Linear equations and matrices

 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

$${\delta }_{ij}=\{\begin{array}{cl}1& \text{if }i=j,\\ 0& \text{otherwise}.\end{array}$$

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. 

$$\begin{array}{rcl}{I}_1& =& (1),\\ I_2& =& \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\\ I_3& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),\end{array}$$

and so on. □

Theorem

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

$$I_{n}X=X=X{I}_m.$$

Proof. Using the definition of ${\delta }_{ij}$, we have

$$(I_{n}X{)}_{ij}=\sum _{k=1}^n{\delta }_{ik}{x}_{kj}=x_{ij}=(X{)}_{ij}$$

and

$$(X{I}_m{)}_{ij}=\sum _{k=1}^m{x}_{ik}{\delta }_{kj}=x_{ij}=(X{)}_{ij}.$$

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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 $c\ne 0$. $(0)$ has no inverse.  □

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

Consider the $2\times 2$ matrix

$$X=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right).$$

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

$$X^{-1}=\left(\begin{array}{cc}d/\mathrm{\Delta }& -b/\mathrm{\Delta }\\ -c/\mathrm{\Delta }& a/\mathrm{\Delta }\end{array}\right).$$

Exercise. Verify this claim by explicitly computing $X{X}^{-1}$. □

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

  The determinant of the $2\times 2$ matrix $X=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, 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:

$$\begin{array}{rcl}ax+by& =& p,\\ cx+dy& =& q.\end{array}$$

Using a matrix and vectors defined as

$$\begin{array}{rcl}A& =& \left(\begin{array}{cc}a& b\\ c& d\end{array}\right),\\ \mathbf{x}& =& \left(\begin{array}{c}x\\ y\end{array}\right),\\ \mathbf{p}& =& \left(\begin{array}{c}p\\ q\end{array}\right),\end{array}$$

we can summarize the above equations as

$$A\mathbf{x}=\mathbf{p}.$$

If $|A|\ne 0$, 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

$$\begin{array}{rcl}2x+y& =& 1,\\ 3x-y& =& 0.\end{array}$$

Let

$$A=\left(\begin{array}{cc}2& 1\\ 3& -1\end{array}\right),$$ so $det(A)=-5$ and $$A^{-1}=\left(\begin{array}{cc}1/5& 1/5\\ 3/5& -2/5\end{array}\right)$$ so that $$\left(\begin{array}{c}x\\ y\end{array}\right)=A^{-1}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{cc}1/5& 1/5\\ 3/5& -2/5\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}\frac{1}{5}\\ \frac{3}{5}\end{array}\right).$$

You should verify this is indeed the solution. □