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Determinants

3.1 Introduction to Determinants

Theorem 1

The determinant of an n×nn \times n matrix AA is given by the formula:

detA=j=1n(1)i+jaijdetAij\det{A} = \sum_{j=1}^{n}(-1)^{i+j}a_{ij}detA_{ij}

Example: Compute detA\det{A}

A=[150241020]detA=1det[4120]+(1)5det[2100]detA=1(˙02)5(0)=2A = \begin{bmatrix} 1 & 5 & 0\\ 2 & 4 & -1\\ 0 & -2 & 0 \end{bmatrix}\\ \det{A} = 1\cdot\det{\begin{bmatrix}4 & -1 \\ -2 & 0\end{bmatrix}} + (-1)5\cdot\det{\begin{bmatrix} 2 & -1 \\ 0 &0\end{bmatrix}}\\ \det{A} = 1\dot(0-2) - 5(0) \\= -2

Theorem 2

If AA is a triangular matrix, then detA\det{A} is the product of the entries on the main diagonal of AA

Questions

4, 10, 22, 28, 34, 42

3.2 Properties of Determinants

Theorem 3

The determinant of a square matrix AA is affected by elementary row operations per the following rules:

Ri+kRjdetARiRjdetARi×kkdetAR_i + kR_j \rightarrow \det{A}\\ R_i \Leftrightarrow R_j \rightarrow -\det{A}\\ R_i\times k \rightarrow k\det{A}

Theorem 4

A square matrix AA is inveritble if and only if detA0\det{A}\neq 0

Theorem 5

If AA is an n×nn \times n matrix, then detA=detA\det{A^\top} = \det{A}

Theorem 6

If AA and BB are n×nn\times n matrices, then detAB=(detA)(detB)\det{AB} = (\det{A})(\det{B})

Questions

Find det[13240125276331072]\det{\begin{bmatrix} 1 & 3 & 2 & -4\\ 0 & 1 & 2 & -5\\ 2 & 7 & 6 & -3\\ -3 & -10 & -7 & 2\\ \end{bmatrix}}

First, we can row-reduce via the following operations:

R32R1,R4+3R1A=[13240125012501110]R3R2,R4+42A=[132401250001000115]R3R4A=[132401250011500010]R_3 -2R_1, R_4 + 3R_1 \\ \rightarrow A = \begin{bmatrix} 1 & 3 & 2 & -4\\ 0 & 1 & 2 & -5\\ 0 & 1 & 2 & 5\\ 0 & -1 & -1 & -10\\ \end{bmatrix}\\ R_3-R_2, R_4+4_2 \\ \rightarrow A = \begin{bmatrix} 1 & 3 & 2 & -4\\ 0 & 1 & 2 & -5\\ 0 & 0 & 0 & 10\\ 0 & 0 & 1 & -15\\ \end{bmatrix}\\ R_3\Leftrightarrow R_4\\ \rightarrow A = \begin{bmatrix} 1 & 3 & 2 & -4\\ 0 & 1 & 2 & -5\\ 0 & 0 & 1 & -15\\ 0 & 0 & 0 & 10\\ \end{bmatrix}\\

The determinant of a triangular matrix is just the product of the diagonal entries. Notice that we swapped two rows, thus we multiply the product by (1)(-1).

detA=(1)11110\det{A} = (-1)\cdot 1 \cdot 1 \cdot 1 \cdot 10

True or False: If detA\det{A} is zero, then two rows or two columns are the same, or a row or a column is zero.

False. By Invertible Matrix Theorem, detA=0\det{A}=0 implies linear dependence. However, this isn't the correct definition.

True or False: detA1=(1)detA\det{A^{-1}} = (-1)\det{A}

False. Most obvious counter-example is the identity matrix II, which serves as its own inverse

detI1=detI\det{I^{-1}} = \det{I}

3.3 Cramer's Rule, Volume, and Linear Transformations

Theorem 7: Cramer’s Rule

Let AA be an invertible n×nn\times n matrix. For any bb in Rn\R^n, the unique solution xx of Ax=bAx=b has entries given by

xi=detAi(b)detAx_i = \frac{\det{A_i(b)}}{\det{A}}

where Ai(b)A_i(b) is AA with the ii-th column replaced with bb.

Example: Use Cramer’s rule to solve the system

3x12x2=65x1+4x2=83x_1 -2x_2 = 6\\ -5x_1 +4x_2 = 8\\
A1(b)=[6284],A2(b)=[3658]x=[detA1(b)/detAdetA2(b)/detA]=[2027]A_1(b)= \begin{bmatrix} 6 & -2\\ 8 & 4 \end{bmatrix}, A_2(b) = \begin{bmatrix} 3 & 6\\ -5 & 8 \end{bmatrix}\\ x = \begin{bmatrix} \det{A_1}(b)/\det{A}\\ \det{A_2}(b)/\det{A} \end{bmatrix} = \begin{bmatrix} 20\\ 27 \end{bmatrix}

Theorem 8: An Inverse Formula

Let AA be an invertible n×nn\times n matrix. Then

A1=1detAadjAA^{-1} = \frac{1}{\det{A}}\text{adj}{A}

where

adjA=[C11Cn1C1nCnn]\text{adj}{A} = \begin{bmatrix} C_{11} & \dots & C_{n1}\\ \vdots & \ddots & \vdots\\ C_{1n} & \dots & C_{nn} \end{bmatrix}\\

and

Cji=(1)j+idetAjiC_{ji} = (-1)^{j+i}\det{A_{ji}}

Example: Find the inverse of the matrix A=[213111142]A=\begin{bmatrix} 2 & 1 & 3\\ 1 & -1 & 1\\ 1 & 4 & -2 \end{bmatrix}

C11=1142,C12=1112,,C33=2111C_{11} = \begin{vmatrix} -1 & 1\\ 4 & -2 \end{vmatrix}, C_{12} = -\begin{vmatrix} 1 & 1\\ 1 & -2 \end{vmatrix}, \dots, C_{33} = \begin{vmatrix} 2 & 1\\ 1 & -1 \end{vmatrix}
A1=1detA[C11Cn1C1nCnn]=14[1265301642]A^{-1} = \frac{1}{\det{A}}\begin{bmatrix} C_{11} & \dots & C_{n1}\\ \vdots & \ddots & \vdots\\ C_{1n} & \dots & C_{nn} \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 12 & 6 & -5\\ 3 & 0 & -1\\ -6 & -4 & 2 \end{bmatrix}\\

Theorem 9

If AA is a 2×22\times 2 matrix, the area of the parallelogram determined by the columns of AA is detA|\det{A}|. This extends to 33 dimensional volume of parallelepipeds. Note that one of the vectors must be at the origin.

Example: Calculate the area of the parallelogram determined by the points (2,2),(0,3),(4,1),(6,4)(- 2, -2), (0, 3), (4, -1), (6, 4)

First, we must transpose the shape such that one of the points is at the origin. So we transpose the whole shape by (0,3)(0,-3).

(2,5),(0,0),(4,4),(6,1)(- 2, -5), (0, 0), (4, -4), (6, 1)

Then, choose any 22 points non-origin vectors that are not on the same yy-coordinate to be the defining vectors and compute the determinant.

2454=4641=2651=28\begin{vmatrix} -2 & 4\\ -5 & -4 \end{vmatrix} = \begin{vmatrix} 4 & 6\\ -4 & 1 \end{vmatrix} = \begin{vmatrix} -2 & 6\\ -5 & 1 \end{vmatrix} = 28

Theorem 10

volume(T(S))=detAvolume(S)volume(T(p+S))=detAvolume(p+S)\text{volume}(T(S)) = |\det{A}|\cdot \text{volume}(S)\\ \text{volume}(T(p+S)) = |\det{A}|\cdot \text{volume}(p+S)

Questions

Compute the solution of the system using Cramer’s rule:

x1+3x2+x3=8x1+2x3=43x1+x2=4x_1 + 3x_2 + x_3 = 8\\ -x_1+ 2x_3 = 4\\ 3x_1 + x_2 = 4

The solution xx to this system Ax=bAx=b is a 33-dimensional vector where each entry is computed using Cramer’s rule:

xi=detAi(b)detAx_i = \frac{\det{A_i(b)}}{\det{A}}
x=[det[ba2a3]det[a1ba3]det[a1a2b]]/detA\vec x = \begin{bmatrix} det\begin{bmatrix} b & a_2 & a_3 \end{bmatrix}\\ det\begin{bmatrix} a_1 & b & a_3 \end{bmatrix}\\ det\begin{bmatrix} a_1 & a_2 & b \end{bmatrix} \end{bmatrix}/\det{A}

Determine the values of the parameter s for which the system has a unique solution, describe the solution.

By Invertible Matrix Theorem and Theorem 4, we know that a matrix is invertible if and only if detA0\det{A} \neq 0. We know the determinant is given by detA=15s260\det{A} = 15s^2-60, which has the roots ±2\pm2. Using Cramer’s rule, we can express the general solution to the system:

x=[x1x2]\vec x = \begin{bmatrix} x_1\\x_2 \end{bmatrix}
x1=3525s15s260=15s1015s260x_1 = \frac{\begin{vmatrix} 3 & 5\\2 & 5s \end{vmatrix}} {15s^2-60} = \frac{15s-10}{15s^2-60}\\
x2=3s312215s260=6s3615s260x_2 = \frac{\begin{vmatrix} 3s & 3\\12 & 2 \end{vmatrix}} {15s^2-60} = \frac{6s-36}{15s^2-60}

Compute the adjugate matrix adjA\text{adj}{A}

A=[112021306]A = \begin{bmatrix} 1 & -1 & 2\\0 & 2 & 1\\ 3 & 0 & 6 \end{bmatrix}

Recall that the adjugate matrix is given by

adjA=[C11Cn1C1nCnn]\text{adj}{A} = \begin{bmatrix} C_{11} & \dots & C_{n1}\\ \vdots & \ddots & \vdots\\ C_{1n} & \dots & C_{nn} \end{bmatrix}\\

where

Cji=(1)j+idetAjiC_{ji} = (-1)^{j+i}\det{A_{ji}}

So we compute each entry individually:

C11=6,C12=3,C13=6C21=6,C22=0,C23=3C31=5,C32=1,C33=2C_{11} = 6, C_{12}=3, C_{13}=-6\\ C_{21}=6, C_{22}=0, C_{23}=-3\\ C_{31}=-5, C_{32}=-1, C_{33}=2

And we create the adjugate matrix in transpose order:

adjA=[665301632]\text{adj}A = \begin{bmatrix} 6 & 6 & -5\\ 3 & 0 & -1\\ -6 & -3 & 2 \end{bmatrix}

Find the area of the parallelogram defined by the following vertices: (0,2)(5,2),(3,1),(2,1)(0,-2) (5, -2), (-3, 1), (2, 1)

By Theorem 9, we know that the area of this parallelogram is given by the determinant of any two non-origin vectors. First, we transpose the shape to the origin, translating by (0,2)(0,2).

(0,0)(5,0),(3,3),(2,3)(0,0) (5, 0), (-3, 3), (2, 3)

Then select any 22 non-origin vectors that are not on the same yy-coordinate and compute their determinant:

5303=5203=15\begin{vmatrix} 5 & -3\\ 0 & 3 \end{vmatrix} = \begin{vmatrix} 5 & 2\\ 0 & 3 \end{vmatrix} = 15

Note that the following two vectors do not work because they share the same yy-coordinate:

3233=15\begin{vmatrix} -3 & 2\\ 3 & 3 \end{vmatrix} = -15

Let T:RmRnT: \R^m \rightarrow \R^n be a linear transformation, and let pp be a vector and SS a set in Rm\R^m. Show that the image of p+Sp+S under TT is the translated set T(p)+T(S)T(p)+T(S) in Rn\R^n.

By definition, p+Sp+S is the set of all vectors of the form p+vp+v. We know that linear transformations are closed under vector addition

T(p+v)=T(p)+T(v)T(p+v) = T(p)+T(v)

Thus, the above condition holds for each vSv\in S.

T(p+S)=T(p)+T(S)T(p+S) = T(p)+T(S)

Compute the area of the image of SS under the mapping xAxx \mapsto Ax

A=[5211],b1=[47],b2=[01]A = \begin{bmatrix} 5 & 2 \\ 1 & 1 \end{bmatrix}, b_1 = \begin{bmatrix} 4 \\ -7 \end{bmatrix}, b_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}

By Theorem 10 we know that the change in volume after the transformation by AA is equal to the determinant of AA.

area(T(S))=detAarea(S)=(52)(40)=12\text{area}(T(S)) = |\det{A}|\cdot \text{area}(S)\\ = (5-2) \cdot (4 -0)\\ = 12