Determinant

The determinant of a Matrix (for a square matrix only) is

2x2

$$\begin{array}{rl}\text{if }& A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\text{, }\\ & \det (A)=|A|=ad-bc\end{array}$$

3x3

It can be computed along any row or column of A, it doesn’t matter what you pick, actually.

Definitions:

At position $(i,j)$ in the matrix (that is row i, column j) we define:

• Minor $(A{)}_{ij}=M_{ij}=$ Determinant of the matrix obtained from $A$ by removing the $i$th row and the $j$th column
• Cofactor: $C_{ij}=(-1{)}^{i+j}{M}_{ij}$

Like a checkerboard,

$$\begin{array}{r}\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right],\left[\begin{array}{cccc}+& -& +& -\\ -& +& -& +\\ +& -& +& -\\ -& +& -& +\end{array}\right]\end{array}$$

Example:

$$\begin{array}{rl}A& =\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]\text{, }\\ & \\ M_{23}& =\left|\begin{array}{cc}1& 2\\ 7& 8\end{array}\right|=1(8)-7(2)\\ M_{23}& =-6\text{,}\\ & \\ C_{23}& =(-1{)}^{2+3}{M}_{23}\\ & =-M_{23}=-(-6)\\ C_{23}& =6\end{array}$$

This should feel familiar to taking the Cross Product

4x4

You gotta take the determinant recursively, so pick determinants with as many zero rows so you don’t have to compute them

Shortcuts

Zero Row/Column:

If you have a row or a column that is zero, the determinant is automatically $0$.

Triangular:

Upper triangular matrices are square matrices with zeros under the main diagonal. Lower triangular matrices are square matrices with zeros above the main diagonal

I guess the “triangle” refers to the numbers that aren’t zero?

$\left[\begin{array}{ccc}\ast & \ast & \ast \\ 0& \ast & \ast \\ 0& 0& \ast \end{array}\right]$ upper ฅ^•ﻌ•^ฅ ˙ᵕ˙ lower $\left[\begin{array}{ccc}\ast & 0& 0\\ \ast & \ast & 0\\ \ast & \ast & \ast \end{array}\right]$

Result: the determinant of a triangular matrix is equal to the multiplication of the entries in the diagonal. Wait! You can totally use Row Operations to make matrices triangular.

However:

• $\text{If }B=A\text{, }{R}_1=2R_1\text{ then }|B|=2|A|$
• $\text{If }B=A\text{, }{R}_1\leftrightarrow R_2\text{ then }|B|=-|A|$
• but adding a multiple of another row does not change the determinant. this has many uses!

If this works backwards too, if the determinant of the matrix is zero, you know its linearly dependent

• At least one column can be expressed as a linear combination of the others
• The matrix transformation maps some non-zero vector to zero (has a non-trivial nullspace)
• The matrix is not invertible (singular)

If a matrix is square and the columns are linearly independent, then the matrix is Invertable, so this is another great way to find if a matrix has an inverse

Also if $A$ and $B$ are square matrices, and $B=A^{-1}$:

$$|AB|=|A||B|$$

If $A$ is invertable than

$$|A^{-1}|=\frac{1}{|A|}$$

This works because if a matrix has a determinant of zero, it can’t have an inverse because its linearly dependent

Applications of the determinant

Solve equations $A\vec{x}=\vec{b}$ when $A$ is invertible:

• Method 1: $\vec{x}=A^{-1}\vec{b}$
• Method 2: “Cramer’s rule”
  • $A{I}_1(\left[\begin{array}{c}x\\ y\\ z\end{array}\right])=A_1(\vec{b})$
  • Thus $|A|x=A_1(\vec{b})$
  • Making $x=\frac{|A_1(\vec{b})|}{|A|}$

Finding $A^{-1}$ using determinants. We know that $A{A}^{-1}$ yields the identity matrix ($I_n$)

$A^{-1}=\frac{1}{|A|}adj(A)$

Adjugate

Find fvol of a parlalellapiped

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Homework 6

Problem 1

$$A=\left[\begin{array}{rrr}1& 1& -1\\ -6& -1& 10\\ 3& 3& 0\end{array}\right]$$

Find the determinant.

$$\begin{array}{rl}\det (A)& =\frac{1}{3}\det (\left[\begin{array}{rrr}3& 3& -3\\ -6& -1& 10\\ 3& 3& 0\end{array}\right])\\ & =\frac{1}{3}\det (\left[\begin{array}{rrr}0& 0& -3\\ -6& -1& 10\\ 3& 3& 0\end{array}\right])\\ & =\frac{1}{3}(\cancel{0\det (k_1})-\cancel{0\det (k_2)}+-3\det (\left[\begin{array}{cc}-6& -1\\ 3& 3\end{array}\right]))\\ & =\frac{1}{3}(-3(-6(\cancel{3})\cancel{--}1(\cancel{3})))\\ & =(\frac{1}{3})(-3)(3(-6+1))\\ & =15\end{array}$$

Use Cramer’s rule to solve to following system:

$$\{\begin{array}{rrrrrrrr}& x_1& +& x_2& -& x_3& =& -5\\ & -6x_1& -& x_2& +& 10x_3& =& -3\\ & 3x_1& +& 3x_2& & & =& 4\end{array}$$

🚨 Cramer’s Rule states, for a 3x3 matrix

$$\begin{array}{rl}& x=\frac{\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|}{\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|},y=\frac{\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|}{\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|},z=\frac{\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|}{\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|}\\ & \\ & x=\frac{\left|\begin{array}{ccc}-5& 1& -1\\ -3& -1& 10\\ 4& 3& 0\end{array}\right|}{15},y=\frac{\left|\begin{array}{ccc}1& -5& -1\\ -6& -3& 10\\ 3& 4& 0\end{array}\right|}{15},z=\frac{\left|\begin{array}{ccc}1& 1& -5\\ -6& -1& -3\\ 3& 3& 4\end{array}\right|}{15}\\ & \\ & x=\frac{195}{15},y=\frac{-175}{15},z=\frac{95}{15}\\ & \\ & x=13,y=\frac{-35}{3},z=\frac{19}{3}\end{array}$$

Problem 2

Let $A=\left[\begin{array}{ccc}-2& 2& 1\\ 3& 1& 3\\ -3& 2& 2\end{array}\right]$

Find the determinant of $A$.

$$\begin{array}{rl}\det (A)& =-2\left|\begin{array}{cc}1& 3\\ 2& 2\end{array}\right|-2\left|\begin{array}{cc}3& 3\\ -3& 2\end{array}\right|+1\left|\begin{array}{cc}3& 1\\ -3& 2\end{array}\right|\\ & =-2(2(1)-2(3))-2(3(2)-(-3)(3))+(3(2)-(-3)(1))\\ & =-2(2-6)-2(6+9)+(6+3)\\ & =8-30+9\\ & =-13\end{array}$$

Find the matrix of cofactors of $A$.

$\text{This is left as an exercise to the reader.}$

Find the adjoint of $A$ (the adjoint of a matrix is the transpose of the matrix containing the cofactors of the matrix.) $\text{This is left as an exercise to the reader.}$

Find the inverse of $A$.

$$\begin{array}{rl}{A}^{-1}& =\frac{1}{\det A}\text{adj}A\\ & =\frac{1}{-13}\left[\begin{array}{ccc}-4& -2& 5\\ -15& -1& 9\\ 9& -2& -8\end{array}\right]\\ & \text{The rest of this is left as an exercise to the reader.}\end{array}$$

Problem 4

Find the volume of the region.

$$⟨\left[\begin{array}{c}-4\\ 9\\ -3\end{array}\right],\left[\begin{array}{c}-6\\ 1\\ -3\end{array}\right],\left[\begin{array}{c}-1\\ 4\\ -2\end{array}\right]⟩$$

Recall that for a region defined by vectors, the volume of that region is the absolute value of the determinant of the matrix with those vectors as column vectors.

$\det (A)=-52\text{, thus the volume }V\text{ is equal to }52\text{.}$

Problem 5

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