7.3 Transpose, column operations, and Cramer's rule

Transpose transfers determinant identities from rows to columns. Column linearity then makes it possible to isolate one coefficient in a linear combination: replacing a column creates repeated columns in every unwanted term. This cancellation gives Cramer’s rule and connects it to the adjoint identity.

Throughout this note, matrices and scalars are real; whenever a determinant is written, the matrix is square.

Transpose does not change determinant

Theorem 7.3.1

(Determinant of a transpose)

For every square matrix $A$,

$$\det(A^T)=\det(A).$$

Although the recursive definition uses the first row, transposition leaves its value unchanged.

Proof

(A factorization proof of transpose invariance)

If $A$ is singular, then $A^T$ is singular as well: transposing either inverse equation would turn an inverse for $A^T$ into an inverse for $A$. Thus both determinants are zero. If $A$ is invertible, write

$$A=E_1E_2\cdots E_k$$

as a product of elementary matrices. A swap or nonzero scaling elementary matrix is unchanged by transposition. The transpose of an addition matrix $I+cE_{ij}$ is $I+cE_{ji}$, which is another addition matrix with determinant one. Therefore $\det(E_r^T)=\det(E_r)$ for every factor, while

$$A^T=E_k^T\cdots E_2^TE_1^T.$$

The determinant product rule gives $\det(A^T)=\det(A)$. The matrix factors reverse order, but their determinant values are scalars, whose product may be reordered. This proof uses the factorization and row-operation results of 7.2. It supplies a derivation of the column expansion stated in 7.1 without using that column formula in the argument.

Theorem 7.3.2

(Cofactor expansion along any column)

For any fixed column $j$ of a square matrix $A=[a_{ij}]$, with $n\ge2$, where $M_{ij}$ is the minor determinant and $C_{ij}=(-1)^{i+j}M_{ij}$ is the cofactor,

$$\det(A)=a_{1j}C_{1j}+a_{2j}C_{2j}+\cdots+a_{nj}C_{nj},$$

Equivalently,

$$\det(A)=\sum_{i=1}^n (-1)^{i+j}a_{ij}M_{ij}.$$

So you may expand along a row or along a column. The right choice is whichever one produces the cleanest minors.

To see why the column formula follows from the row formula, fix column $j$. Write $A(i|j)$ for the matrix obtained by deleting row $i$ and column $j$. The entry $a_{ij}$ becomes the entry in row $j$ of $A^T$, and deleting that row and column gives

$$(A^T)(j\mid i)=(A(i\mid j))^T$$

The two submatrices have the same determinant by transpose invariance. Therefore row-$j$ expansion of $A^T$ is

$$\det(A^T)=\sum_{i=1}^n(-1)^{j+i}a_{ij}\det((A^T)(j\mid i)) =\sum_{i=1}^n(-1)^{i+j}a_{ij}\det(A(i\mid j))$$

which equals $\det(A)$. This is the promised column-$j$ expansion.

Worked example 7.3.3

(A column expansion with many zeros)

Let

$$A= \begin{bmatrix} 1&7&0\\ 6&9&8\\ 0&1&5 \end{bmatrix}.$$

The third column has one zero, so expand along that column:

$$\det(A)= 0\cdot C_{13} +8C_{23} +5C_{33}.$$

That gives

$$\det(A) =-8 \begin{vmatrix} 1&7\\ 0&1 \end{vmatrix} +5 \begin{vmatrix} 1&7\\ 6&9 \end{vmatrix} =-8(1)+5(9-42)=-173.$$

The same answer would come from row expansion, but the chosen column makes the arithmetic shorter.

Column operations obey the same pattern

Because transpose preserves determinant, every row-operation rule has a column version.

Theorem 7.3.4

(How column operations change determinant)

Let $B$ be obtained from a square matrix $A$ by one elementary column operation.

1. Swapping two distinct columns $p$ and $q$ multiplies the determinant by $-1$.
2. Multiplying column $p$ by $\beta$ multiplies the determinant by $\beta$. For a reversible elementary operation, $\beta\neq0$.
3. Replacing column $q$ by $C_q+cC_p$, with $p\neq q$, leaves the determinant unchanged.

These are the row rules applied to $A^T$, followed by transpose invariance. The source and target columns in rule 3 must be different. The same transfer gives multilinearity in one column and the fact that two equal columns force a zero determinant. Those two properties will do the cancellation work in Cramer's rule.

Worked example 7.3.5

(Use column operations to create zeros)

Consider

$$C= \begin{bmatrix} 3&2&-1\\ 4&1&6\\ -3&-1&2 \end{bmatrix}.$$

Apply the column operations $C_1\leftarrow C_1-C_2$ and then $C_1\leftarrow C_1+C_3$, using the updated first column at each step:

$$\begin{bmatrix} 3&2&-1\\ 4&1&6\\ -3&-1&2 \end{bmatrix} \longrightarrow \begin{bmatrix} 1&2&-1\\ 3&1&6\\ -2&-1&2 \end{bmatrix}.$$

After the first step, the second step gives

$$\begin{bmatrix} 1&2&-1\\ 3&1&6\\ -2&-1&2 \end{bmatrix} \longrightarrow \begin{bmatrix} 0&2&-1\\ 9&1&6\\ 0&-1&2 \end{bmatrix}.$$

These are column-addition operations, so the determinant is unchanged. Expand along the first column:

$$\det(C)= 9(-1)^{2+1} \begin{vmatrix} 2&-1\\ -1&2 \end{vmatrix} =-9(4-1)=-27.$$

The negative cofactor sign at position $(2,1)$ is essential. Choose row or column operations according to which creates useful zeros, and use the updated matrix at each step.

Adjoint matrices package all cofactors at once

The cofactor of one entry helps in one expansion. The adjoint matrix collects all cofactors into one object.

Definition 7.3.6

(Adjoint matrix)

For an $n\times n$ matrix $A=[a_{ij}]$ with $n\ge2$, let $C_{ij}$ denote the cofactor of $a_{ij}$, and first form the cofactor matrix

$$\begin{bmatrix} C_{11}&C_{12}&\cdots&C_{1n}\\ C_{21}&C_{22}&\cdots&C_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ C_{n1}&C_{n2}&\cdots&C_{nn} \end{bmatrix}.$$

The adjoint matrix is the transpose of that cofactor matrix:

$$\operatorname{adj}(A)= \begin{bmatrix} C_{11}&C_{21}&\cdots&C_{n1}\\ C_{12}&C_{22}&\cdots&C_{n2}\\ \vdots&\vdots&\ddots&\vdots\\ C_{1n}&C_{2n}&\cdots&C_{nn} \end{bmatrix}.$$

For the $1\times1$ base case, define $\det([a])=a$ and $\operatorname{adj}([a])=[1]$. Then $[a][1]=[a]=\det([a])I_1$. The cofactor identity proof below handles $n\ge2$.

Theorem 7.3.7

(Adjoint identity and inverse formula)

For every square matrix $A$, including singular matrices,

$$A\,\operatorname{adj}(A)=\det(A)I.$$

If $\det(A)\neq0$, then $A$ is invertible and

$$A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A).$$

Proof

(The diagonal and off-diagonal cases)

The following proof assumes $n\ge2$; the $1\times1$ base case was defined above.

The $(i,j)$ entry of $A \operatorname{adj}(A)$ is

$$\bigl(A\operatorname{adj}(A)\bigr)_{ij}=\sum_{k=1}^n a_{ik}C_{jk}.$$

If $i=j$, this is cofactor expansion along row $i$, so it equals $\det(A)$. If $i\neq j$, form $B$ by replacing row $j$ of $A$ with row $i$. The matrix $B$ has two equal rows, hence $\det(B)=0$. Expanding $B$ along its replaced row $j$ uses exactly the cofactors $C_{jk}$ from $A$: deleting row $j$ removes the replacement. Therefore

$$0=\det(B)=\sum_{k=1}^n a_{ik}C_{jk}.$$

The diagonal entries are $\det(A)$ and every off-diagonal entry is zero. When $\det(A)\neq0$, divide by that scalar and use the square-matrix inverse criterion. The identity itself remains valid when $A$ is singular; only the inverse formula requires division.

Worked example 7.3.8

(Recover the 2×2 inverse formula from the adjoint)

Let

$$A= \begin{bmatrix} a&b\\ c&d \end{bmatrix}, \qquad \det(A)=ad-bc\neq0.$$

The cofactor matrix is

$$\begin{bmatrix} d&-c\\ -b&a \end{bmatrix},$$

so

$$\operatorname{adj}(A)= \begin{bmatrix} d&-b\\ -c&a \end{bmatrix}.$$

Therefore

$$A^{-1}= \frac{1}{ad-bc} \begin{bmatrix} d&-b\\ -c&a \end{bmatrix}.$$

The familiar inverse formula is really the adjoint identity written in the $2\times2$ case. Multiplication displays the cancellation:

$$A\operatorname{adj}(A)=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}=(ad-bc)I_2.$$

The transpose in the adjoint is essential. If the untransposed cofactor matrix $\begin{bmatrix}d&-c\\-b&a\end{bmatrix}$ were used instead, the product would generally have entries $ad-b^2$, $ab-ac$, $cd-bd$, and $ad-c^2$, not a scalar on the diagonal with zero off-diagonal entries.

Cramer's rule solves one coordinate at a time

Theorem 7.3.9

(Cramer's rule)

Let $A\in M_n(\mathbb{R})$ satisfy $\det(A)\ne0$, and let $b\in\mathbb{R}^n$. For each $j$, let $M_j$ be the matrix obtained by replacing the $j$th column of $A$ with $b$.

If $x$ is the unique solution of

$$Ax=b,$$

then

$$x_j=\frac{\det(M_j)}{\det(A)}.$$

Each coordinate is a determinant ratio, expressing its dependence on the coefficient matrix and the right-hand side.

Proof

(Fixed-column linearity isolates x_j)

Write the columns of $A$ as $a_1,\ldots,a_n$. Since $\det(A)\neq0$, the system has a unique solution and

$$b=x_1a_1+\cdots+x_na_n.$$

In $M_j$, keep every column except column $j$ fixed and substitute this sum only into column $j$. By linearity in that column,

$$\det(M_j)=\sum_{k=1}^n x_k\det(a_1,\ldots,a_{j-1},a_k,a_{j+1},\ldots,a_n).$$

When $k\neq j$, the determinant has two equal columns, so that term is zero. The remaining term is $x_j \det(A)$. Hence

$$\det(M_j)=x_j\det(A), \qquad x_j=\frac{\det(M_j)}{\det(A)}.$$

The order of the columns never changes, which is why no additional sign appears.

Why replacing one column isolates one unknown

The visual below follows the determinant logic behind Cramer's rule: replace one column by the right-hand side vector, use column linearity, and watch repeated columns force the extra terms to vanish.

For $n\ge2$, the established inverse formula gives a second, compatible reading of the same coordinate formula. Since $\det(A)\ne0$, we have $\operatorname{adj}(A)=\det(A)A^{-1}$. Applying $A^{-1}$ to $Ax=b$ and then substituting this identity gives

$$\det(A)x=\operatorname{adj}(A)b, \qquad x_j=\frac{\sum_{i=1}^n C_{ij}b_i}{\det(A)}$$

The numerator is exactly $\det(M_j)$: expand $M_j$ down its replaced column $j$. Deleting that column removes $b$ and leaves the same minor as in $A$, so the cofactor attached to $b_i$ is $C_{ij}$. Thus the adjoint route and the fixed-column linearity route meet at the same signed sum. For $n=1$, write $A=[a]$ and $b=[b_1]$ with $a\ne0$; directly, $x_1=b_1/a$.

Worked example 7.3.10

(A 3×3 adjoint and replacement determinants)

Let

$$A=\begin{bmatrix}1&2&3\\1&0&1\\1&1&-1\end{bmatrix}, \qquad b=\begin{bmatrix}2\\3\\1\end{bmatrix}.$$

Expansion along the first row gives

$$\det(A)=1(-1)-2(-2)+3(1)=6.$$

The cofactor matrix is

$$\begin{bmatrix}-1&2&1\\5&-4&1\\2&2&-2\end{bmatrix}.$$

Transposing it gives

$$\operatorname{adj}(A)=\begin{bmatrix}-1&5&2\\2&-4&2\\1&1&-2\end{bmatrix}.$$

Replacing columns one, two, and three by $b$ gives

$$M_1=\begin{bmatrix}2&2&3\\3&0&1\\1&1&-1\end{bmatrix},\quad M_2=\begin{bmatrix}1&2&3\\1&3&1\\1&1&-1\end{bmatrix},\quad M_3=\begin{bmatrix}1&2&2\\1&0&3\\1&1&1\end{bmatrix}.$$

Their first-row expansions are

$$\det(M_1)=-2+8+9=15,\qquad \det(M_2)=-4+4-6=-6,\qquad \det(M_3)=-3+4+2=3.$$

Thus the solution is

$$(x_1,x_2,x_3)=\left(\frac{15}{6},\frac{-6}{6},\frac36\right) =\left(\frac52,-1,\frac12\right).$$

Substitution gives $5/2-2+3/2=2$, $5/2+1/2=3$, and $5/2-1-1/2=1$, so all three equations agree with $b$.

The numerical data also provide an independent audit of the signs and of the adjoint route. Multiplying the displayed adjoint by $b$ gives

$$\operatorname{adj}(A)b=\begin{bmatrix}15\\-6\\3\end{bmatrix}$$

so the adjoint identity predicts $x=\operatorname{adj}(A)b/\det(A)$, exactly the vector obtained from the three replacement determinants. For example, the first entry is $-1(2)+5(3)+2(1)=15$; the signs in the cofactor matrix are already included before this multiplication.

Worked example 7.3.11

(Solve a 2×2 system with Cramer's rule)

Solve

$$\begin{cases} 2x_1+x_2=5,\\ x_1+3x_2=7. \end{cases}$$

Write

$$A= \begin{bmatrix} 2&1\\ 1&3 \end{bmatrix}, \qquad b= \begin{bmatrix} 5\\ 7 \end{bmatrix}.$$

First compute

$$\det(A)=2\cdot3-1\cdot1=5.$$

Replace the first column by $b$:

$$M_1= \begin{bmatrix} 5&1\\ 7&3 \end{bmatrix}, \qquad \det(M_1)=15-7=8.$$

Replace the second column by $b$:

$$M_2= \begin{bmatrix} 2&5\\ 1&7 \end{bmatrix}, \qquad \det(M_2)=14-5=9.$$

So

$$x_1=\frac{8}{5}, \qquad x_2=\frac{9}{5}.$$

Substitution checks the result: $2(8/5)+9/5=5$ and $8/5+3(9/5)=7$.

Common mistake

Common mistake

(Cramer's rule is not a universal system solver)

Cramer's rule requires a square coefficient matrix and a nonzero determinant. If the system is rectangular or singular, the rule is not available. Even when it applies, it is usually less efficient than Gaussian elimination for large systems.

Counterexample mode

Zero replacement determinants do not classify singular systems

Let $A=\operatorname{diag}(1,0,0)$, a nonzero singular matrix. With $b=e_1=(1,0,0)^T$, the equations are $x_1=1$, $0=0$, and $0=0$, so the solutions are $(1,s,t)$. With $b=e_2=(0,1,0)^T$, the second equation is $0=1$, so the system is inconsistent.

Write the replacement column triples explicitly. For $b=e_1$, $(M_1)=(e_1,0,0)$, $(M_2)=(e_1,e_1,0)$, and $(M_3)=(e_1,0,e_1)$; every determinant is zero because of a zero column or repeated columns. For $b=e_2$, $(M_1)=(e_2,0,0)$, $(M_2)=(e_1,e_2,0)$, and $(M_3)=(e_1,0,e_2)$; every determinant is zero because each triple has a zero column, and $M_1$ has two zero columns. Thus all six formal ratios are $0/0$ and undefined. Require $\det(A)\ne0$ before using Cramer's rule. For a singular system, row-reduce the augmented matrix and inspect consistency and free variables.

Quick check

Quick check

(Does transpose change the determinant of a square matrix?)

State the theorem directly.

Solution · Answer

No. $\det(A^T)=\det(A)$.

Quick check

(What column operation leaves the determinant unchanged?)

Think of the column analogue of a type-III row operation.

Solution · Answer

Adding a scalar multiple of one distinct source column to another target column leaves the determinant unchanged.

Quick check

(When may Cramer's rule be used?)

Identify the structural hypotheses on A.

Solution · Answer

When the coefficient matrix is square and invertible, equivalently when its determinant is nonzero.

Exercises

Quick check

(Use a column expansion to compute $\det\begin{bmatrix}1&0&2\\3&4&5\\0&0&6\end{bmatrix}$.)

Choose the column with the most zeros.

Solution · Guided solution

Expand along the second column:

$$\det= 4(-1)^{2+2} \begin{vmatrix} 1&2\\ 0&6 \end{vmatrix} =4(6)=24.$$

Quick check

(For $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$, write down $\operatorname{adj}(A)$.)

Compute the four cofactors first, then transpose the cofactor matrix.

Solution · Guided solution

The cofactors are $C_{11}=4$, $C_{12}=-3$, $C_{21}=-2$, and $C_{22}=1$. Hence

$$\operatorname{adj}(A)= \begin{bmatrix} 4&-2\\ -3&1 \end{bmatrix}.$$

Quick check

(Use Cramer's rule to find x_1 for the system $x_1+x_2=3$, $2x_1-x_2=0$.)

You only need $\det(A)$ and $\det(M_1)$.

Solution · Guided solution

Here

$$A= \begin{bmatrix} 1&1\\ 2&-1 \end{bmatrix}, \qquad b= \begin{bmatrix} 3\\ 0 \end{bmatrix}.$$

Then

$$\det(A)=-1-2=-3, \qquad M_1= \begin{bmatrix} 3&1\\ 0&-1 \end{bmatrix}, \qquad \det(M_1)=-3.$$

So

$$x_1=\frac{\det(M_1)}{\det(A)}=1.$$

Related notes

Return to 7.2 Row operations, products, and invertibility if the row-operation bookkeeping is still shaky.

Keep 3.3 Transpose, symmetric, and skew-symmetric matrices nearby, because the transpose theorem here extends that earlier structural chapter.

For system language, connect this note back to 1.1 Equations and solution sets and 2.3 Gaussian elimination and RREF.