Row and column expansion of a determinant

Matrices: Matrices

Row and column expansion of a determinant

Recursive calculation of the determinant of the 3x3 matrix A recursive calculation of the determinant of a \(3\times 3\) matrix follows from the following rearrangement of the terms in \(\text{det}(A)\): \[\begin{aligned}\text{det}(A) &= a_{11}(a_{22}a_{33}-a_{23}a_{32}) - a_{12}(a_{21}a_{33}-a_{23}a_{31})+ a_{13}(a_{21}a_{32}-a_{22}a_{31})\\ \\ &= a_{11}\cdot \text{det}\!\matrix{a_{22} & a_{23}\\ a_{32} & a_{33}} - a_{12}\cdot \text{det}\!\matrix{a_{21} & a_{23}\\ a_{31} & a_{33}} + a_{13}\cdot \text{det}\!\matrix{a_{21} & a_{22}\\ a_{31} & a_{32}}\end{aligned}\] The last expression is a linear combination of three determinants of \(2\times 2\) submatrices where the coefficients are chosen with alternating signs in the first row and the \(2\times 2\) matrices are identical to the matrices obtained by deleting the first row and the column in which the respective coefficient is: \[a_{11}\cdot \text{det}\!\matrix{\color{red}{a_{11}} & \color{red}{a_{12}} & \color{red}{a_{13}}\\ \color{red}{a_{21}} & a_{22} & a_{23} \\ \color{red}{a_{31}} & a_{32} & a_{33}} - a_{12}\cdot\text{det}\!\matrix{\color{red}{a_{11}} & \color{red}{a_{12}} & \color{red}{a_{13}}\\ a_{21} & \color{red}{a_{22}} & a_{23} \\ a_{31} & \color{red}{a_{32}} & a_{33}} + a_{13}\cdot\text{det}\!\matrix{\color{red}{a_{11}} & \color{red}{a_{12}} & \color{red}{a_{13}}\\ a_{21} & a_{22} & \color{red}{a_{23}} \\ a_{31} & a_{32} & \color{red}{a_{33}}}\]

This recursive method can be generalized to a square matrix of arbitrary size, and can also be done with other rows or even columns in the matrix.

Laplace expansion of the determinant of a matrix Let \(A\) be an \(n\times n\) matrix. For each index \((i,j)\) you can construct the \((n-1)\times(n-1)\) matrix \(M_{ij}\) by deleting in \(A\) the \(i\)th row and \(j\)th column. The determinant \(\det(M_{ij})\) is called the (\(i,j\))-minor of the element \(a_{ij}\) of \(A\). The cofactor of the element \(a_{ij}\), denoted by \(A_{ij}\), is the signed (\(i,j\))-minor: \[A_{ij}=(-1)^{i+j}\cdot\det(M_{ij})\] Note that the minus signs in the minors are according to the following alternating scheme: \[\matrix{+& - & + & - & \cdots\\ - & + & - & + & \cdots\\ + & - & + & - & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots}\] Then we have the following two formulas for the determinant, which represent an 'expansion' along a row or along a column: \[\begin{aligned}\det(A) &= a_{i1}A_{i1}+a_{i2}A_{i2}+\cdots a_{in}A_{in}= \sum_{j=1}^n a_{ij}A_{ij}\qquad\text{voor elke }i\\ \det(A) &= a_{1j}A_{1j}+a_{2j}A_{2j}+\cdots a_{in}A_{nj} = \sum_{i=1}^n a_{ij}A_{ij}\qquad\text{for each}j\end{aligned}\] One refers to these formulas as the Laplace expansion of \(\det(A)\).

Selection of row or column for expansion The above row or column expansions of a determinant are of course very useful when a row or column has a lot of zeroes: expansion along that row or column then provides few smaller determinants.

Example Consider \[A=\left(\,\begin{array}{rrr} -1 & 0 & 2\\ 2 & 1 & 3\\ 1 & 3 & -1 \end{array}\,\right) \] In the first row occurs a \(0\) and so expansion of the determinant along this row is promising. Then the required submatrices are \[ M_{11}=\left(\,\begin{array}{rr} 1 & 3\\ 3 & -1 \end{array}\,\right)\hbox{ en } M_{13}=\left(\,\begin{array}{rr} 2 & 1\\ 3 & 3 \end{array}\,\right) \]For the computation of the determinant of #A# via expansion along the first row, the next two subdeterminants are needed: \[\begin{array}{rcl}\det( M_{11})&=&\left|\,\begin{array}{rr} 1 & 3\\ 3 & -1 \end{array}\,\right| =-10\\ \det( M_{13}) &=&\left|\,\begin{array}{rr} 2 & 1\\ 1 & 3 \end{array}\,\right| = 5 \end{array}\] Using the formula for expansion along the first row we find \[\begin{array}{rcl}
\det (A)&=&\sum_{j=1}^3 (-1)^{1+j}a_{1j}\det (M_{1j})\\ & =& (-1)^{1+1}a_{11}\det (M_{11}) - (-1)^{1+2}a_{12}\det (M_{12}) + (-1)^{1+3}a_{13}\det (A_{13})\\ & =& a_{11}\cdot\det (M_{11}+ a_{12}\cdot\det (M_{12})+ a_{13}\cdot \det (M_{13} \\ &=& -1\cdot (-10)- 0\cdot \det (M_{12}) +2\cdot 5\\ &=& 20\end{array}
\]

Upper triangular matrix An extreme example can be found in the determination of the determinant of a so-called triangular matrix: above (or below) the main diagonal are only zeros. The determinant is then equal to the product of the diagonal elements. The following upper triangular matrix is expanded along the first column and you continue to do this in the next steps: \[
\begin{array}{l l}
\det (A) & =\left|\,\begin{array}{ccccc}
a_{11} & \ast & \ldots & \ldots & \ast\\
0 & a_{22} & & & \vdots\\
\vdots & \ddots & \ddots & & \vdots\\
\vdots & \ddots & \ddots & a_{n-1,n-1} & \ast\\
0 & \ldots & \ldots & 0 & a_{nn}
\end{array}\,\right|\ \\\\
& =\ a_{11}\cdot \left|\,\begin{array}{ccccc}
a_{22} & \ast & \ldots & \ldots & \ast\\
0 & \ddots & & & \vdots\\
\vdots & \ddots & \ddots & & \vdots\\
\vdots & \ddots & \ddots & a_{n-1,n-1} & \ast\\
0 & \ldots & \ldots & 0 & a_{nn}
\end{array}\,\right|\\\\
& =\ a_{11}\cdot a_{22}\cdot \left|\,\begin{array}{ccc}
a_{33} & \ldots & \ast\\
& \ddots & \vdots\\
0 & & a_{nn}
\end{array}\,\right|\\\\ &=\ a_{11}\cdot a_{22}\cdots\ \left|\,\begin{array}{cc}
a_{n-1,n-1} & \ast\\
0 & a_{nn}
\end{array}\,\right|\\\\
& =\ a_{11}\cdot a_{22}\,\cdots\, a_{n-1,n-1}\cdot a_{nn}
\end{array}
\]

Compute \(\text{det}\!\matrix{0 & 0 & 1 \\ -1 & -1 & 2 \\ 1 & 1 & 0 \\ }\).

\[\begin{aligned}\text{det}\matrix{0 & 0 & 1 \\ -1 & -1 & 2 \\ 1 & 1 & 0 \\ } &= 0\cdot\text{det}\!\matrix{-1 & 2 \\ 1 & 0 \\ } -0\cdot\text{det}\!\matrix{-1 & 2 \\ 1 & 0 \\ } + 1\cdot\text{det}\!\matrix{-1 & -1 \\ 1 & 1 \\ }\\
&= (0\cdot -2) - (0\cdot-2) + (1\cdot 0) \\ &= 0\end{aligned}\] Here we have assumed that the calculation of the determinant of a \(2\times 2\) matrix can be done via mental arithmetic. For example: \(\text{det}\!\matrix{-1 & 2 \\ 1 & 0 \\ }=(-1\cdot0)-(2\cdot 1) = -2\).

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