Tutorials › Linear Algebra › Minors of Matrix Entries

Determinants by Cofactor Expansion · Tutorial 212 of 1000

Minors of Matrix Entries

Learn how to form a minor by deleting the correct row and column, preserve the remaining entry order, and evaluate the resulting determinants.

Beginner 9 min read

What You'll Learn

  • Define the minor associated with an entry of a square matrix.
  • Identify which rows and columns remain after deletion.
  • Calculate minors of entries in a \(3\times3\) matrix using \(2\times2\) determinants.
  • Set up \(4\times4\) minors as \(3\times3\) matrices.
  • Evaluate \(3\times3\) determinants needed for \(4\times4\) minors.
  • Distinguish a minor from the entry it is associated with.

What a Minor Records

Each entry of a square matrix has a minor associated with it. To find that minor, remove the entire row containing the entry and the entire column containing it. The entries left over form a smaller square matrix; the minor is the determinant of that smaller matrix. This deletion rule is simple, but the row and column indices must both be tracked carefully.

For an \(n\times n\) matrix with \(n\ge2\), the minor \(M_{ij}\) is associated with the entry in row \(i\), column \(j\). It is a number, not a matrix entry copied from the original matrix. The matrix that remains after deletion is the matrix whose determinant gives that number. The remaining rows and columns stay in their original order.

Definition: Let \(\mathbf{A}=[a_{ij}]\in\mathbb{R}^{n\times n}\), where \(n\ge2\). Delete row \(i\) and column \(j\) from \(\mathbf{A}\), keeping the other rows and columns in their original order, to obtain an \((n-1)\times(n-1)\) matrix. The minor of the entry \(a_{ij}\), denoted \(M_{ij}\), is the determinant of that remaining matrix.

The indices in \(M_{ij}\) tell us what to delete: the first index identifies the row and the second identifies the column. For example, \(M_{23}\) means delete row 2 and column 3. It does not mean take the entry in row 2, column 3, nor does it mean delete only that entry.

Minors in a \(3\times3\) Matrix

Deleting one row and one column from a \(3\times3\) matrix leaves a \(2\times2\) matrix. Its determinant can be evaluated using the formula from the earlier material on determinants of \(2\times2\) matrices: \(\det\begin{bmatrix}a&b\\c&d\end{bmatrix}=ad-bc\). Thus, every \(3\times3\) minor requires two steps: identify the four remaining entries in their original order, then apply this formula.

Result: In a \(3\times3\) matrix, let the two rows left after deleting row \(i\) be \(r<s\), and let the two columns left after deleting column \(j\) be \(p<q\). Then \(M_{ij}=a_{rp}a_{sq}-a_{rq}a_{sp}\).

Proof. Deleting row \(i\) leaves exactly two rows, which we call \(r\) and \(s\) in their original order. Deleting column \(j\) leaves exactly two columns, called \(p\) and \(q\) in their original order. The remaining matrix is therefore \(\begin{bmatrix}a_{rp}&a_{rq}\\a_{sp}&a_{sq}\end{bmatrix}\). Applying the \(2\times2\) determinant formula gives \(a_{rp}a_{sq}-a_{rq}a_{sp}\), as claimed. \(\square\)

Worked Example: Two Minors of a \(3\times3\) Matrix

For \(\mathbf{A}=\begin{bmatrix}2&-1&3\\0&4&1\\5&2&-2\end{bmatrix}\), find \(M_{12}\) and \(M_{31}\).

For \(M_{12}\), delete row 1 and column 2. The rows left are rows 2 and 3, and the columns left are columns 1 and 3, so

$$ M_{12} =\det\begin{bmatrix}0&1\\5&-2\end{bmatrix} =(0)(-2)-(1)(5)=-5. $$

For \(M_{31}\), delete row 3 and column 1. The rows left are rows 1 and 2, and the columns left are columns 2 and 3:

$$ M_{31} =\det\begin{bmatrix}-1&3\\4&1\end{bmatrix} =(-1)(1)-(3)(4)=-1-12=-13. $$

Check: In the first remaining matrix the diagonal product is \(0\), and the other product is \(5\), giving \(0-5=-5\). In the second, the products are \(-1\) and \(12\), giving \(-1-12=-13\). The rows and columns used in each calculation are precisely those not deleted.

For a \(3\times3\) matrix, the general rule can also be written out entry by entry. Each expression uses the four entries left after the specified row and column are removed:

$$ \begin{aligned} M_{11}&=a_{22}a_{33}-a_{23}a_{32},& M_{12}&=a_{21}a_{33}-a_{23}a_{31},& M_{13}&=a_{21}a_{32}-a_{22}a_{31},\\ M_{21}&=a_{12}a_{33}-a_{13}a_{32},& M_{22}&=a_{11}a_{33}-a_{13}a_{31},& M_{23}&=a_{11}a_{32}-a_{12}a_{31},\\ M_{31}&=a_{12}a_{23}-a_{13}a_{22},& M_{32}&=a_{11}a_{23}-a_{13}a_{21},& M_{33}&=a_{11}a_{22}-a_{12}a_{21}. \end{aligned} $$

For instance, \(M_{23}\) deletes row 2 and column 3, leaving rows 1 and 3 and columns 1 and 2. Its remaining matrix is \(\begin{bmatrix}a_{11}&a_{12}\\a_{31}&a_{32}\end{bmatrix}\), so \(M_{23}=a_{11}a_{32}-a_{12}a_{31}\). This entry-by-entry form is a useful check on indexing, but the deletion procedure is the reliable way to construct each minor.

Minors in a \(4\times4\) Matrix

Deleting one row and one column from a \(4\times4\) matrix leaves a \(3\times3\) matrix. The minor is the determinant of that matrix, not the matrix itself. To calculate numerical \(4\times4\) minors, we need to evaluate the resulting \(3\times3\) determinants. For a matrix \(\begin{bmatrix}a&b&c\\d&e&f\\g&h&k\end{bmatrix}\), the six-term formula is \(aek+bfg+cdh-ceg-bdk-afh\).

Result: For a \(3\times3\) matrix \(\begin{bmatrix}a&b&c\\d&e&f\\g&h&k\end{bmatrix}\), its determinant is \(aek+bfg+cdh-ceg-bdk-afh\).

Proof. In the permutation definition of a determinant, each term chooses one entry from each row and each column. For three rows, the six possible column orders are \((1,2,3),(2,3,1),(3,1,2),(3,2,1),(2,1,3),(1,3,2)\). The first three orders have positive sign and give the products \(aek\), \(bfg\), and \(cdh\). The last three have negative sign and give \(ceg\), \(bdk\), and \(afh\). Adding the positive products and subtracting the negative products gives the displayed formula. \(\square\)

Worked Example: A Minor in a \(4\times4\) Matrix

Let \(\mathbf{B}=\begin{bmatrix}1&2&0&1\\3&1&4&2\\0&2&1&1\\2&0&3&1\end{bmatrix}\). Find \(M_{24}\).

Delete row 2 and column 4. The rows left, in order, are rows 1, 3, and 4; the columns left, in order, are columns 1, 2, and 3. Therefore the remaining matrix is \(\begin{bmatrix}1&2&0\\0&2&1\\2&0&3\end{bmatrix}\). Applying the six-term formula with \(a=1,b=2,c=0,d=0,e=2,f=1,g=2,h=0,k=3\) gives

$$ \begin{aligned} M_{24} &=(1)(2)(3)+(2)(1)(2)+(0)(0)(0)\\ &\quad -(0)(2)(2)-(2)(0)(3)-(1)(1)(0)\\ &=6+4+0-0-0-0=10. \end{aligned} $$

Check: The three positive products are \(6,4,0\), and each of the three subtracted products is \(0\). Their signed total is \(10\). The calculation uses the matrix formed by deleting row 2 and column 4, rather than the original entry \(b_{24}=2\).

Worked Example: A Negative \(4\times4\) Minor

Using the same matrix \(\mathbf{B}\), find \(M_{31}\).

Delete row 3 and column 1. The remaining rows are rows 1, 2, and 4, while the remaining columns are columns 2, 3, and 4. Thus the submatrix is \(\begin{bmatrix}2&0&1\\1&4&2\\0&3&1\end{bmatrix}\). Substituting \(a=2,b=0,c=1,d=1,e=4,f=2,g=0,h=3,k=1\) into the six-term formula gives

$$ \begin{aligned} M_{31} &=(2)(4)(1)+(0)(2)(0)+(1)(1)(3)\\ &\quad -(1)(4)(0)-(0)(1)(1)-(2)(2)(3)\\ &=8+0+3-0-0-12=-1. \end{aligned} $$

Check: The positive products total \(8+0+3=11\), and the subtracted products total \(0+0+12=12\). Their difference is \(11-12=-1\). Again, row 3 and column 1 are absent from the submatrix, and all other rows and columns retain their original order.

A Reliable Deletion Procedure

When a minor is requested, it helps to separate the selection of the submatrix from its determinant calculation. This avoids mixing up the original entry \(a_{ij}\) with the number \(M_{ij}\). Use the following sequence:

1
Read the indices.
For \(M_{ij}\), mark row \(i\) and column \(j\) of the original square matrix.
2
Delete both.
Remove the complete marked row and the complete marked column. Do not remove only their intersection.
3
Copy the remaining entries in order.
Keep the surviving rows in their original top-to-bottom order and the surviving columns in their original left-to-right order.
4
Evaluate the smaller determinant.
For a \(3\times3\) original matrix, use the \(2\times2\) determinant formula. For a \(4\times4\) original matrix, the remaining matrix is \(3\times3\).

This procedure also gives a useful size check. A minor from a \(3\times3\) matrix must come from a \(2\times2\) matrix, so it is calculated as a \(2\times2\) determinant. A minor from a \(4\times4\) matrix must come from a \(3\times3\) matrix. If the submatrix has the wrong size, at least one deletion was not carried out correctly.

Common Mistake: Deleting Only the Entry

Consider \(\mathbf{C}=\begin{bmatrix}2&1&4\\3&5&0\\1&2&6\end{bmatrix}\), and suppose we need \(M_{12}\). A wrong approach is to remove only the entry \(c_{12}=1\) and then choose nearby entries to form a \(2\times2\) determinant. For example, \(\det\begin{bmatrix}2&4\\3&0\end{bmatrix}=-12\) is not \(M_{12}\): that matrix has not been formed by deleting row 1 and column 2.

The correct deletion removes all of row 1 and all of column 2. The remaining entries are from rows 2 and 3 and columns 1 and 3:

$$ M_{12} =\det\begin{bmatrix}3&0\\1&6\end{bmatrix} =(3)(6)-(0)(1)=18. $$

The wrong calculation produced \(-12\), while the correct minor is \(18\). The submatrix itself is the check: its entries must come only from the rows and columns that survive both deletions.

Why Minors Are Useful

Minors turn a determinant problem into a smaller determinant problem. For the examples here, the definition reduces \(3\times3\) matrices to \(2\times2\) determinants and \(4\times4\) matrices to \(3\times3\) determinants. This reduction is the reason minors are introduced before determinant calculations involving larger matrices. At this stage, the essential skill is to form the right smaller matrix and evaluate its determinant accurately.

Keep three distinctions in mind. The entry \(a_{ij}\) is a scalar in the original matrix. The submatrix used for \(M_{ij}\) is formed after deleting row \(i\) and column \(j\). The minor \(M_{ij}\) is the determinant of that submatrix. Deletion itself does not insert an extra sign or alter any surviving entry; it only selects the smaller matrix in its original order.

Check Your Understanding

For each question, first identify the surviving rows and columns, then evaluate the determinant of the resulting matrix.

  1. For \(\begin{bmatrix}1&3&2\\0&4&-1\\5&2&1\end{bmatrix}\), find \(M_{13}\).
  2. For the same matrix, find \(M_{21}\).
  3. For \(\begin{bmatrix}1&0&2&1\\2&3&1&0\\0&1&4&2\\1&2&0&3\end{bmatrix}\), write the \(3\times3\) matrix used to calculate \(M_{32}\).
  4. In your own words, explain why \(M_{ij}\) is not the same as \(a_{ij}\).
  5. What size matrix remains after deleting one row and one column from a \(4\times4\) matrix?

Test yourself: Linear Algebra final exam practice

Ready to check what you know? Work a full-length practice final, then open the complete worked solutions.

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers the full Calculus AB & BC curriculum (the core of Calculus I and II): full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon