\[\newcommand{\arr}[1]{\underline{\underline{#1}}}\]

\[\newcommand{\vec}[1]{\underline{#1}}\]

\[\require{mhchem}\]

Using matrix inverses, and matrix determinants

Using the inverse of A to solve a system of linear equations

• If you have the inverse of a matrix \(\arr{A}\) representative of a linear system of equations, you can use it to find \(\vec{x}\):

\[\begin{align*} \arr{A}\vec{x}=\vec{b}\\ \arr{A}^{-1}\arr{A}\vec{x}=\arr{A}^{-1}\vec{b}\\ \vec{x}=\arr{A}^{-1}\vec{b} \end{align*}\]

now \(\vec{x}\) is just the product.

• Fun fact: An n\(\times\)n system \(\arr{A}\vec{x}=\vec{b}\) has a unique solution if and only if \(\arr{A}\) is nonsingular (i.e. has an inverse). *Other properties:

\[\begin{align*} (\arr{A}\arr{C})^{-1}=\arr{C}^{-1}\arr{A}^{-1}\\ (\arr{A}\arr{B}\arr{C})^{-1}=\arr{C}^{-1}\arr{B}^{-1}\arr{A}^{-1} \end{align*}\]

• Be careful:
  if \(\arr{A}=\begin{bmatrix}1&0&-2\\ 2&-1&3\\ 4&0&2\end{bmatrix}\) then \(\arr{A}^{-1}=\begin{bmatrix}2/10&0&2/10\\ -8/10&-1&7/10\\ \bf{-4/10}&\bf{0}&\bf{1/10}\end{bmatrix}\)

if \(\arr{B}=\begin{bmatrix}1&0&-2\\ 4&0&2\\ 2&-1&3\end{bmatrix}\)(switched rows), then \(\arr{B}^{-1}\neq\begin{bmatrix}2/10&0&2/10\\ \bf{-4/10}&\bf{0}&\bf{1/10}\\ -8/10&-1&7/10\end{bmatrix}\) but \(\arr{B}^{-1}=\begin{bmatrix}2/10&2/10&0\\ -8/10&7/10&-1\\ -4/10&1/10&0\end{bmatrix}\)

Notice the columns were switched here.

Determinants

• Every square matrix is associated with a real number (a scalar) called the determinant.

• The value of the determinant will tell us whether or not the matrix is singular.

• Generalized definition:

(4)\[\begin{equation} D = |\arr{A}| = \det(\arr{A})=\begin{cases} a_{11} & \text{if $n=1$}.\\ a_{11}A_{11} + a_{12}A_{12} + ... + a_{1n}A_{1n} & \text{if n>1}. \end{cases} \end{equation}\]

where \(A_{1k} = (-1)^{1+k} \det(\arr{M_{1k}})\), for \(k=1,2,...n\)
Here, \(A_{1k}\) is a “cofactor” and \(\det(\arr{M_{1k}})\) is the “minor” of \(a_{1k}\).
Also, \(\arr{M_{1k}}\) = submatrix formed from \(\arr{A}\) when deleting row \(1\) and column \(k\).

OK, so this makes no sense without some examples. Let’s consider matrices of different sizes.

Case 1: 1\(\times\)1 Matrix

\(|A| = a_{11}=a\)

• \(\arr{A}\) is non-singular if \(a\neq0\) and singular if \(a=0\).
  Ex: \(\arr{A}=[5] \hspace{0.5cm} \rightarrow{} \det(\arr{A})=5\) (non singular)
  \(\hspace{0.6cm} \arr{B}=[0] \hspace{0.5cm} \rightarrow{} \det(\arr{B})=0\) (singular)

Let’s think about a physical system \(ax=5\). If \(a\) is zero, we have a problem! If it’s non-zero, we can solve.

Case 2: 2\(\times\)2 matrix

Let \(\arr{A} = \begin{bmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix}\)

• \(\arr{A}\) will be non-singular only if we can convert it to \(\arr{I}\) using elementary row operations. This property is called “row equivalence”.

• Let’s consider how we would convert \(\arr{A}\) to \(\arr{I}\).
  If \(a_{11}\neq0\) then we can perform the following row operations:

1. Multiply the second row of \(\arr{A}\) by \(a_{11}\) \((R_2=a_{11}R_2)\)

\[\begin{align*} \arr{A}=\begin{bmatrix}a_{11} & a_{12}\\ a_{11}a_{21} & a_{11}a_{22}\end{bmatrix} \end{align*}\]

1. Subtract \(a_{21} \times \)Row 1 from Row 2 \((R_2 = R_2 - a_{21} R_1)\)\

\[\begin{align*} \arr{A}=\begin{bmatrix}a_{11} & a_{12}\\ 0 & a_{11}a_{22} - a_{12}a_{21}\end{bmatrix} \end{align*}\]

• Since \(a_{11}\neq0\), \(\arr{A}\) will be row equivalent to \(\arr{I}\) IFF:\

\[\begin{align*} a_{11}a_{22}-a_{12}a_{21} \neq 0 \hspace{2cm} (*) \end{align*}\]

• If \(a_{11}=0\), then we can switch the two rows:

\[\begin{align*} \arr{A} = \begin{bmatrix} a_{21} & a_{22}\\ 0 & a_{12} \end{bmatrix} \end{align*}\]

• Now, \(\arr{A}\) will be row equivalent to \(\arr{I}\) if \(a_{21}\neq0\) and \(a_{12}\neq0\)

• This is the same as saying \(a_{11}a_{21} \neq 0\) which is equivalent to (*) for \(a_{11}=0\):

\[\begin{align*} a_{11}a_{22} - a_{12}a_{21} = a_{12}a_{21}\neq0 \end{align*}\]

• Great! These cases of \(a_{11}\neq0\) and \(a_{11}=0\) are consistent and we can define
  \(det(\arr{A}) = |\arr{A}| = a_{11}a_{22} - a_{12} a_{21}\)

(5)\[\begin{equation} \begin{cases} =0 & \text{when $\arr{A}$ is singular}\\ \neq 0 & \text{when $\arr{A}$ is non-singular} \end{cases} \end{equation}\]

Case 3: 3\(\times\) 3 Matrix

• We will not derive the determinant for a 3 \(\times\) 3 matrix in class (see text).

• How to find it?

\[\begin{align*} \det(\arr{A}) = \left|\begin{array}{} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{array}\right| \end{align*}\]

• Working with Row 1, decompose as follows:

(6)\[\begin{align} \det(\arr{A}) &= a_{11} \left|\begin{array}{} a_{22}&a_{23}\\ a_{32}&a_{33} \end{array}\right| -a_{12} \left|\begin{array}{} a_{21}&a_{23}\\ a_{31}&a_{33} \end{array}\right| + a_{13} \left|\begin{array}{} a_{21}&a_{22}\\ a_{31}&a_{33} \end{array}\right|\\ &= a_{11}(a_{22}a_{33}-a_{23}a_{32}) - a_{12}(a_{21}a_{33}-a_{23}a_{31}) + a_{13}(a_{21}a_{32}-a_{22}a_{31}) \end{align}\]

• Looks complicated but it’s just a number. Each determinant is a “minor” determinant of the submatrix formed by eliminating row \(j\) and column \(k\).\

Example

\(Ex: \arr{A} = \left|\begin{array}{} 2&5&4\\ 3&1&2\\ 5&4&6\end{array}\right|\)

What is \(\det(\arr{A})\)?

(7)\[\begin{align} \det(\arr{A}) &= 2 \left|\begin{array}{} 1&2\\ 4&6\end{array}\right| - 5 \left|\begin{array}{} 3&2\\ 5&6\end{array}\right| +4 \left|\begin{array}{} 3&1\\ 5&4\end{array}\right|\\ &=2(6-8) - 5(18-10) + 4(12-5)\\ &=-4 -40 + 28\\ \det(\arr{A})&=-16 \end{align}\]

Row choice for determinants

You don’t need to work with Row 1. You can use any column or row you’d like but must switch signs for even number rows or columns.

Ex: Choose column 2
\(\arr{A} = \left|\begin{array}{} 2&5&4\\ 3&1&2\\ 5&4&6\end{array}\right|\)

(8)\[\begin{align} \det(\arr{A}) &= -5 \left|\begin{array}{} 3&2\\ 5&6\end{array}\right| +1 \left|\begin{array}{} 2&4\\ 5&6\end{array}\right| -4 \left|\begin{array}{} 2&4\\ 3&2\end{array}\right|\\ &=-5(18-10) +(12-20) - 4(4-12)\\ &=-40 -8 + 32\\ &=-16 \end{align}\]

In-class problem

Calculate the determinant of

\[\begin{align*} \begin{bmatrix} -1&2&3\\ 0&5&3\\ 3&1&1 \end{bmatrix} \end{align*}\]

General definition of the determinant (any size matrix)

OK, so let’s revisit the general determinant definition:

(9)\[\begin{equation} D = |\arr{A}| = \det(\arr{A})=\begin{cases} a_{11} & \text{if $n=1$}.\\ a_{11}A_{11} + a_{12}A_{12} + ... + a_{1n}A_{1n} & \text{if n>1}. \end{cases} \end{equation}\]

where \(A_{1k} = (-1)^{1+k} \det(\arr{M_{1k}})\), for \(k=1,2,...n\)

• So, for a 3 \(\times\)3 matrix, this would be

(10)\[\begin{align} |\arr{A}| & = a_{11}(-1)^2\det(\arr{M_{11}}) + a_{12}(-1)^3\det(\arr{M_{12}}) + a_{13}(-1)^4\det(\arr{M_{13}})\\ & =a_{11}\det(\arr{M_{11}}) - a_{12}\det(\arr{M_{12}}) + a_{13}\det(\arr{M_{13}}) \end{align}\]

• This generalized definition is for when we work off of Row 1 only.

• We can further generalize

(11)\[\begin{align} |\arr{A}| = a_{j1}A_{j1} + a_{j2}A_{j2} + ... + a_{jn}A_{jn} && \text{for n > 1} \end{align}\]

where \(A_{jk} = (-1)^{j+k}\det(\arr{M_{jk}})\) (careful with the signs!)
\(A_{jk}\) is the “cofactor” and \(\det(\arr{M_{jk}})\) is the “minor”

• An n \(\times\) n matrix has \(n^2\) cofactors and \(n^2\) minors.

(12)\[\begin{align} \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix} & \rightarrow 3 \times 3 & \text{will have 9 cofactors and 9 minors}\\ \\ \begin{array}{} A_{11}=+\det(\arr{M_{11}}) && A_{12}=-\det(\arr{M_{12}}) && A_{13}=+\det(\arr{M_{13}})\\ A_{21}=-\det(\arr{M_{21}}) && A_{22}=+\det(\arr{M_{22}}) && A_{23}=-\det(\arr{M_{23}})\\ A_{31}=+\det(\arr{M_{31}}) && A_{32}=-\det(\arr{M_{32}}) && A_{33}=+\det(\arr{M_{33}}) \end{array} \end{align}\]

• Note the “checker board” of the signs:
  \begin{array}{}

• & - & +\

• & + & -\

• & - & + \end{array}

• In general, working with Row 1 is easiest, but the super-generalized formula (allowing you to work with any row or column) can be useful depending on the matrix.
  Ex: \begin{bmatrix} 9 & 2 & \bf{4}\ 5 & 3 & \bf{0}\ 1 & 6 & \bf{0} \end{bmatrix}
  Which column or row to choose? Column 3, which has most zeros.

(13)\[\begin{align} |\arr{A}| &= a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}\\ & = a_{13} (-1)^{1+3}\det(\arr{M_{13}})\\ & = 4[(5)(6)-(3)(1)]\\ & = 108 \end{align}\]

Question: What if you have a row or column of all zeros?

\(\underline{ANS}:\) Determinant is zero. We can “choose” all-zero row or column and then all \(a_{jk}\)’s will be zero.
Makes sense right? When \(|\arr{A}|=0, \arr{A}\) is singular, or not invertible i.e. we can’t convert the matrix to \(\arr{I}\) if there is a row or column of all zeros.

Use of determinant in solving systems of equations

We will use the determinant a lot next class. Let’s clarify a couple of uses now.

We’re interested in a solution to the system \(\arr{A}\vec{x}=\vec{b}\). There are two situations:

• Non-homogeneous system of equations (\(\vec{b}\neq\vec{0}\))

  • If \(\det(\arr{A})\neq 0\) or

  • rank(\(\arr{A}\))=n, or

  • \(\arr{A}\) is invertible,

  • THEN a unique solution exists

• Homogeneous system of equations (\(\vec{b}=\vec{0}\))

  • If \(\det(\arr{A})\neq 0\)

    • There is only the trivial solution \(\vec{x}=\vec{0}\)

  • If \(\det(\arr{A})= 0\)

    • There is both the trivial solution \(\vec{x}=\vec{0}\) AND a series of non-trivial solutions

Numerical calculation of determinant

Easy! np.linalg.det

Calculate the determinant of

\[\begin{align*} \begin{bmatrix} -1&2&3\\ 0&5&3\\ 3&1&1 \end{bmatrix} \end{align*}\]

import numpy as np
A = np.array([[-1,2,3],[0,5,3],[3,1,1]])
np.linalg.det(A)

-28.99999999999999