9.12 Matrices tutorial

Linear Systems

A linear system of equations in \(n\) variables is a set of \(m\) equations of the form:
\(a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_n = b_1\) \(a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n = b_2\) \(\vdots\) \(a_{m1} x_1 + a_{m2} x_2 + \cdots + a_{mn} x_n = b_m\)

(S)

Where \(a_{ij}, b_i\) with \(1 \leq i \leq m, 1 \leq j \leq n\) are coefficients.

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Solution

A solution for the linear system \((S)\) is a collection of \(n\) numbers \(s_1, s_2, \dots, s_n\) such that \(a_{11} s_1 + a_{12} s_2 + \cdots + a_{1n} s_n = b_1\) \(a_{21} s_1 + a_{22} s_2 + \cdots + a_{2n} s_n = b_2\) \(\vdots\) \(a_{m1} s_1 + a_{m2} s_2 + \cdots + a_{mn} s_n = b_m\)

Example

\(s_1=-1,s_2=3\) is a solution of the linear system: \(3x_1 + 2x_2 = 3\), \(-x_1 + x_2 = 4\),

Since \(3(-1) + 2(3) = 3\), \(-(-1) + 3 = 4\).

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Some definitions

A general solution of a linear system is the set of all the solutions.

Two linear systems are equivalent if they have the same general solution.

A linear system is called consistent if it has at least one solution, and is called inconsistent if it has no solution.

Remark

Given a linear system, exactly one of the following holds: (1) The linear system has a unique solution (consistent). (2) The linear system has infinite solutions (consistent). (3) The linear system has no solution (inconsistent).

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Given a linear system

\(a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1\)​ \(a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2\), \(\vdots\), \(a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = b_m\),

we associate two matrices, the coefficient matrix: \(\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}_{m \times n}\)​

and the augmented matrix: \(\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} & b_1 \\ a_{21} & a_{22} & \cdots & a_{2n} & b_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} & b_m \end{pmatrix}_{m \times (n+1)}\)​

Elementary row operations

Given a matrix, it is possible to perform the following operations called elementary row operations:

1. Interchange two rows, \(r_i \leftrightarrow r_j\).

2. Multiply a row by a nonzero constant, \(cr_i\) with \(c \neq 0\).

3. Add a multiple from one row to another, \(cr_i + r_j\).

Example

\(\begin{pmatrix}6 & 2 & 1 & -1\\ 4 & 2 & 2 & 2\\ -2 & 2 & 1 & 7\end{pmatrix}\rightarrow r_1\leftrightarrow r_2\rightarrow\begin{pmatrix}4 & 2 & 2 & 2\\ 6 & 2 & 1 & -1\\ -2 & 2 & 1 & 7\end{pmatrix}\rightarrow\frac12r_1\\\rightarrow\begin{pmatrix}2 & 1 & 1 & 1\\ 6 & 2 & 1 & -1\\ -2 & 2 & 1 & 7\end{pmatrix}\rightarrow-3r_1+r_2\rightarrow\begin{pmatrix}2 & 1 & 1 & 1\\ 0 & -1 & -2 & -4\\ -2 & 2 & 1 & 7\end{pmatrix}\)​

Remark

The importance of elementary row operations is that when applied to the augmented matrix of a linear system, they produce an equivalent system.

Leading entry

The first non-zero element in each row in a matrix is called the leading entry.

Row echelon form

A matrix is said to be in row echelon form if it satisfies: (1) Every row of zeros is at the bottom. (2) The leading entry in each non-zero row is located in a column to the left of the leading entry below it.

Example:

\[ A = \begin{pmatrix} 4 & 3 & 2 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 1 \end{pmatrix}, B = \begin{pmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, C = \begin{pmatrix} 0 & 2 & 1 \\ 1 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix} \]

Then \(A\) and \(B\) are in row echelon form but \(C\) is not.

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Gauss elimination method for solving linear systems

A linear system can be solved by following these steps:

(1) Write the augmented matrix of the linear system.

(2) Use elementary row operations to convert the augmented matrix to a matrix in row echelon form.

(3) Use backward substitution to solve the linear system corresponding to the matrix in row echelon form (which is equivalent to the initial system).

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Example

Solve the following linear system using the Gauss elimination method.

\(\begin{aligned}x_2 - x_3 &= 3 \\ -2x_1 + 4x_2 - x_3 &= 1 \\ -2x_1 + 5x_2 - 4x_3 &= -2\end{aligned}\)

Solution:

\(\begin{pmatrix} 0 & 1 & -1 & 3 \\ -2 & 4 & -1 & 1 \\ -2 & 5 & -4 & -2 \end{pmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{pmatrix} -2 & 4 & -1 & 1 \\ 0 & 1 & -1 & 3 \\ -2 & 5 & -4 & -2 \end{pmatrix} \xrightarrow{-R_1 + R_3} \begin{pmatrix} -2 & 4 & -1 & 1 \\ 0 & 1 & -1 & 3 \\ 0 & 1 & -3 & -3 \end{pmatrix} \xrightarrow{-R_2 + R_3} \begin{pmatrix} -2 & 4 & -1 & 1 \\ 0 & 1 & -1 & 3 \\ 0 & 0 & -2 & -6 \end{pmatrix}\)

\(\begin{aligned} -2x_1 + 4x_2 - x_3 &= 1 \ (E1) \\ x_2 - x_3 &= 3 \ (E2) \\ -2x_3 &= -6 \ (E3) \end{aligned}\)

From \((E_3)\) we have \(x_3 = 3\). Substituting \(x_3 = 3\) in \((E_2)\) we have \(x_2 - 3 = 3 \implies x_2 = 6\).

Substituting \(x_2 = 6\) and \(x_3 = 3\) in \((E_1)\) we have \(-2x_1 + 24 - 3 = 1 \implies -2x_1 = -20 \implies x_1 = 10\).

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One more

\(\begin{aligned} x_1 + 2x_2 &= -7 \\ -x_1 - x_2 &= 1 \\ 2x_1 + x_2 &= 5 \end{aligned}\)

Solution:

\(\begin{pmatrix} 1 & 2 & -7 \\ -1 & -1 & 1 \\ 2 & 1 & 5 \end{pmatrix} \xrightarrow{R_1 + R_2, \ -2R_1 + R_3} \begin{pmatrix} 1 & 2 & -7 \\ 0 & 1 & -6 \\ 0 & -3 & 19 \end{pmatrix} \xrightarrow{3R_2 + R_3} \begin{pmatrix} 1 & 2 & -7 \\ 0 & 1 & -6 \\ 0 & 0 & 1 \end{pmatrix}\)

From the last row we have the equation \(0 = 0x_1 + 0x_2 = 1\) which is not possible, and therefore the linear system has no solution.

Theorem

A linear system is consistent, if and only if, the row echelon form of the augmented matrix has no row of the form \((0 \ 0 \ \dots \ 0 \ \ c)\) with \(c \neq 0\).

Upper and lower triangular matrix

A matrix \(A = (a_{ij})_{n \times n}\) is called an upper triangular matrix if \(a_{ij} = 0\) for \(i > j\), that is, if all entries below the main diagonal are 0.

\(A = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ 0 & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_{nn} \end{pmatrix}\)

A matrix \(A = (a_{ij})_{n \times n}\) is called a lower triangular matrix if \(a_{ij} = 0\) for \(i < j\), that is, if all entries above the main diagonal are 0.

\(A = \begin{pmatrix} a_{11} & 0 & \dots & 0 \\ a_{21} & a_{22} & \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{pmatrix}\)

1. Find an upper triangular matrix \(A\) such that \(A^3 = \begin{pmatrix} 8 & -57 \\ 0 & 27 \end{pmatrix}\)

   \(A=\begin{pmatrix}2 & -3\\ 0 & 3\end{pmatrix}\)​

2. Let \(A = \begin{pmatrix} -5 & -3 \\ 2 & 1 \end{pmatrix}\). Find \(A^{-1}\), \((A^T)^{-1}\) and \((7A)^{-1}\).

   \(A^{-1}=\begin{pmatrix}1 & 3\\ -2 & -5\end{pmatrix}\), \(\left(A^{T}\right)^{-1}=\begin{pmatrix}1 & -2\\ 3 & -5\end{pmatrix}\).

   \((7A)^{-1}=\begin{pmatrix}\frac17 & \frac37\\ -\frac27 & -\frac57\end{pmatrix}\)​

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3. Let \(A\) and \(B\) be square matrices of the same size. Show that \(AB = BA\), if and only if, \((A + B)(A - B) = A^2 - B^2\).

   \(\Rightarrow\) Suppose that \(AB = BA\), then:
   \((A + B)(A - B) = A^2 - AB + BA - B^2 = A^2 - B^2\)

   \(\Leftarrow\) Suppose that \((A + B)(A - B) = A^2 - B^2\), then:
   ​\(A^2 - AB + BA - B^2 = A^2 - B^2\)
   ​\(-AB + BA = 0\)
   ​\(AB = BA\)

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4. Let \(A\) and \(B\) be square matrices of the same size. Show that \((AB)^T + \left[ A(B^T - B) \right]^T = BA^T\).

   \(B^{T}A^{T}+(B^{T}-B)^{T}A^{T}\\=B^{T}A^{T}+BA^{T}-B^{T}A^{T}\\=BA^{T}\)

5. A matrix \(A \in M_{n \times n}(\mathbb{K})\) is called orthogonal if \(A^T = A^{-1}\). Prove that the matrix \(A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\) is orthogonal.

   \(A^T = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}, \quad A^{-1} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\)

6. A matrix \(A \in M_{n \times n}(\mathbb{K})\) is called idempotent if \(A^2 = A\). If \(A = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\), prove that \(A\) and \(I_2 - A\) are idempotent.

   \(A^2 = \frac{1}{4} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \checkmark\)

   \(I_2-A=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}-\frac12\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}=\frac12\begin{pmatrix}1 & -1\\ -1 & 1\end{pmatrix}\checkmark\)

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