Determinants and cofactor expansion

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.

Why one scalar is useful

Elimination detects invertibility by reducing a square matrix to pivot equations. A determinant records the same distinction in one scalar: its value is nonzero exactly when the matrix is invertible. To construct this scalar and justify how it changes under elimination, begin with signed expansions into smaller square matrices.

The construction is recursive. A determinant of size $n$ is expressed using determinants of size $n-1$, and the recursion stops at the $1\times1$ case. The sign attached to each smaller determinant matters: omitting one minus sign can change the final value even when every minor has been computed correctly.

Concept lensStructural

One scalar has two useful readings

The recursive expansion is an algorithmic way to calculate a number: choose an entry, delete its row and column, and continue on the smaller determinant. The same number also has a structural role: zero rows, repeated rows, and triangular shape force predictable values. These structural rules follow from the expansion. A cofactor is a signed scalar, not a new row or column of the matrix.

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 $n\ge2$.

Expand along a convenient row or column

Theorem

Expansion along an arbitrary row or column

For a square matrix $A=[a_{ij}]$ of size $n\ge2$, and every fixed row index $i$,

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

The corresponding column statement is: for every fixed column index $j$,

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

Thus the first-row formula is the recursive definition, while an arbitrary-row or arbitrary-column formula is a theorem saying that the same determinant is independent of the chosen expansion line. The general proof is omitted here; the consequences below use this theorem precisely as stated.

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.

Worked examples

Worked example

The 2×2 formula is the first recursive case

Let

$$A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}.$$

Deleting row $1$ and column $1$ leaves $[a_{22}]$, while deleting row $1$ and column $2$ leaves $[a_{21}]$. Their minors are $M_{11}=a_{22}$ and $M_{12}=a_{21}$, so the checkerboard signs give

$$C_{11}=a_{22},\qquad C_{12}=-a_{21}.$$

The definition now yields

$$\det(A)=a_{11}a_{22}-a_{12}a_{21}.$$

The $ad-bc$ rule is therefore the $2\times2$ instance of the recursive definition, not an unrelated formula to be memorised separately.

Worked example

A 3×3 expansion with the signs made explicit

Let

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

The first-row entries are $3,2,-1$, and the first-row signs are $+,-,+$. Writing the third contribution as $+(-1)M_{13}$ avoids losing the plus before the negative entry:

$$\det(B)=3\begin{vmatrix}1&6\\-1&2\end{vmatrix} -2\begin{vmatrix}4&6\\-3&2\end{vmatrix} +(-1)\begin{vmatrix}4&1\\-3&-1\end{vmatrix}.$$

The three minors are, respectively,

$$1\cdot2-6(-1)=8,\qquad 4\cdot2-6(-3)=26,$$

and

$$4(-1)-1(-3)=-1.$$

Consequently,

$$\det(B)=3(8)-2(26)+(-1)(-1)=24-52+1=-27.$$

There are two separate signs in the last term: the checkerboard sign is positive, and the matrix entry itself is $-1$.

Worked example

A sparse 4×4 matrix: finish every surviving minor

Consider the matrix

$$C=\begin{bmatrix} 1&9&7&7\\ 0&5&2&5\\ 1&9&8&0\\ 1&9&8&3 \end{bmatrix}.$$

The second row has one zero, and its checkerboard signs are $-,+,-,+$. Expand along that row:

$$\det(C)=-0M_{21}+5M_{22}-2M_{23}+5M_{24}.$$

The required minors are

$$M_{22}=\det\begin{bmatrix}1&7&7\\1&8&0\\1&8&3\end{bmatrix}=3,$$$$M_{23}=\det\begin{bmatrix}1&9&7\\1&9&0\\1&9&3\end{bmatrix}=0, \qquad M_{24}=\det\begin{bmatrix}1&9&7\\1&9&8\\1&9&8\end{bmatrix}=0.$$

Expanding each of these $3\times3$ minors along its first row gives

$$M_{22}=1(8\cdot3-0\cdot8)-7(1\cdot3-0\cdot1)+7(1\cdot8-8\cdot1) =24-21+0=3,$$$$M_{23}=1(9\cdot3-0\cdot9)-9(1\cdot3-0\cdot1)+7(1\cdot9-9\cdot1) =27-27+0=0,$$$$M_{24}=1(9\cdot8-8\cdot9)-9(1\cdot8-8\cdot1)+7(1\cdot9-9\cdot1) =0-0+0=0.$$

Thus the signed second-row sum is

$$\det(C)=5(3)-2(0)+5(0)=15.$$

The zero in the first position removes a branch, but it is still necessary to keep the minus sign on the $2M_{23}$ term and to verify the remaining minors.

Worked example

Triangular determinants collapse to the diagonal

For

$$T=\begin{bmatrix}2&3&4\\0&5&7\\0&0&-1\end{bmatrix},$$

the triangular theorem gives

$$\det(T)=2\cdot5\cdot(-1)=-10.$$

The induction proof explains why the entries above the diagonal do not appear: the last row leaves only $-1$ in its expansion, and its retained minor is the upper-triangular matrix $\begin{bmatrix}2&3\\0&5\end{bmatrix}$. Expanding that minor leaves $2$ and then $5$. A diagonal matrix is the special case in which all off-diagonal entries are already zero, so $\det(I_n)=1$.

Counterexample mode

Determinants are not additive in the whole matrix

It is false in general that $\det(A+B)=\det(A)+\det(B)$. With $I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}$,

$$\det(I_2+I_2)=\det(2I_2)=4, \qquad \det(I_2)+\det(I_2)=1+1=2.$$

The valid repair is multilinearity in one fixed row. For a row $r=(r_1,r_2)$, let $F(r)=\det\begin{bmatrix}r_1&r_2\\1&0\end{bmatrix}$, and let $u=(1,2)$ and $v=(0,1)$. Then

$$F(2u-v)=\det\begin{bmatrix}2&3\\1&0\end{bmatrix}=-3,$$

while

$$2F(u)-F(v)=2\det\begin{bmatrix}1&2\\1&0\end{bmatrix} -\det\begin{bmatrix}0&1\\1&0\end{bmatrix}=-3.$$

The other row stayed fixed in both terms. That fixed-row condition is the content of multilinearity.

Common mistakes

Common mistake

Mixing up submatrix, minor, and cofactor

The submatrix $A(i\mid j)$ is a matrix. The minor $M_{ij}$ is the determinant of that submatrix. The cofactor is $C_{ij}=(-1)^{i+j}M_{ij}$. If you write the checkerboard sign in the expansion and also use a cofactor that already contains it, you have multiplied the sign twice.

Common mistake

A determinant belongs to a square matrix

The determinant is defined here only for square matrices. A $2\times3$ matrix does not have a determinant in this construction, because deleting one row and one column from a rectangular matrix does not produce the recursive square problem required by the definition.

Exercises

Checkpoint

Compute $\det\begin{bmatrix}1&2\\3&4\end{bmatrix}$ from the recursive definition.

Write the two first-row cofactors, including the checkerboard sign, before multiplying by the entries.

Solution · Answer 1

The cofactors from the first row are $C_{11}=4$ and $C_{12}=-3$. Hence

$$\det\begin{bmatrix}1&2\\3&4\end{bmatrix}=1(4)+2(-3)=-2.$$

Checkpoint

Compute $\det\begin{bmatrix}2&0&1\\0&3&0\\4&0&5\end{bmatrix}$ by expanding along the best row or column.

Choose the second row, which has two zero entries, and keep its middle sign.

Solution · Answer 2

Expand along the second row:

$$\det=3(-1)^{2+2}\begin{vmatrix}2&1\\4&5\end{vmatrix} =3(10-4)=18.$$

Checkpoint

Write the second-row cofactor expansion and justify its signs.

$$C=\begin{bmatrix}1&9&7&7\\0&5&2&5\\1&9&8&0\\1&9&8&3\end{bmatrix}.$$

Keep each minor indexed by its position in the original matrix.

Solution · Answer 3

For row $2$, the signs are $-,+,-,+$. The expansion is therefore $-0M_{21}+5M_{22}-2M_{23}+5M_{24}$; the middle minus sign is attached to the entry in column $3$.

Checkpoint

Explain why a square matrix with two identical rows has determinant zero.

Use the row-swap sign change and the fact that swapping identical rows leaves the matrix unchanged.

Solution · Answer 4

Swapping the two equal rows changes the determinant to its negative, but the matrix itself is unchanged. Thus $\det(A)=-\det(A)$, so over the real numbers $2\det(A)=0$ and $\det(A)=0$.

Checkpoint

What is $\det\begin{bmatrix}4\end{bmatrix}$?

Use the $1\times1$ base case of the recursive definition.

Solution · Answer 5

It is $4$, because the determinant of a $1\times1$ matrix equals its only entry.

Checkpoint

Why is it wise to expand along a row with many zeros?

Count how many nonzero terms remain in the expansion.

Solution · Answer 6

Every zero entry contributes zero times its cofactor, so fewer nonzero terms and fewer smaller determinants need to be evaluated.

Checkpoint

If a diagonal matrix has diagonal entries $2$, $-1$, and $5$, what is its determinant?

Apply the triangular determinant theorem.

Solution · Answer 7

The determinant is $2(-1)(5)=-10$.

Checkpoint

Explain why a matrix with one zero row cannot be invertible.

Connect the zero-row theorem with the determinant criterion for invertibility in the next note.

Solution · Answer 8

A zero row forces determinant $0$. The next note shows that an invertible matrix has nonzero determinant, so a matrix with a zero row cannot be invertible.

Related notes

Read 2.1 Matrix basics to review square matrices and indexing language.

Continue with 7.2 Row operations, products, and invertibility to see how determinants behave under elimination.

Keep 5.1 Invertible matrices nearby, because determinants soon become one more equivalent test for invertibility.