-13-6+13=-6.$

Therefore: $y=\frac{\det (A_y)}{\det (A)}=\frac{-6}{-4}=\frac{3}{2}.$

Full-solution check (not required, but reassuring — WHY: an independent route should reproduce $y$). Solving all three equations simultaneously gives $x=\frac{5}{2},y=\frac{3}{2},z=\frac{3}{2}.$ Verify each equation with these values:

• Eq 1: $\frac{5}{2}+2(\frac{3}{2})+\frac{3}{2}=\frac{5}{2}+3+\frac{3}{2}=7$… that is $\frac{5}{2}+\frac{6}{2}+\frac{3}{2}=\frac{14}{2}=7\ne 6$, so these values are wrong — a warning that guessing round numbers is dangerous. Let us solve properly.

Solve properly (elimination). From Eq 1, $x=6-2y-z$. Substitute into Eq 2 and Eq 3:

• Eq 2: $2(6-2y-z)+y+3z=13\Rightarrow 12-4y-2z+y+3z=13\Rightarrow -3y+z=1.$
• Eq 3: $(6-2y-z)+3y+2z=13\Rightarrow 6+y+z=13\Rightarrow y+z=7.$

Subtract the first reduced equation from the second: $(y+z)-(-3y+z)=7-1\Rightarrow 4y=6\Rightarrow y=\frac{3}{2}.$ Then $z=7-y=7-\frac{3}{2}=\frac{11}{2}$, and $x=6-2(\frac{3}{2})-\frac{11}{2}=6-3-\frac{11}{2}=3-\frac{11}{2}=-\frac{5}{2}.$

The true solution is $x=-\frac{5}{2},y=\frac{3}{2},z=\frac{11}{2}$. Check Eq 2: $2(-\frac{5}{2})+\frac{3}{2}+3(\frac{11}{2})=-5+\frac{3}{2}+\frac{33}{2}=-5+\frac{36}{2}=-5+18=13.✓$ Check Eq 3: $-\frac{5}{2}+3(\frac{3}{2})+2(\frac{11}{2})=-\frac{5}{2}+\frac{9}{2}+11=\frac{4}{2}+11=2+11=13.✓$

Both routes agree: $\boxed{y=\frac{3}{2}}$. Cramer's rule delivered the right value directly, and the full elimination confirms it.

The determinant does not depend on which row/column you expand along. ::: True — it is a fixed property of the matrix; expansion choice only changes the arithmetic route. A row that is a multiple of another row forces the determinant to be… ::: $0$ (the box is flat, zero volume). To solve for the $k$-th variable by Cramer's rule you replace… ::: the $k$-th column of $A$ with the right-hand-side vector $\mathbf{b}$. The sign attached to entry $(i,j)$ is… ::: $(-1{)}^{i+j}$, depending only on position, never on the entry's value.