MATH 240

Wed. October 2nd, 2019

Determinants

Let A=(abbc)A=\begin{pmatrix} a&b\\b&c \end{pmatrix} be a 2×22\times 2 matrix. If ad=bc≠0ad=bc\neq 0, then AA is invertible. Further, if ad−bc=0ad-bc=0, then AA is not invertible.

This means that there is one number associated with each 2×22\times 2 matrix that determines whether it is invertible or not. This number turns out to exist for any n×nn\times n matrix as well.

For any n×nn\times n square matrix AA, let AijA_{ij} denote the n−1×n−1n-1\times n-1 submatrix of AA obtained by removing the iith row and the jjth column of AA.

Ex.
A=(4137221432134457)A12=(214313457)\begin{aligned} A&=\begin{pmatrix} 4&1&3&7\\ 2&2&1&4\\ 3&2&1&3\\ 4&4&5&7 \end{pmatrix}\\ A_{12}&=\begin{pmatrix} 2&1&4\\ 3&1&3\\ 4&5&7 \end{pmatrix} \end{aligned}

For n>2n>2, the determinant of an n×nn\times n matrix AA whose entries are denoted as aija_{ij} is given by
detA=a11detA11−a12detA12+...+(−1)n+1a1ndetA1n,\text{det}A=a_{11}\text{det}A_{11}-a_{12}\text{det}A_{12}+...+(-1)^{n+1}a_{1n}\text{det}A_{1n}, or, more compactly,
detA=∑j=1n (−1)1+j a1j detA1j.\text{det}A=\sum_{j=1}^n \ (-1)^{1+j} \ a_{1j} \ \text{det}A_{1j}.

Solving for the Determinant

Ex. A=(15024−10−20)A=\begin{pmatrix} 1&5&0\\2&4&-1\\0&-2&0 \end{pmatrix}

detA=(−1)2 a11 detA11+(−1)3 a12 detA12+(−1)4 a13 detA13=1 det(4−1−20)−5 det(2−100)+0 det(240−2)=1(0−2)−5(0−0)+0=−2.\begin{aligned} \text{det}A&=(-1)^2 \ a_{11} \ \text{det}A_{11}+(-1)^3 \ a_{12} \ \text{det}A_{12}+(-1)^4 \ a_{13} \ \text{det}A_{13}\\ &=1 \ \text{det}\begin{pmatrix} 4&-1\\-2&0 \end{pmatrix}-5 \ \text{det}\begin{pmatrix} 2&-1\\0&0 \end{pmatrix}+0 \ \text{det}\begin{pmatrix} 2&4\\0&-2 \end{pmatrix}\\ &=1(0-2)-5(0-0)+0\\ &=-2. \end{aligned}

Cofactors

Given an n×nn\times n matrix AA, let Cij=(−1)i+j detAijC_{ij}=(-1)^{i+j} \ \text{det}A_{ij}. The determinant of AA now becomes
detA=a11C11+a12C12+...+a1nC1n.\text{det}A=a_{11}C_{11}+a_{12}C_{12}+...+a_{1n}C_{1n}. These CC-matrices are known as cofactors of AA.

Theorem: The determinant of a n×nn\times n matrix AA can be computed by a cofactor expansion
detA=aj1Cj1+aj2Cj2+...+ajnCjn\text{det}A=a_{j1}C_{j1}+a_{j2}C_{j2}+...+a_{jn}C_{jn}
or
detA=a1jC1j+a2jC2j+...+anjCnj.\text{det}A=a_{1j}C_{1j}+a_{2j}C_{2j}+...+a_{nj}C_{nj}.

This means the determinant can be computed across any row or column, rather than just the first row as shown previously. Because of this, we can speed up the expensive computation required to find the determinant by computing across a row or column with many zeroes.

Ex. We can more quickly find the determinant of the matrix from the previous example by computing along the third row instead of the first. A=(15024−10−20)A=\begin{pmatrix} 1&5&0\\2&4&-1\\0&-2&0 \end{pmatrix}

detA=(−1)4 a31 detA31+(−1)5 a32 detA32+(−1)6 a33 detA33=0 det(504−1)+2 det(102−1)+0 det(1524)=0+2(−1−0)+0=−2.\begin{aligned} \text{det}A&=(-1)^4 \ a_{31} \ \text{det}A_{31}+(-1)^5 \ a_{32} \ \text{det}A_{32}+(-1)^6 \ a_{33} \ \text{det}A_{33}\\ &=0 \ \text{det}\begin{pmatrix} 5&0\\4&-1 \end{pmatrix}+2 \ \text{det}\begin{pmatrix} 1&0\\2&-1 \end{pmatrix}+0 \ \text{det}\begin{pmatrix} 1&5\\2&4 \end{pmatrix}\\ &=0+2(-1-0)+0\\ &=-2. \end{aligned}

Determinant of Diagonal Matrices

Definition. A diagonal matrix is a matrix where all of the non-diagonal entries are zero.

For a diagonal matrix AA, the determinant of AA is the product of its diagonal entries.

Ex.
A=(1000002000003000004000005)detA=(1)(2)(3)(4)(5)=120.\begin{aligned} A&=\begin{pmatrix} 1&0&0&0&0\\0&2&0&0&0\\0&0&3&0&0\\0&0&0&4&0\\0&0&0&0&5 \end{pmatrix}\\ \text{det}A&=(1)(2)(3)(4)(5)\\ &=120. \end{aligned}