Row operations, products, and invertibility

Proof of the finite real matrix product laws

Let $m,n,p,q$ be positive integers, let $A\in\mathbb R^{m\times n}$, $B\in\mathbb R^{n\times p}$, and $C\in\mathbb R^{p\times q}$. All the following arguments use the row-by-column definition and finite real sums. For any output position $(i,\ell)$,

$$\begin{aligned} ((AB)C)_{i\ell} &=\sum_{k=1}^{p}\left(\sum_{j=1}^{n}a_{ij}b_{jk}\right)c_{k\ell}\\ &=\sum_{j=1}^{n}a_{ij}\left(\sum_{k=1}^{p}b_{jk}c_{k\ell}\right) =(A(BC))_{i\ell}. \end{aligned}$$

The middle equality only reorders two finite sums and uses real distributivity. Both products are $m\times q$, so equality at every position proves associativity. No factor has changed its order.

For $D\in\mathbb R^{n\times p}$, the same definition gives

$$[A(B+D)]_{ik}=\sum_{j=1}^{n}a_{ij}(b_{jk}+d_{jk}) =(AB)_{ik}+(AD)_{ik}.$$

For $E\in\mathbb R^{m\times n}$, it gives

$$[(A+E)B]_{ik}=\sum_{j=1}^{n}(a_{ij}+e_{ij})b_{jk} =(AB)_{ik}+(EB)_{ik}.$$

Thus both distributive laws hold with the displayed matching sizes. For every real scalar $c$,

$$[(cA)B]_{ik}=\sum_j(ca_{ij})b_{jk} =c\sum_j a_{ij}b_{jk} =\sum_j a_{ij}(cb_{jk})=[A(cB)]_{ik},$$

proving $(cA)B=c(AB)=A(cB)$ by entrywise equality. In particular, for $x,y\in\mathbb R^n$ and real $s,t$,

$$[A(sx+ty)]_i=\sum_j a_{ij}(sx_j+ty_j) =s(Ax)_i+t(Ay)_i.$$

This proves the matrix-vector linearity used when comparing solutions of the same linear system. All identities concern compatible finite real matrices; they do not assert commutativity or cancellation of arbitrary matrix factors.

Build the elementary matrix from the identity

Suppose an elementary row operation $\rho$ is meant to act on matrices with $p$ rows. Start with the identity matrix $I_p$, apply the same elementary row operation to $I_p$, and call the result $E_\rho$. All scalars are taken from the underlying field of the matrices; in particular, every nonzero scaling factor $c$ has a reciprocal $1/c$ in that field.

Definition

Elementary row-operation matrix

The elementary row-operation matrix associated with an elementary row operation $\rho$ on $p$-row matrices is the $p \times p$ matrix obtained by applying $\rho$ to $I_p$.

This matrix is also called an elementary matrix. The longer name keeps its connection with the specified row operation visible.

Equivalently,

$$E_\rho = \rho(I_p).$$

The reason this definition is useful is the following theorem.

Theorem

Elementary row operations are left multiplication

Let $A$ be any matrix with $p$ rows. If $\rho$ is one elementary row operation and $\rho(A)$ denotes the result of applying it to $A$, then

$$\rho(A) = E_\rho A.$$

So applying that elementary row operation to $A$ is the same as multiplying $A$ on the left by the elementary row-operation matrix obtained from the same operation on $I_p$.

The multiplication must be on the left. Row operations change rows by mixing rows. Left multiplication forms new rows of $A$ as linear combinations of old rows of $A$. Right multiplication would instead mix columns.

The dimensions make the same point. If $A$ is $p \times n$, then $E_\rho$ is $p \times p$, so $E_\rho A$ is defined and remains $p \times n$. There is no assumption that $A$ is square. By contrast, a column-operation matrix acting on these same $n$ columns would have size $n \times n$ and multiply on the right.

Why left multiplication performs the operation

Proof

Verify the three elementary cases row by row

Write the rows of an arbitrary $p \times n$ matrix $A$ as $a_1,a_2,\ldots,a_p$, and let $e_k^{\mathsf T}$ denote row $k$ of $I_p$. The basic identity

$$e_k^{\mathsf T}A=a_k$$

says that a row of a left multiplier selects the corresponding row of $A$. Now check the three elementary operations.

1. For $R_i\leftrightarrow R_j$, the matrix $E_\rho$ has row $i$ equal to $e_j^{\mathsf T}$, row $j$ equal to $e_i^{\mathsf T}$, and every other row $k$ equal to $e_k^{\mathsf T}$. Thus the rows of $E_\rho A$ are the rows of $A$ with precisely rows $i$ and $j$ exchanged.

2. For $R_i\leftarrow cR_i$, where $c\ne0$, row $i$ of $E_\rho$ is $c e_i^{\mathsf T}$ and all other rows are unchanged. Hence row $i$ of the product is

   $$(c e_i^{\mathsf T})A=c a_i,$$

   while row $k$ remains $a_k$ for $k\ne i$.

3. For $R_j\leftarrow R_j+cR_i$, where $i\ne j$, row $j$ of $E_\rho$ is $e_j^{\mathsf T}+c e_i^{\mathsf T}$ and every other row is unchanged. Therefore the new row $j$ is

   $$(e_j^{\mathsf T}+c e_i^{\mathsf T})A=a_j+c a_i,$$

   exactly as prescribed.

These cases exhaust the elementary row operations, so they prove $E_\rho A=\rho(A)$ for every compatible $A$. The proof does not claim that an arbitrary manipulation of rows has such an elementary matrix; $\rho$ must be one of the three operations just checked.

Acting on an augmented matrix

Suppose $Ax=b$ is represented by the augmented matrix $[A\mid b]$. An elementary row operation must act on the full augmented matrix, not separately on whichever entries happen to be convenient. The multiplication identity is

$$E_\rho[A\mid b]=[E_\rho A\mid E_\rho b].$$

Thus the coefficients and the right-hand side undergo the same combination of rows. A vector $x$ satisfies $Ax=b$ if and only if it satisfies $E_\rho Ax=E_\rho b$: the forward implication follows by multiplying the equation by $E_\rho$, and the reverse implication follows by applying the reverse elementary operation. This is the precise reason the solution set is preserved.

Changing only $A$ while leaving $b$ fixed, or changing only $b$, generally produces a different system. The augmentation bar is therefore a visual separator, not a boundary at which a row operation may stop.

A sequence of elementary row operations becomes one matrix product

The real payoff is not just representing one elementary row operation. A whole sequence of elementary row operations becomes one product of elementary row-operation matrices.

Theorem

A sequence of elementary row operations as a product

Suppose

$$A_1 \xrightarrow{\rho_1} A_2 \xrightarrow{\rho_2} A_3 \xrightarrow{\rho_3} \cdots \xrightarrow{\rho_k} A_{k+1}.$$

Then

$$A_{k+1} = E_{\rho_k}E_{\rho_{k-1}}\cdots E_{\rho_2}E_{\rho_1}A_1.$$

Here every $\rho_j$ is an elementary row operation on matrices with the same number of rows. The product represents the whole sequence, although the product itself need not be a single elementary row-operation matrix.

The order in this formula is important. The first row operation appears closest to $A_1$, because it is applied first:

$$A_2 = E_{\rho_1}A_1, \qquad A_3 = E_{\rho_2}A_2 = E_{\rho_2}E_{\rho_1}A_1.$$

Worked example

Two row operations combined

Let

$$A = \begin{bmatrix} 1 & 0 & 1\\ 0 & 2 & 1\\ 1 & 0 & 0 \end{bmatrix}.$$

Apply

$$\rho_1:\ R_2 \leftarrow R_2 + R_1, \qquad \rho_2:\ R_1 \leftarrow R_1 + 2R_2.$$

The corresponding row-operation matrices are

$$E_{\rho_1} = \begin{bmatrix} 1 & 0 & 0\\ 1 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}, \qquad E_{\rho_2} = \begin{bmatrix} 1 & 2 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}.$$

After both operations, the result is

$$E_{\rho_2}E_{\rho_1}A = \begin{bmatrix} 3&2&0\\ 1&1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&1\\ 0&2&1\\ 1&0&0 \end{bmatrix} = \begin{bmatrix} 3&4&5\\ 1&2&2\\ 1&0&0 \end{bmatrix}.$$

Instead of multiplying $E_{\rho_2}E_{\rho_1}$ directly, you can also obtain this combined matrix by applying the same two elementary row operations to $I_3$ in the same order. The final numerical matrix also confirms that the second operation uses the already-updated row $2$.

Reverse row operations and inverses

Every elementary row operation has a reverse operation:

• the reverse of $R_j \leftarrow R_j + cR_i$ is $R_j \leftarrow R_j - cR_i$;
• the reverse of $R_i \leftarrow cR_i$, with $c \ne 0$, is $R_i \leftarrow (1/c)R_i$;
• the reverse of a row swap is the same row swap.

This gives a precise matrix statement.

Theorem

Elementary row-operation matrices are invertible

Every elementary row-operation matrix is invertible. Its inverse is the elementary row-operation matrix corresponding to the reverse elementary row operation.

Proof

Why the reverse matrix is a two-sided inverse

Let $E_\rho$ represent an elementary row operation $\rho$, and let $E_{\rho^{-1}}$ represent the reverse elementary operation. For every matrix $X$ with $p$ rows, applying $\rho$ and then reversing it restores $X$. By the left-multiplication theorem,

$$E_{\rho^{-1}}E_\rho X=X.$$

This statement holds for every such $X$; in particular, put $X=I_p$ to obtain $E_{\rho^{-1}}E_\rho=I_p$. Reversing first and then applying $\rho$ similarly gives $E_\rho E_{\rho^{-1}}=I_p$. Hence

$$E_\rho^{-1}=E_{\rho^{-1}}.$$

Concretely, a swap is its own inverse, scaling by $c\ne0$ is undone by scaling by $1/c$, and adding $c$ times row $i$ to row $j$ is undone by adding $-c$ times row $i$ to row $j$.

For example, if

$$E = \begin{bmatrix} 1 & 0 & 0\\ 3 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$

performs $R_2 \leftarrow R_2 + 3R_1$, then

$$E^{-1} = \begin{bmatrix} 1 & 0 & 0\\ -3 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$

performs $R_2 \leftarrow R_2 - 3R_1$.

This is the algebraic reason elementary row operations are reversible, and it explains why row reduction is so closely connected to invertible matrices.

Proof

Undo the last operation first

Let $E_1,\ldots,E_k$ encode a chronological chain on matrices with $p$ rows, so its combined multiplier is $J=E_k\cdots E_1$. Each inverse $E_j^{-1}$ represents the reverse of the corresponding elementary operation. To recover the input, the first move must undo $E_k$, because that was the last change made to the rows. The remaining reverse moves undo $E_{k-1},\ldots,E_1$ in that order. Consequently the reverse multiplier is

$$K=E_1^{-1}E_2^{-1}\cdots E_k^{-1}.$$

There are two orders to distinguish: the time order of the reverse operations starts with $E_k^{-1}$, while the written product places that first action nearest the matrix being restored. For three steps, associativity gives

$$KJ=E_1^{-1}E_2^{-1}(E_3^{-1}E_3)E_2E_1 =E_1^{-1}(E_2^{-1}E_2)E_1=I_p.$$

The other product $JK$ cancels adjacent inverse pairs starting with $E_1E_1^{-1}$ and also equals $I_p$. The same successive cancellation works for any finite chain. No factors have been commuted, and the argument does not require the matrix being transformed to be square.

Apply this reading to the earlier two-operation example. Its combined multiplier is $E_{\rho_2}E_{\rho_1}$, so its inverse is $E_{\rho_1}^{-1}E_{\rho_2}^{-1}$. Starting from the final matrix, first subtract twice the current second row from the first, recovering $[1\ 0\ 1]$ as the first row. Then subtract that recovered first row from the second, recovering $[0\ 2\ 1]$. Reversing the chain means restoring the intermediate states, which explains the order without a memorized rule.

Finite elimination terminates and preserves the solution set

Let $A\in\mathbb R^{m\times n}$, $b\in\mathbb R^m$, with positive finite integers $m,n$. The permitted row operations are an exchange of two distinct rows, multiplication of a row by a nonzero scalar, and addition of a scalar multiple of one row to a different row. Their inverses are respectively the same exchange, multiplication by the reciprocal, and subtraction of that multiple. Applying an operation to the entire augmented array therefore preserves exactly the solutions of $Ax=b$: the new equations are consequences of the old, and its inverse gives the reverse implication. Each operation is also multiplication on the left by the result of applying it to the identity matrix, an invertible elementary matrix.

Scan the coefficient columns from left to right, maintaining the next unused row. If the current column has a nonzero entry in an unused row, exchange that row into the next position, divide by its nonzero pivot, and subtract multiples of it from every other row to clear that column. Advance the unused-row index. If every unused entry in the column is zero, skip the column. Previously chosen pivot columns remain unit columns: the current unused rows have zeros in them, so the new pivot row and every subsequent subtraction preserve those zeros and earlier pivots. There are only $n$ column scans and at most $\min(m,n)$ pivots, each with finitely many row operations. Thus the algorithm terminates. At its end, the nonzero coefficient rows have increasing pivot columns and zeros below and above each pivot; any remaining coefficient row is zero. This proves the reduced echelon form used below, without assuming it will eventually appear.

Classification and the homogeneous freedom

Write $r$ for the number of pivots in the coefficient part. A zero coefficient row with a nonzero transformed right side is the impossible equation $0=c$, so the system has no solution. Otherwise assign arbitrary real values to all $n-r$ nonpivot variables. In each pivot row, its pivot variable equals the transformed right side minus the displayed linear combination of these free variables; no other pivot variable occurs. These formulas both construct a solution and recover every solution. If $r=n$ the solution is unique. If $r<n$, changing a free variable changes the vector, giving infinitely many solutions.

For $Az=0$ the free variables can all be zero, giving the zero solution. If $r<n$, set one free variable to one and the others to zero, and use the pivot formulas. This constructs a nonzero homogeneous solution. If $r=n$, the formulas force every homogeneous variable to zero. For a consistent system, fixing one solution $x_p$, subtraction proves that every solution has the form $x_p+z$ with $Az=0$, and substitution proves the converse.

For square $A$, if every column has a pivot, the coefficient form is $I$. The product $E$ of the finitely many elementary matrices satisfies $EA=I$. Since $E$ is invertible, $A=E^{-1}$ and therefore also $AE=I$: $E$ is a two-sided inverse. If a pivot is missing, the nonzero homogeneous solution constructed above rules out any inverse. This also justifies applying the same row operations to $[A\mid I]$ to obtain $[I\mid A^{-1}]$ in the invertible case.

Build the determinant from smaller matrices

Definition

Submatrix, minor, and cofactor

Let $A=[a_{ij}]$ be an $n\times n$ matrix with $n\ge 2$. For indices $1\le i,j\le n$, delete row $i$ and column $j$ simultaneously. The resulting $(n-1)\times(n-1)$ matrix is the submatrix $A(i\mid j)$.

The minor belonging to $a_{ij}$ is the scalar

$$M_{ij}=\det\bigl(A(i\mid j)\bigr).$$

The cofactor belonging to $a_{ij}$ is the signed scalar

$$C_{ij}=(-1)^{i+j}M_{ij}.$$

The signs form a checkerboard whose upper-left sign is positive:

$$\begin{bmatrix} +&-&+&\cdots\\ -&+&-&\cdots\\ +&-&+&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{bmatrix}.$$

The notation $A(i\mid j)$ names a matrix, $M_{ij}$ is a determinant of that matrix, and $C_{ij}$ is that scalar after the sign has been attached. Keeping these three objects distinct prevents a common double-sign error.

Definition

Recursive definition of the determinant

For a $1\times1$ matrix $[a_{11}]$, define

$$\det([a_{11}])=a_{11}.$$

For an $n\times n$ matrix with $n>1$, define its determinant by expansion along the first row:

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

This is a definition, not yet a theorem about other rows or columns. Each minor in the sum has size $n-1$, so repeated use of the definition eventually reaches the $1\times1$ base case. No $0\times0$ determinant convention is needed: the submatrix and minor definitions are used only for $n\ge2$.

Arbitrary cofactor expansion from the recursive determinant

Let $A$ be real $n\times n$, $n\ge1$, with determinant defined by the one-by-one base case and first-row recursion. Let $S_n$ be the permutations of $1,\ldots,n$. Define the inversion number $\operatorname{inv}(\sigma)$ as the number of pairs $r<s$ with $\sigma(r)>\sigma(s)$, and put $\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}$.

Suppose $\sigma(i)=j$ and delete position $i$ and value $j$, keeping the relative orders of the remaining positions and values; the resulting permutation is $\tau\in S_{n-1}$. If $u$ entries before position $i$ are smaller than $j$, the inversions removed are $(i-1-u)+(j-1-u)=i+j-2-2u$. Consequently

$$\operatorname{sgn}(\sigma)=(-1)^{i+j}\operatorname{sgn}(\tau).$$

Define $P_n(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{r=1}^n a_{r,\sigma(r)}$. For $n=1$, it is the recursive determinant. For $n>1$, grouping by $\sigma(1)=j$ and using the sign identity gives $P_n(A)=\sum_j(-1)^{1+j}a_{1j}P_{n-1}(A(1\mid j))$. Induction therefore proves $P_n(A)=\det A$.

Now fix any row $i$ and group the same finite permutation sum by $\sigma(i)=j$. Deletion is a bijection to all permutations of the corresponding minor, so

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

For a fixed column $j$, each permutation has a unique position $i$ with $\sigma(i)=j$. Grouping by that position gives

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

Both expansion formulas concern $n\ge2$, so no empty determinant convention was introduced. This proves the general expansion theorem without assuming alternating row rules or the determinant product theorem.

Why the structural rules follow

Zero rows and zero columns

Theorem

A zero row or zero column gives determinant zero

If every entry in one row or one column of a square matrix is zero, then its determinant is zero.

Theorem

A row or column with one possible nonzero entry

For a square matrix of size $n\ge2$, suppose all entries in row $k$ are zero except possibly $c_{k\ell}$. Then

$$\det(C)=c_{k\ell}(-1)^{k+\ell}\det\bigl(C(k\mid\ell)\bigr).$$

The same formula holds when all entries in column $\ell$ are zero except possibly $c_{k\ell}$. The formula includes the case $c_{k\ell}=0$.

For $n=1$, a matrix with a zero row is $[0]$, whose determinant is $0$ by the base definition. For $n\ge2$, expand along that row using the arbitrary-row theorem. Every term has a zero entry as its factor, so every term is zero and hence $\det(A)=0$.

The zero-column statement can be proved directly by induction, which makes the inheritance of the zero column explicit. For $n=1$, the only $1\times1$ matrix with a zero column is $[0]$, and its determinant is $0$. Assume the statement for $n=k$. Let $C$ be a $(k+1)\times(k+1)$ matrix with zero column $\ell$. Expand along the first row. The term from column $\ell$ is zero. In every other term, delete row $1$ and a column $j\ne\ell$; the original zero column survives as a zero column of the resulting $k\times k$ submatrix (with its index shifted when $j\lt\ell$). By the induction hypothesis each such submatrix has determinant zero. Hence every term is zero, proving $\det(C)=0$.

For the one-support theorem, the row case follows by expanding along row $k$: all terms except the $\ell$-term vanish. For the column case, expand along row $k$. The $\ell$-term remains. For each $j\ne\ell$, the submatrix $C(k\mid j)$ inherits a zero column from column $\ell$ of $C$, so its determinant is zero by the zero-column result. Therefore only $c_{k\ell}(-1)^{k+\ell}\det(C(k\mid\ell))$ remains. This is the precise reason that a sparse row or column collapses the calculation to one minor.

Triangular matrices

Theorem

Triangular determinant theorem

If $T$ is upper triangular or lower triangular, then

$$\det(T)=t_{11}t_{22}\cdots t_{nn}.$$

We prove the upper-triangular statement by induction on $n$. For $n=1$, it is the determinant definition. Suppose it holds for upper-triangular $k\times k$ matrices, and let $T$ be upper triangular of size $k+1$. Its last row has zero entries except possibly $t_{k+1,k+1}$. Expanding along that last row gives

$$\det(T)=t_{k+1,k+1}\det\bigl(T(k+1\mid k+1)\bigr),$$

because the sign is $(-1)^{2k+2}=1$. The retained submatrix is still upper triangular, with diagonal $t_{11},\ldots,t_{kk}$. By induction its determinant is $t_{11}\cdots t_{kk}$, giving the required product through $t_{k+1,k+1}$. For a lower-triangular matrix, expand along the first row, whose only possible nonzero entry is $t_{11}$; the retained submatrix is lower triangular with diagonal $t_{22},\ldots,t_{nn}$, and the same induction proves the result.

Multilinearity in one fixed row

Theorem

Multilinearity in one fixed row

Fix every row except row $p$. If the new row is $\alpha u+\beta v$, then

$$\det(\text{same rows except }\alpha u+\beta v) =\alpha\det(\text{same rows except }u) +\beta\det(\text{same rows except }v).$$

For $n=1$, the base definition gives

$$\det([\alpha a+\beta b])=\alpha a+\beta b =\alpha\det([a])+\beta\det([b]).$$

For $n\ge2$, fix every row except row $p$. Let matrices $A,B,C$ agree in all those fixed rows, and suppose their $p$th rows satisfy $c_p=\alpha a_p+\beta b_p$. For each column $j$, deleting row $p$ gives the same submatrix:

$$A(p\mid j)=B(p\mid j)=C(p\mid j).$$

Expanding $\det(C)$ along row $p$ and using the entrywise relation $c_{pj}=\alpha a_{pj}+\beta b_{pj}$ gives

$$\begin{aligned} \det(C) &=\sum_{j=1}^{n}(\alpha a_{pj}+\beta b_{pj})(-1)^{p+j}\det(C(p\mid j))\\ &=\alpha\det(A)+\beta\det(B). \end{aligned}$$

The equality works precisely because every other row is held fixed; changing two rows at once is a different statement.

Adjacent swaps and repeated rows

Theorem

Repeated rows force determinant zero

If a square matrix has two identical rows, then its determinant is $0$.

Let $A$ have adjacent rows $r_q,r_{q+1}$, and let $B$ be obtained by swapping those two physical rows. Expand $\det(A)$ along row $q$. Expand $\det(B)$ along row $q+1$, where the row $r_q$ has moved. For each column $j$, the retained submatrix in the two expansions is the same matrix: deleting the position occupied by $r_q$ leaves the other row in the same order. However, the signs differ:

$$(-1)^{q+1+j}=-(-1)^{q+j}.$$

The entries and minors therefore match term by term with the opposite sign, so $\det(B)=-\det(A)$. Swapping rows in positions $p\lt q$ can be realised by $2(q-p)-1$ adjacent swaps, an odd number, and hence also changes the sign.

If rows $p$ and $q$ of $A$ are identical, the matrix obtained by swapping them is still $A$. The swap result then gives $\det(A)=-\det(A)$. Since the entries are real, this implies $2\det(A)=0$ and therefore $\det(A)=0$. This proof does not rely on a later row-replacement rule.

Indexing and strategy

For each term, keep three objects visible: the selected entry, its submatrix after deleting one row and one column, and the minor obtained by taking that submatrix determinant. Write the checkerboard sign from the original row and column positions before evaluating the minor. When a minor is expanded again, its local indices guide the inner sign; they do not replace the outer sign.

The arbitrary-row and arbitrary-column theorem makes the choice of expansion line strategic. Choose a line with many zeros or simple surviving minors, and state that choice before calculating. A zero entry removes a branch immediately, whereas a zero minor must be established by its own calculation. The recursive first-row formula defines the determinant; arbitrary-line expansion is the theorem that permits this computational choice.

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.