Linear equations and matrices

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 We see that simultaneous linear equations can be concisely expressed by using a matrix. Furthermore, we can find a general way to solve such equations based on matrix algebra. Here, we mostly deal with \(2\times 2\) matrices, but the same principle applies to matrices of any size.



Definition (Kronecker's delta)

Kronecker's delta is the symbol \(\delta_{ij}\) such that

\[\delta_{ij} = \left\{ \begin{array}{cl} 1 & \text{if $i = j$},\\ 0 & \text{otherwise}. \end{array}\right. \]

Definition (Identity matrix)

The \(n\times n\) identity matrix is \(I_n = (\delta_{ij})\).

The identity matrices have their diagonal elements that are all 1 and off-diagonal elements that are all 0.

Example. 

\[\begin{eqnarray} I_1 &=& (1),\\ I_2 &=& \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix},\\ I_3 &=& \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}, \end{eqnarray} \]

and so on. □

Theorem

Let \(X=(x_{ij})\) be an \(n\times m\) matrix. Then

\[I_nX = X = XI_m.\]

Proof. Using the definition of \(\delta_{ij}\), we have

\[(I_nX)_{ij} = \sum_{k=1}^{n}\delta_{ik}x_{kj} = x_{ij} = (X)_{ij}\]

and

\[(XI_m)_{ij} = \sum_{k=1}^{m}x_{ik}\delta_{kj} = x_{ij} = (X)_{ij}.\]

■

Definition (Inverse matrix)

Let \(X \in M_n\). The matrix \(Y \in M_n\) is said to be an inverse matrix of \(X\) if \(XY = I_n\). A matrix that has an inverse is said to be invertible.

It will be proved that an inverse matrix \(Y\) of \(X\) is unique. That is, if there were two inverse matrices of \(X\), say \(Y\) and \(Z\), then \(Y = Z\). Based on this fact, we write \(X^{-1}\) to denote the inverse matrix of \(X\). Furthermore, it turns out that \(X^{-1}\) is the unique matrix such that \(X^{-1}X = I_n\).

Example. The inverse of \((c)\in M_1\) is \((1/c)\) if \(c\neq 0\). \((0)\) has no inverse.  □

Example. Consider \(M_2\). The ``zero'' matrix \(\begin{pmatrix}0 & 0\\0&0\end{pmatrix}\) has no inverse. (Why?) □

Consider the \(2\times 2\) matrix

\[X = \begin{pmatrix} a & b\\c&d\end{pmatrix}.\]

We want to find its inverse. Let \[\Delta  = ad-bc.\] If \(\Delta = 0\), \(X\) is not invertible (i.e., \(X^{-1}\) does not exist). If \(\Delta \neq 0\), then \(X\) is invertible and

\[X^{-1} = \begin{pmatrix} d/\Delta & -b/\Delta\\ -c/\Delta & a/\Delta \end{pmatrix}. \]

Exercise. Verify this claim by explicitly computing \(XX^{-1}\). □

Definition (Determinant (\(2\times 2\)))

  The determinant of the \(2\times 2\) matrix \(X = \begin{pmatrix}a&b\\c& d\end{pmatrix}\), denoted as \(\det(X)\) or \(|X|\), is defined by

    \[\det (X) = ad -bc.\]

We will study more general determinants in detail in later posts.

Consider the following simultaneous linear equations:

\[ \begin{eqnarray} ax + by &=& p,\\ cx + dy &=& q.\end{eqnarray} \]

Using a matrix and vectors defined as

\[\begin{eqnarray} A &=&\begin{pmatrix} a & b\\ c& d\end{pmatrix},\\ \mathbf{x} &=&\begin{pmatrix}x\\y\end{pmatrix},\\ \mathbf{p} &=&\begin{pmatrix}p\\q\end{pmatrix},\end{eqnarray} \]

we can summarize the above equations as

\[A\mathbf{x} = \mathbf{p}.\]

If \(|A| \neq 0\), multiplying \(A^{-1}\) on the both sides from the left, we have

\[A^{-1}(A\mathbf{x}) = A^{-1}\mathbf{p}.\] But \(A^{-1}(A\mathbf{x})= (A^{-1}A)\mathbf{x} = I_2\mathbf{x} = \mathbf{x}\) so that \[\mathbf{x} = A^{-1}\mathbf{p},\]

and we have just solved the linear equations.

Example. Consider solving

\[\begin{eqnarray} 2x + y &=& 1,\\ 3x - y &=& 0. \end{eqnarray}\]

Let

\[A = \begin{pmatrix} 2 & 1\\ 3 & -1 \end{pmatrix},\] so \(\det(A) = -5\) and \[A^{-1} = \begin{pmatrix} 1/5 & 1/5\\ 3/5 & -2/5 \end{pmatrix} \] so that \[\begin{pmatrix} x\\ y \end{pmatrix} = A^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1/5 & 1/5\\ 3/5 & -2/5 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{1}{5}\\ \frac{3}{5} \end{pmatrix}.\]
You should verify this is indeed the solution. □


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