The matrix \(A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix}\)
Terms
| Term | This example | Why |
|---|---|---|
| determinant | $\det A = -3$ | The determinant of the matrix \[A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix}\] is the expansion along the first row, using the cofactors: \[\det A = 1 \cdot C_{11} + 2 \cdot C_{12} + 3 \cdot C_{13} = 1 \cdot 2 + 2 \cdot 2 + 3 \cdot (-3) = -3.\] |
| minor | $M_{11} = 2$, $M_{12} = -2$, $M_{13} = -3$ | The minors for the first row of the matrix \[A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix}\] are the determinants of the $2 \times 2$ matrices left after deleting row $1$ and one column: \[M_{11} = \det \begin{bmatrix} 5 & 6 \\ 8 & 10 \end{bmatrix} = 2, \quad M_{12} = \det \begin{bmatrix} 4 & 6 \\ 7 & 10 \end{bmatrix} = -2, \quad M_{13} = \det \begin{bmatrix} 4 & 5 \\ 7 & 8 \end{bmatrix} = -3.\] |
| cofactor | $C_{11} = 2$, $C_{12} = 2$, $C_{13} = -3$ | The cofactors for the first row of the matrix \[A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix}\] are the minors $2$, $-2$, and $-3$ with the signs $+$, $-$, and $+$, so $C_{12} = -M_{12} = 2$, while $C_{11} = M_{11}$ and $C_{13} = M_{13}$. |
Discussion
For this example, the last entry in $A$ was chosen to be $10$ instead of $9$ so that the determinant is not $0$. The matrix with the entries $1$ through $9$ in order is singular, because its third row is twice its second row minus its first row, so its determinant is $0$. <!– For that matrix, the first-row expansion would be
\[1 \cdot (-3) + 2 \cdot 6 + 3 \cdot (-3) = 0,\]which hides how the minors and cofactors combine into the determinant. –>