Row operations, products, and invertibility

The previous note defined the determinant by cofactor expansion. That definition is fundamental, but expanding every large determinant is rarely the best calculation. Row reduction gives a second route: replace a matrix by a simpler one, record the exact determinant multiplier at every step, and read the final triangular determinant from its diagonal.

The row-operation multipliers lead to the following identities:

$$\text{row rules}\;\Longrightarrow\;\det(EX)=\det(E)\det(X) \;\Longrightarrow\;\det(AB)=\det(A)\det(B) \;\Longrightarrow\;\det(A)\ne0\Longleftrightarrow A\text{ is invertible}.$$

Every matrix in this note is square when a determinant is written. For a product, the factors have the same size, so that the product is again square.

Track how elimination changes the determinant

Row reduction changes entries, but the determinant does not respond to all changes in the same way. A row swap changes its sign; scaling one row multiplies it by the same scalar; adding a multiple of a distinct row leaves it unchanged.

Concept lensAlgorithmic

Read elimination as a determinant ledger

At each arrow, write the operation and its determinant multiplier beside the matrix. The three entries are $-1$, $\beta$, and $1$ for a swap, a nonzero scaling, and a row replacement respectively. The diagonal product belongs to the final triangular matrix; the ledger tells you how to transfer that value back to the starting matrix.

Reversible operations and their matrices

Definition

Elementary row operations and elementary matrices

Work over the real numbers. Let $A$ be an $n\times n$ matrix. An elementary row operation is one of the following, with the indicated row indices distinct when two rows occur:

1. $R_p\leftrightarrow R_q$: interchange rows $p$ and $q$;
2. $R_p\leftarrow\beta R_p$, where $\beta\ne0$;
3. $R_q\leftarrow R_q+\alpha R_p$.

The corresponding elementary matrix $E$ is obtained by applying that operation to $I_n$. Left multiplication gives the same operation on every $n\times m$ matrix $X$: the transformed matrix is $EX$. The restriction $\beta\ne0$ is essential for an elementary operation because it makes the operation reversible.

The determinant scaling rule itself still holds for $\beta=0$: a zero row gives determinant zero. However, zero scaling is not reversible and therefore is not an elementary factor in a product that is being undone. Keeping these two facts separate prevents a zero factor from being cancelled later.

The multiplier for each row operation

Theorem

The three row-operation determinant rules

For an $n\times n$ matrix $A$, with $1\le p,q\le n$ and $p\ne q$ whenever two row indices occur:

1. If $B$ results from $A$ by $R_p\leftrightarrow R_q$, then $\det(B)=-\det(A)$.
2. If $B$ results from $A$ by $R_p\leftarrow\beta R_p$, then $\det(B)=\beta\det(A)$, including the identity at $\beta=0$.
3. If $B$ results from $A$ by $R_q\leftarrow R_q+\alpha R_p$ with $p\ne q$, then $\det(B)=\det(A)$.

The first rule is the alternating property of the determinant. For the second, hold every row except row $p$ fixed and use linearity in that row. If $a_p$ is replaced by $\beta a_p$, the determinant is multiplied by $\beta$.

For a replacement, the row indices must be distinct. If $R_p\leftarrow R_p+\alpha R_p$, the new row is $(1+\alpha)a_p$, so the determinant is multiplied by $1+\alpha$.

Proof

Derive the row-replacement rule from multilinearity

Write the rows of $A$ as $a_1,\ldots,a_n$, and let $C$ be obtained by replacing row $q$ by $a_q+\alpha a_p$, where $p\ne q$. Construct $A^\star$ by replacing row $q$ of $A$ by $a_p$. The $q$th row of $C$ is

$$a_q+\alpha a_p=1\cdot a_q+\alpha\cdot a_p.$$

All other rows of $C$, $A$, and $A^\star$ agree. Therefore multilinearity in row $q$ gives

$$\det(C)=\det(A)+\alpha\det(A^\star).$$

Rows $p$ and $q$ of $A^\star$ are identical, so the alternating property gives $\det(A^\star)=0$. Hence $\det(C)=\det(A)$. This proof also shows exactly which distinct-row assumption makes the extra term vanish.

Theorem

The mixed-operation ledger

Suppose a sequence of elementary operations changes $A$ into $U$. If the sequence contains $s$ swaps and scales rows by nonzero factors $\beta_1,\ldots,\beta_q$, then

$$\det(U)=(-1)^s\left(\prod_{j=1}^{q}\beta_j\right)\det(A).$$

Consequently,

$$\det(A)=\frac{\det(U)}{(-1)^s\prod_{j=1}^{q}\beta_j}.$$

Row replacements do not appear in the product because their multiplier is $1$.

If $U$ is a square row-echelon matrix, it is upper triangular and $\det(U)$ is the product of its diagonal entries. A zero row makes this product zero, so the ledger immediately gives $\det(A)=0$ because its denominator contains only nonzero factors. If this row-echelon matrix has no zero row, each of its $n$ diagonal positions is a pivot, so the diagonal product is nonzero. An arbitrary upper triangular matrix need not have this property: for example, $\begin{bmatrix}0&1\\0&1\end{bmatrix}$ is upper triangular but has zero determinant. The row-echelon qualification is therefore essential.

A disciplined row-reduction algorithm

Before starting, write the matrix size and mark the roles of the rows. At each stage, use only a legal elementary operation: the two row indices in a swap or replacement must be distinct, and every scaling factor recorded in the ledger must be nonzero. Then follow this sequence:

1. Reduce the entries below each pivot, using row replacements whenever they suffice. Record a swap immediately when a pivot must be moved.
2. If a pivot row is normalized, record its factor $\beta$ with its sign and position in the sequence. The final diagonal product belongs to the scaled matrix, so this factor will be divided out.
3. Stop at an upper triangular matrix. If a zero row appears, stop there: no later division by a supposed pivot is justified, and the determinant is already certified to be zero.
4. Multiply the diagonal entries of the final triangular matrix. Finally apply $\det(A)=\det(U)/((-1)^s\prod\beta_j)$, rather than trying to remember the corrections mentally.

This algorithm permits different valid elimination choices. Two reduction paths may produce different triangular matrices and different intermediate pivots, but their ledgers must return the same determinant. That agreement is a useful check on arithmetic, while the theorem explains why it is guaranteed. The same discipline also applies to an augmented matrix: when solving $Ax=b$, an elementary operation acts on $[A\mid b]$ as a whole, whereas determinant bookkeeping concerns the square coefficient matrix $A$.

From elementary factors to a product theorem

For one elementary operation, let $E$ be its matrix. When $X$ is also square of the same size, $EX$ is square. Since $E$ is the result of applying the operation to $I_n$, the row rules say $\det(E)$ is respectively $-1$, $\beta$, or $1$. Applying the same rule to $X$ gives

$$\det(EX)=\det(E)\det(X).$$

If $E_s\cdots E_1$ is a sequence, induction on the number of factors gives

$$\det(E_s\cdots E_1X)= \left(\prod_{j=1}^{s}\det(E_j)\right)\det(X).$$

Now suppose $A$ is invertible. The earlier row-reduction theorem supplies a factorization $A=E_s\cdots E_1$. Applying the last formula first to $I_n$ and then to $B$ yields

$$\det(A)=\prod_j\det(E_j)\ne0, \qquad \det(AB)=\left(\prod_j\det(E_j)\right)\det(B).$$

Thus $\det(AB)=\det(A)\det(B)$ in the invertible case.

For the singular case, row-reduce $A$ to an echelon form $R$. The invertibility dictionary from the earlier notes says that a singular square matrix has a zero row in $R$. Let $E_s\cdots E_1A=R$ and set $c=\prod_j\det(E_j)$. Every factor is nonzero because every operation is reversible. The zero row of $R$ remains a zero row in $RB$, so

$$c\det(A)=\det(R)=0, \qquad c\det(AB)=\det(RB)=0.$$

Since $c\ne0$, cancellation gives $\det(A)=0$ and $\det(AB)=0$. The equality again follows, now with both sides zero.

The notation $E_s\cdots E_1A$ also records the order of operations. The rightmost factor $E_1$ acts first, then $E_2$, and so on. No commutativity of matrices is being used in this argument. The factors are all $n\times n$, while $B$ is an $n\times n$ matrix, so each intermediate product is defined. This dimensional condition is part of the theorem, not a cosmetic convention.

Theorem

Determinant of a product

If $A$ and $B$ are square matrices of the same size, then

$$\det(AB)=\det(A)\det(B).$$

This statement was proved by the two cases above. It is stronger than a calculation trick: although $AB$ and $BA$ can be different matrices, scalar multiplication commutes, so

$$\det(AB)=\det(A)\det(B)=\det(B)\det(A)=\det(BA).$$

Determinant and invertibility

Theorem

Determinant criterion for invertibility

For a square matrix $A$,

$$A\text{ is invertible}\quad\Longleftrightarrow\quad\det(A)\ne0.$$

The proof of the product theorem already establishes both directions: an invertible matrix factors into elementary matrices with nonzero determinants, while a singular matrix reduces to an echelon matrix with a zero row. The nonzero reduction multiplier transfers that zero determinant to the original matrix.

Worked examples

Worked example

Row replacement only: no determinant factor accumulates

Let

$$A=\begin{bmatrix}1&2&3\\2&5&7\\3&8&10\end{bmatrix}.$$

Use replacements in the stated order:

$$\begin{bmatrix}1&2&3\\2&5&7\\3&8&10\end{bmatrix} \xrightarrow{R_2\leftarrow R_2-2R_1,\;R_3\leftarrow R_3-3R_1} \begin{bmatrix}1&2&3\\0&1&1\\0&2&1\end{bmatrix} \xrightarrow{R_3\leftarrow R_3-2R_2} \begin{bmatrix}1&2&3\\0&1&1\\0&0&-1\end{bmatrix}=U.$$

Each operation adds a multiple of one distinct row to another, so each ledger entry is $1$. The final matrix is upper triangular. Therefore

$$\det(A)=\det(U)=1\cdot1\cdot(-1)=-1.$$

The order matters for producing the displayed states: the second replacement uses the already updated row $R_2$. It changes no determinant factor.

Worked example

Swap, scaling, and replacement in one ledger

Let

$$B=\begin{bmatrix}0&1\\4&3\end{bmatrix}.$$

Continue all the way to the identity:

$$\begin{bmatrix}0&1\\4&3\end{bmatrix} \xrightarrow{R_1\leftrightarrow R_2} \begin{bmatrix}4&3\\0&1\end{bmatrix} \xrightarrow{R_1\leftarrow\frac14R_1} \begin{bmatrix}1&\frac34\\0&1\end{bmatrix} \xrightarrow{R_1\leftarrow R_1-\frac34R_2} I_2.$$

The three multipliers are $-1$, $\frac14$, and $1$. Hence

$$1=\det(I_2)=(-1)\left(\frac14\right)\det(B), \qquad\text{so}\qquad \det(B)=-4.$$

The triangular intermediate matrix has determinant $4$, which is consistent with the first line $4=-\det(B)$. The final replacement removes the remaining off-diagonal entry while preserving the already corrected determinant value $1$ of the scaled matrix.

Worked example

A parameter value, a minor calculation, and a kernel check

Let

$$A(x)=\begin{bmatrix} 2&1&0&1\\ 0&1&1&1\\ 1&0&0&x\\ 0&2&3&1 \end{bmatrix}.$$

Expanding along the first column, with the cofactor sign at position $(3,1)$ equal to $+1$, gives

$$\det(A)=2 \begin{vmatrix}1&1&1\\0&0&x\\2&3&1\end{vmatrix} +\begin{vmatrix}1&0&1\\1&1&1\\2&3&1\end{vmatrix}.$$

For the first minor, expand along its second row. Only the third entry can be nonzero, and its cofactor sign is $(-1)^{2+3}=-1$:

$$\begin{vmatrix}1&1&1\\0&0&x\\2&3&1\end{vmatrix} =-x\begin{vmatrix}1&1\\2&3\end{vmatrix}=-x(3-2)=-x.$$

For the second minor, expand along its first row:

$$\begin{vmatrix}1&0&1\\1&1&1\\2&3&1\end{vmatrix} =\begin{vmatrix}1&1\\3&1\end{vmatrix} +\begin{vmatrix}1&1\\2&3\end{vmatrix} =(-2)+1=-1.$$

Consequently

$$\det(A)=-2x-1,$$

so $A(x)$ is singular exactly when $x=-\frac12$. At that value, direct row equations give a nonzero kernel vector:

$$A\begin{bmatrix}1\\-4\\2\\2\end{bmatrix} =\begin{bmatrix}2-4+2\\-4+2+2\\1-1\\-8+6+2\end{bmatrix} =\begin{bmatrix}0\\0\\0\\0\end{bmatrix}.$$

This verifies singularity independently. For every other $x$, the determinant is nonzero, so the determinant criterion says the homogeneous system has only the zero solution.

Theorem

Inverse determinants and integral powers

If $A$ is invertible, then

$$\det(A^{-1})=\frac1{\det(A)}.$$

For every positive integer $m$ and every square $A$,

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

If $A$ is invertible, the same identity holds for every integer $m$, including negative $m$.

Indeed, $A^{-1}A=I_n$ and multiplicativity give $\det(A^{-1})\det(A)=1$. Positive powers follow by finite repeated application of the product theorem. We define $A^0=I_n$ for every square matrix, including singular matrices. When $A$ is singular, this says $\det(A^0)=1$; we do not assert the scalar expression $(\det A)^0$, since that would invite the undefined case $0^0$. For negative $m$, $A^m$ is defined only when $A$ is invertible, and the result follows from $A^m=(A^{-1})^{-m}$ and the inverse formula.

Common mistakes

Common mistake

Apply row operations to the entire augmented matrix

Row operations preserve the solution set only when they are applied to the entire augmented matrix $[A\mid b]$, including the right-hand side. It is not a valid unqualified statement that row-equivalent coefficient matrices have the same solution set. For example, $x=1$ is represented by $[1\mid1]$, whereas $2x=1$ is represented by $[2\mid1]$; changing the coefficient from $1$ to $2$ while keeping the right-hand side fixed changes the solution. For a homogeneous system, the augmented column is zero and remains zero, which is why row operations on $A$ alone are harmless in that special setting.

Other frequent errors are forgetting that a scaling factor must be divided out, counting a swap only once when it occurs twice in different stages, and using $R_p\leftarrow R_p+\alpha R_p$ as though it were a determinant-preserving row replacement. The last operation is actually scaling by $1+\alpha$.

Exercises

Checkpoint

If one row of A is multiplied by 5, what happens to $\det(A)$?

Use the row-scaling rule directly.

Solution · Answer

The determinant is multiplied by $5$.

Checkpoint

If A has a zero row after row reduction, what can you conclude about $\det(A)$?

Read the determinant of the echelon form and then use the nonzero ledger multiplier to return to the original matrix.

Solution · Answer

$\det(A)=0$. The echelon determinant is zero, and the nonzero swap/scaling ledger factor cannot change that conclusion when it is cancelled.

Checkpoint

Can a square matrix with determinant $0$ be invertible?

Use the equivalence theorem.

Solution · Answer

No. A square matrix is invertible exactly when its determinant is nonzero.

Checkpoint

Use row reduction to compute $\det\begin{bmatrix}1&1&0\\2&3&1\\0&1&4\end{bmatrix}$.

Try to avoid row scaling so the bookkeeping stays short.

Solution · Guided solution

Subtract $2R_1$ from $R_2$, then subtract the new $R_2$ from $R_3$:

$$\begin{bmatrix}1&1&0\\2&3&1\\0&1&4\end{bmatrix} \longrightarrow \begin{bmatrix}1&1&0\\0&1&1\\0&1&4\end{bmatrix} \longrightarrow \begin{bmatrix}1&1&0\\0&1&1\\0&0&3\end{bmatrix}.$$

Only row replacements were used, so $\det=1\cdot1\cdot3=3$.

Checkpoint

Suppose $\det(A)=4$ and $\det(B)=-3$ for two $3\times3$ matrices. What is $\det(AB)$?

Use multiplicativity.

Solution · Guided solution

$$\det(AB)=\det(A)\det(B)=4(-3)=-12.$$

Checkpoint

If $\det(A)=6$, what is $\det(A^{-1})$?

Assume A is invertible.

Solution · Guided solution

$$\det(A^{-1})=\frac1{\det(A)}=\frac16.$$

Related notes

Return to 7.1 Determinants and cofactor expansion if the basic definition still feels too symbolic.

Continue with 7.3 Transpose, column operations, and Cramer's rule to see the column viewpoint and the adjoint formula.

Keep 5.1 Invertible matrices in view, because this note upgrades that chapter with a determinant criterion.