MATH 240

Fri. October 4th, 2019

Triangular Matrices

Definition. A matrix is upper triangular if everything below the main diagonal is 0. (Entries on and above the main diagonal may or may not be zeroes).

Similarly, a matrix is lower triangular if everything above the main diagonal is 0. (Entries on and below the main diagonal may or may not be zeroes).

Determinant of Triangular Matrices

Let AA be a triangular matrix.
detA=a11a22...ann=∏i=1naii.\text{det}A=a_{11}a_{22}...a_{nn}=\prod_{i=1}^na_{ii}. This is the same computation used to find the determinant of a diagonal matrix.

Simplifying Determinant Computations

Theorem. Let AA be a square matrix.

  1. If a multiple of one row of AA is added to another row to form a matrix BB, then detB=detA\text{det}B=\text{det}A.
  2. If two rows of AA are interchanged to produce BB, then detB=−detA\text{det}B=-\text{det}A.
  3. If one row of AA is multiplied by kk to produce BB, then detB=k detA\text{det}B=k\ \text{det}A

By row-reducing a matrix into an upper triangular matrix and keeping these rules in mind, we can find its determinant much more quickly.

Ex.
A=(1−4−2−289−170)detA=∣1−4−2−289−170∣=∣1−4−2005−170∣=−∣1−4−2−170005∣=−∣1−4−203−2005∣=−(1)(3)(5)=−15.\begin{aligned} A&=\begin{pmatrix} 1&-4&-2\\-2&8&9\\-1&7&0 \end{pmatrix}\\ \text{det}A&=\begin{vmatrix} 1&-4&-2\\-2&8&9\\-1&7&0 \end{vmatrix}\\ &=\begin{vmatrix} 1&-4&-2\\0&0&5\\-1&7&0 \end{vmatrix}\\ &=-\begin{vmatrix} 1&-4&-2\\-1&7&0\\0&0&5 \end{vmatrix}\\ &=-\begin{vmatrix} 1&-4&-2\\0&3&-2\\0&0&5 \end{vmatrix}\\ &=-(1)(3)(5)\\ &=-15. \end{aligned}

B=(1−4−2−170−289)detB=∣1−4−2−170−289∣=−∣1−4−2−289−170∣=−detA=−(−15)=15.\begin{aligned} B&=\begin{pmatrix} 1&-4&-2\\-1&7&0\\-2&8&9 \end{pmatrix}\\ \text{det}B&=\begin{vmatrix} 1&-4&-2\\-1&7&0\\-2&8&9 \end{vmatrix}\\ &=-\begin{vmatrix} 1&-4&-2\\-2&8&9\\-1&7&0 \end{vmatrix}\\ &=-\text{det}A\\ &=-(-15)\\ &=15. \end{aligned}

More Theorems Related to the Determinant

Theorem. Let AA be a square matrix.

  1. AA is invertible if and only if detA≠0\text{det}A\neq 0.
  2. detAT=detA\text{det}A^T=\text{det}A.
  3. detAB=detA detB\text{det}AB=\text{det}A\ \text{det}B.

Rotation Matrices

A=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)B=(cos⁡ψ−sin⁡ψsin⁡ψcos⁡ψ)AB=(cos⁡θcos⁡ψ−sin⁡θsin⁡ψ−cos⁡θsin⁡ψ−sin⁡θcos⁡ψsin⁡θcos⁡ψ+cos⁡θsin⁡ψ−sin⁡θsin⁡ψ+cos⁡θcos⁡ψ)=(cos⁡(θ+ψ)−sin⁡(θ+ψ)sin⁡(θ+ψ)cos⁡(θ+ψ))\begin{aligned} A&=\begin{pmatrix} \cos{\theta}&-\sin{\theta}\\ \sin{\theta}&\cos{\theta} \end{pmatrix}\\ B&=\begin{pmatrix} \cos{\psi}&-\sin{\psi}\\ \sin{\psi}&\cos{\psi} \end{pmatrix}\\ AB&=\begin{pmatrix} \cos{\theta}\cos{\psi}-\sin{\theta}\sin{\psi}& -\cos{\theta}\sin{\psi}-\sin{\theta}\cos{\psi}\\ \sin{\theta}\cos{\psi}+\cos{\theta}\sin{\psi}& -\sin{\theta}\sin{\psi}+\cos{\theta}\cos{\psi} \end{pmatrix}\\ &=\begin{pmatrix} \cos{(\theta+\psi)}& -\sin{(\theta+\psi)}\\ \sin{(\theta+\psi)}& \cos{(\theta+\psi)} \end{pmatrix} \end{aligned} This result maxes sense because multiplying the matrices is essentially combining the two rotations, which produces a new rotation matrix of the two angles added together.
detA=cos⁡2θ+sin⁡2θ=1detB=cos⁡2ψ+sin⁡2ψ=1detAB=detA detB=1\begin{aligned} \text{det}A&=\cos^2{\theta}+\sin^2{\theta}=1\\ \text{det}B&=\cos^2{\psi}+\sin^2{\psi}=1\\ \text{det}AB&=\text{det}A\ \text{det}B=1 \end{aligned}