8.3 Matrix Arithmetic

In Section, we used a special class of matrices, the augmented matrices, to assist us in solving systems of linear equations. In this section, we study matrices as mathematical objects of their own accord, temporarily divorced from systems of linear equations. To do so conveniently requires some more notation. When we write $A={\left[a_{ij}\right]}_{m\times n}$, we mean $A$ is an $m$ by $n$ matrix¹ and $a_{ij}$ is the entry found in the $i$th row and $j$th column. Schematically, we have

$$\begin{array}{ccc}\hfill \hfill & \hfill \hfill & \hfill \text{begin{tabular}c\$j\$countscolumns\hfill \\ \hfill fromlefttoright\mathrm{\backslash end\{tabular\}}\}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \stackrel{}{\to }\hfill & \hfill \hfill & \hfill \hfill \\ \hfill A\hfill & \hfill =\hfill & \hfill \left[\begin{array}{cccc}\hfill a_{\text{tiny\$1\$}\text{tiny\$1\$}}& \hfill a_{\text{tiny\$1\$}\text{tiny\$2\$}}& \hfill \cdots & \hfill a_{\text{tiny\$1\$}\text{tiny\$n\$}}\\ \hfill a_{\text{tiny\$2\$}\text{tiny\$1\$}}& \hfill a_{\text{tiny\$2\$}\text{tiny\$2\$}}& \hfill \cdots & \hfill a_{\text{tiny\$2\$}n}\\ \hfill ⋮& \hfill ⋮& \hfill & \hfill ⋮\\ \hfill a_{m\text{tiny\$1\$}}& \hfill a_{m\text{tiny\$2\$}}& \hfill \cdots & \hfill a_{mn}\end{array}\right]\downarrow \hfill & \hfill \text{begin{tabular}c\$i\$countsrows\hfill \\ \hfill fromtoptobottom\mathrm{\backslash end\{tabular\}}\}\hfill \end{array}$$

With this new notation we can define what it means for two matrices to be equal.

Essentially, two matrices are equal if they are the same size and they have the same numbers in the same spots.² For example, the two $2\times 3$ matrices below are, despite appearances, equal.

$$\left[\begin{array}{ccc}\hfill 0& \hfill -2& \hfill 9\\ \hfill 25& \hfill 117& \hfill -3\end{array}\right]=\left[\begin{array}{ccc}\hfill \ln (1)& \hfill \sqrt[3]{-8}& \hfill e^{2\ln (3)}\\ \hfill {125}^{2/3}& \hfill 3^2\cdot 13& \hfill \log (0.001)\end{array}\right]$$

Now that we have an agreed upon understanding of what it means for two matrices to equal each other, we may begin defining arithmetic operations on matrices. Our first operation is addition.

As an example, consider the sum below.

$$\left[\begin{array}{cc}\hfill 2& \hfill 3\\ \hfill 4& \hfill -1\\ \hfill 0& \hfill -7\end{array}\right]+\left[\begin{array}{cc}\hfill -1& \hfill 4\\ \hfill -5& \hfill -3\\ \hfill 8& \hfill 1\end{array}\right]=\left[\begin{array}{cc}\hfill 2+(-1)& \hfill 3+4\\ \hfill 4+(-5)& \hfill (-1)+(-3)\\ \hfill 0+8& \hfill (-7)+1\end{array}\right]=\left[\begin{array}{cc}\hfill 1& \hfill 7\\ \hfill -1& \hfill -4\\ \hfill 8& \hfill -6\end{array}\right]$$

It is worth the reader's time to think what would have happened had we reversed the order of the summands above. As we would expect, we arrive at the same answer. In general, $A+B=B+A$ for matrices $A$ and $B$, provided they are the same size so that the sum is defined in the first place. This is the commutative property of matrix addition. To see why this is true in general, we appeal to the definition of matrix addition. Given $A={\left[a_{ij}\right]}_{m\times n}$ and $B={\left[b_{ij}\right]}_{m\times n}$,

$$A+B={\left[a_{ij}\right]}_{m\times n}+{\left[b_{ij}\right]}_{m\times n}={\left[a_{ij}+b_{ij}\right]}_{m\times n}={\left[b_{ij}+a_{ij}\right]}_{m\times n}={\left[b_{ij}\right]}_{m\times n}+{\left[a_{ij}\right]}_{m\times n}=B+A$$

where the second equality is the definition of $A+B$, the third equality holds by the commutative law of real number addition, and the fourth equality is the definition of $B+A$. In other words, matrix addition is commutative because real number addition is. A similar argument shows the associative property of matrix addition also holds, inherited in turn from the associative law of real number addition. Specifically, for matrices $A$, $B$, and $C$ of the same size, $(A+B)+C=A+(B+C)$. In other words, when adding more than two matrices, it doesn't matter how they are grouped. This means that we can write $A+B+C$ without parentheses and there is no ambiguity as to what this means.³ These properties and more are summarized in the following theorem.

The identity property is easily verified by resorting to the definition of matrix addition; just as the number $0$ is the additive identity for real numbers, the matrix comprised of all $0$'s does the same job for matrices. To establish the inverse property, given a matrix $A={\left[a_{ij}\right]}_{m\times n}$, we are looking for a matrix $B={\left[b_{ij}\right]}_{m\times n}$ so that $A+B=0_{m\times n}$. By the definition of matrix addition, we must have that $a_{ij}+b_{ij}=0$ for all $i$ and $j$. Solving, we get $b_{ij}=-a_{ij}$. Hence, given a matrix $A$, its additive inverse, which we call $-A$, does exist and is unique and, moreover, is given by the formula: $-A={\left[-a_{ij}\right]}_{m\times n}$. The long and short of this is: to get the additive inverse of a matrix, take additive inverses of each of its entries. With the concept of additive inverse well in hand, we may now discuss what is meant by subtracting matrices. You may remember from arithmetic that $a-b=a+(-b)$; that is, subtraction is defined as `adding the opposite (inverse).' We extend this concept to matrices. For two matrices $A$ and $B$ of the same size, we define $A-B=A+(-B)$. At the level of entries, this amounts to

$$A-B=A+(-B)={\left[a_{ij}\right]}_{m\times n}+{\left[-b_{ij}\right]}_{m\times n}={\left[a_{ij}+\left(-b_{ij}\right)\right]}_{m\times n}={\left[a_{ij}-b_{ij}\right]}_{m\times n}$$

Thus to subtract two matrices of equal size, we subtract their corresponding entries. Surprised?

Our next task is to define what it means to multiply a matrix by a real number. Thinking back to arithmetic, you may recall that multiplication, at least by a natural number, can be thought of as `rapid addition.' For example, $2+2+2=3\cdot 2$. We know from algebra⁴ that $3x=x+x+x$, so it seems natural that given a matrix $A$, we define $3A=A+A+A$. If $A={\left[a_{ij}\right]}_{m\times n}$, we have

$$3A=A+A+A={\left[a_{ij}\right]}_{m\times n}+{\left[a_{ij}\right]}_{m\times n}+{\left[a_{ij}\right]}_{m\times n}={\left[a_{ij}+a_{ij}+a_{ij}\right]}_{m\times n}={\left[3a_{ij}\right]}_{m\times n}$$

In other words, multiplying the matrix in this fashion by $3$ is the same as multiplying each entry by $3$. This leads us to the following definition.

One may well wonder why the word `scalar' is used for `real number.' It has everything to do with `scaling' factors.⁶ A point $P(x,y)$ in the plane can be represented by its position matrix, $P$:

$$(x,y)\leftrightarrow P=\left[\begin{array}{c}\hfill x\\ \hfill y\end{array}\right]$$

Suppose we take the point $(-2,1)$ and multiply its position matrix by $3$. We have

$$3P=3\left[\begin{array}{c}\hfill -2\\ \hfill 1\end{array}\right]=\left[\begin{array}{c}\hfill 3(-2)\\ \hfill 3(1)\end{array}\right]=\left[\begin{array}{c}\hfill -6\\ \hfill 3\end{array}\right]$$

which corresponds to the point $(-6,3)$. We can imagine taking $(-2,1)$ to $(-6,3)$ in this fashion as a dilation by a factor of $3$ in both the horizontal and vertical directions. Doing this to all points $(x,y)$ in the plane, therefore, has the effect of magnifying (scaling) the plane by a factor of $3$.

As did matrix addition, scalar multiplication inherits many properties from real number arithmetic. Below we summarize these properties.

As with the other results in this section, Theorem can be proved using the definitions of scalar multiplication and matrix addition. For example, to prove that $k(A+B)=kA+kB$ for a scalar $k$ and $m\times n$ matrices $A$ and $B$, we start by adding $A$ and $B$, then multiplying by $k$ and seeing how that compares with the sum of $kA$ and $kB$.

$$k(A+B)=k\left({\left[a_{ij}\right]}_{m\times n}+{\left[b_{ij}\right]}_{m\times n}\right)=k{\left[a_{ij}+b_{ij}\right]}_{m\times n}={\left[k\left(a_{ij}+b_{ij}\right)\right]}_{m\times n}={\left[k{a}_{ij}+k{b}_{ij}\right]}_{m\times n}$$

As for $kA+kB$, we have

$$kA+kB=k{\left[a_{ij}\right]}_{m\times n}+k{\left[b_{ij}\right]}_{m\times n}={\left[k{a}_{ij}\right]}_{m\times n}+{\left[k{b}_{ij}\right]}_{m\times n}={\left[k{a}_{ij}+k{b}_{ij}\right]}_{m\times n}\mathrm{✓}$$

which establishes the property. The remaining properties are left to the reader. The properties in Theorems and establish an algebraic system that lets us treat matrices and scalars more or less as we would real numbers and variables, as the next example illustrates.

While the solution to the previous example is written in excruciating detail, in practice many of the steps above are omitted. We have spelled out each step in this example to encourage the reader to justify each step using the definitions and properties we have established thus far for matrix arithmetic. The reader is encouraged to solve the equation in Example Example 1 as they would any other linear equation, for example: $3a-(2+5a)=-4+\frac{1}{3}(9)$.

We now turn our attention to matrix multiplication - that is, multiplying a matrix by another matrix. Based on the `no surprises' trend so far in the section, you may expect that in order to multiply two matrices, they must be of the same size and you find the product by multiplying the corresponding entries. While this kind of product is used in other areas of mathematics,⁷ we define matrix multiplication to serve us in solving systems of linear equations. To that end, we begin by defining the product of a row and a column. We motivate the general definition with an example. Consider the two matrices $A$ and $B$ below.

$$\begin{array}{cc}A=\left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill -1\\ \hfill -10& \hfill 3& \hfill 5\end{array}\right]& B=\left[\begin{array}{cccc}\hfill 3& \hfill 1& \hfill 2& \hfill -8\\ \hfill 4& \hfill 8& \hfill -5& \hfill 9\\ \hfill 5& \hfill 0& \hfill -2& \hfill -12\end{array}\right]\end{array}$$

Let $R1$ denote the first row of $A$ and $C1$ denote the first column of $B$. To find the `product' of $R1$ with $C1$, denoted $R1\cdot C1$, we first find the product of the first entry in $R1$ and the first entry in $C1$. Next, we add to that the product of the second entry in $R1$ and the second entry in $C1$. Finally, we take that sum and we add to that the product of the last entry in $R1$ and the last entry in $C1$. Using entry notation, $$R1\cdot C1=a_{11}{b}_{11}+a_{12}{b}_{21}+a_{13}{b}_{31}=(2)(3)+(0)(4)+(-1)(5)=6+0+(-5)=1$$. We can visualize this schematically as follows

$$\left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill -1\\ \hfill -10& \hfill 3& \hfill 5\end{array}\right]\left[\begin{array}{cccc}\hfill 3& \hfill 1& \hfill 2& \hfill -8\\ \hfill 4& \hfill 8& \hfill -5& \hfill 9\\ \hfill 5& \hfill 0& \hfill -2& \hfill -12\end{array}\right]$$

$$\begin{array}{ccccc}\hfill \underbrace{\begin{array}{cc}\hfill \overrightarrow{\begin{array}{ccc}\hfill \boxed{\text{2}}\hfill & \hfill \phantom{-}0\hfill & \hfill -1\hfill \end{array}}& \begin{array}{c}\hfill \boxed{\text{3}}\hfill \\ \hfill 4\hfill \\ \hfill 5\hfill \end{array}\downarrow \hfill \end{array}}\hfill & \hfill \hfill & \hfill \underbrace{\begin{array}{cc}\hfill \overrightarrow{\begin{array}{ccc}\hfill 2\hfill & \hfill \phantom{-}\boxed{\text{0}}\hfill & \hfill -1\hfill \end{array}}& \begin{array}{c}\hfill 3\hfill \\ \hfill \boxed{\text{4}}\hfill \\ \hfill 5\hfill \end{array}\downarrow \hfill \end{array}}\hfill & \hfill \hfill & \hfill \underbrace{\begin{array}{cc}\hfill \overrightarrow{\begin{array}{ccc}\hfill 2\hfill & \hfill \phantom{-}0\hfill & \hfill \boxed{\text{\$-1\$}}\hfill \end{array}}& \begin{array}{c}\hfill 3\hfill \\ \hfill 4\hfill \\ \hfill \boxed{\text{5}}\hfill \end{array}\downarrow \hfill \end{array}}\hfill \\ \hfill a_{\text{tiny\$11\$}}{b}_{\text{tiny\$11\$}}\hfill & \hfill +\hfill & \hfill a_{\text{tiny\$12\$}}{b}_{\text{tiny\$21\$}}\hfill & \hfill +\hfill & \hfill a_{\text{tiny\$13\$}}{b}_{\text{tiny\$31\$}}\hfill \\ \hfill (2)(3)\hfill & \hfill +\hfill & \hfill (0)(4)\hfill & \hfill +\hfill & \hfill (-1)(5)\hfill \end{array}$$

To find $R2\cdot C3$ where $R2$ denotes the second row of $A$ and $C3$ denotes the third column of $B$, we proceed similarly. We start with finding the product of the first entry of $R2$ with the first entry in $C3$ then add to it the product of the second entry in $R2$ with the second entry in $C3$, and so forth. Using entry notation, we have $$R2\cdot C3=a_{21}{b}_{13}+a_{22}{b}_{23}+a_{23}{b}_{33}=(-10)(2)+(3)(-5)+(5)(-2)=-45$$. Schematically,

$$\left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill -1\\ \hfill -10& \hfill 3& \hfill 5\end{array}\right]\left[\begin{array}{cccc}\hfill 3& \hfill 1& \hfill 2& \hfill -8\\ \hfill 4& \hfill 8& \hfill -5& \hfill 9\\ \hfill 5& \hfill 0& \hfill -2& \hfill -12\end{array}\right]$$

$$\begin{array}{ccccc}\hfill \underbrace{\begin{array}{cc}\hfill \overrightarrow{\begin{array}{ccc}\hfill \boxed{\text{\$-10\$}}\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}}& \begin{array}{c}\hfill \boxed{\text{hphantom{\$-\$}}2\}\hfill \\ \hfill -5\hfill \\ \hfill -2\hfill \end{array}\downarrow \hfill \end{array}}\hfill & \hfill \hfill & \hfill \underbrace{\begin{array}{cc}\hfill \overrightarrow{\begin{array}{ccc}\hfill -10\hfill & \hfill \boxed{\text{3}}\hfill & \hfill 5\hfill \end{array}}& \begin{array}{c}\hfill \phantom{-}2\hfill \\ \hfill \boxed{\text{\$-5\$}}\hfill \\ \hfill -2\hfill \end{array}\downarrow \hfill \end{array}}\hfill & \hfill \hfill & \hfill \underbrace{\begin{array}{cc}\hfill \overrightarrow{\begin{array}{ccc}\hfill -10\hfill & \hfill 3\hfill & \hfill \boxed{\text{\$5\$}}\hfill \end{array}}& \begin{array}{c}\hfill \phantom{-}2\hfill \\ \hfill -5\hfill \\ \hfill \boxed{\text{\$-2\$}}\hfill \end{array}\downarrow \hfill \end{array}}\hfill \\ \hfill a_{\text{tiny\$21\$}}{b}_{\text{tiny\$13\$}}=(-10)(2)=-20\hfill & \hfill +\hfill & \hfill a_{\text{tiny\$22\$}}{b}_{\text{tiny\$23\$}}=(3)(-5)=-15\hfill & \hfill +\hfill & \hfill a_{\text{tiny\$23\$}}{b}_{\text{tiny\$33\$}}=(5)(-2)=-10\hfill \end{array}$$

Generalizing this process, we have the following definition.

Note that in order to multiply a row by a column, the number of entries in the row must match the number of entries in the column. We are now in the position to define matrix multiplication.

There are a number of subtleties in Definition which warrant closer inspection. First and foremost, Definition tells us that the $ij$-entry of a matrix product $AB$ is the $i$th row of $A$ times the $j$th column of $B$. In order for this to be defined, the number of entries in the rows of $A$ must match the number of entries in the columns of $B$. This means that the number of columns of $A$ must match⁸ the number of rows of $B$. In other words, to multiply $A$ times $B$, the second dimension of $A$ must match the first dimension of $B$, which is why in Definition, $A_{m\times \underset{¯}{n}}$ is being multiplied by a matrix $B_{\underset{¯}{n}\times r}$. Furthermore, the product matrix $AB$ has as many rows as $A$ and as many columns of $B$. As a result, when multiplying a matrix $A_{\underset{¯}{m}\times n}$ by a matrix $B_{n\times \underset{¯}{r}}$, the result is the matrix $A{B}_{\underset{¯}{m}\times \underset{¯}{r}}$. Returning to our example matrices below, we see that $A$ is a $2\times \underset{¯}{3}$ matrix and $B$ is a $\underset{¯}{3}\times 4$ matrix. This means that the product matrix $AB$ is defined and will be a $2\times 4$ matrix.

$$\begin{array}{cc}A=\left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill -1\\ \hfill -10& \hfill 3& \hfill 5\end{array}\right]& B=\left[\begin{array}{cccc}\hfill 3& \hfill 1& \hfill 2& \hfill -8\\ \hfill 4& \hfill 8& \hfill -5& \hfill 9\\ \hfill 5& \hfill 0& \hfill -2& \hfill -12\end{array}\right]\end{array}$$

Using $Ri$ to denote the $i$th row of $A$ and $Cj$ to denote the $j$th column of $B$, we form $AB$ according to Definition.

$$\begin{array}{ccccc}\hfill AB& =& \left[\begin{array}{cccc}\hfill R1\cdot C1& \hfill R1\cdot C2& \hfill R1\cdot C3& \hfill R1\cdot C4\\ \hfill R2\cdot C1& \hfill R2\cdot C2& \hfill R2\cdot C3& \hfill R2\cdot C4\end{array}\right]\hfill & =& \left[\begin{array}{cccc}\hfill 1& \hfill 2& \hfill 6& \hfill -4\\ \hfill 7& \hfill 14& \hfill -45& \hfill 47\end{array}\right]\hfill \end{array}$$

Note that the product $BA$ is not defined, since $B$ is a $3\times \underset{¯}{4}$ matrix while $A$ is a $\underset{¯}{2}\times 3$ matrix; $B$ has more columns than $A$ has rows, and so it is not possible to multiply a row of $B$ by a column of $A$. Even when the dimensions of $A$ and $B$ are compatible such that $AB$ and $BA$ are both defined, the product $AB$ and $BA$ aren't necessarily equal.⁹ In other words, $AB$ may not equal $BA$. Although there is no commutative property of matrix multiplication in general, several other real number properties are inherited by matrix multiplication, as illustrated in our next theorem.

The one property in Theorem which begs further investigation is, without doubt, the multiplicative identity. The entries in a matrix where $i=j$ comprise what is called the main diagonal of the matrix. The identity matrix has $1$'s along its main diagonal and $0$'s everywhere else. A few examples of the matrix $I_k$ mentioned in Theorem are given below. The reader is encouraged to see how they match the definition of the identity matrix presented there.

$$\begin{array}{ccccc}[1]& \left[\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]& \left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]& \left[\begin{array}{cccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1\end{array}\right]& \\ I_1& I_2& I_3& I_4& \end{array}$$

The identity matrix is an example of what is called a square matrix as it has the same number of rows as columns. Note that to in order to verify that the identity matrix acts as a multiplicative identity, some care must be taken depending on the order of the multiplication. For example, take the matrix $2\times 3$ matrix $A$ from earlier

$$A=\left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill -1\\ \hfill -10& \hfill 3& \hfill 5\end{array}\right]$$

In order for the product $I_{k}A$ to be defined, $k=2$; similarly, for $A{I}_k$ to be defined, $k=3$. We leave it to the reader to show $I_{2}A=A$ and $A{I}_3=A$. In other words,

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]\left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill -1\\ \hfill -10& \hfill 3& \hfill 5\end{array}\right]& =& \left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill -1\\ \hfill -10& \hfill 3& \hfill 5\end{array}\right]\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill -1\\ \hfill -10& \hfill 3& \hfill 5\end{array}\right]\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]& =& \left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill -1\\ \hfill -10& \hfill 3& \hfill 5\end{array}\right]\hfill \end{array}$$

While the proofs of the properties in Theorem are computational in nature, the notation becomes quite involved very quickly, so they are left to a course in Linear Algebra. The following example provides some practice with matrix multiplication and its properties. As usual, some valuable lessons are to be learned.

Example Example 2 illustrates some interesting features of matrix multiplication. First note that in part 1, neither $A$ nor $B$ is the zero matrix, yet the product $AB$ is the zero matrix. Hence, the the zero product property enjoyed by real numbers and scalar multiplication does not hold for matrix multiplication. Parts 2 and 3 introduce us to polynomials involving matrices. The reader is encouraged to step back and compare our expansion of the matrix product $\left(M-2I_4\right)\left(M+3I_4\right)$ in part 3 with the product $(x-2)(x+3)$ from real number algebra. The exercises explore this kind of parallel further.

As we mentioned earlier, a point $P(x,y)$ in the $xy$-plane can be represented as a $2\times 1$ position matrix. We now show that matrix multiplication can be used to rotate these points, and hence graphs of equations.

When we started this section, we mentioned that we would temporarily consider matrices as their own entities, but that the algebra developed here would ultimately allow us to solve systems of linear equations. To that end, consider the system

$$\{\begin{array}{ccc}\hfill 3x-y+z& =& 8\hfill \\ \hfill x+2y-z& =& 4\hfill \\ \hfill 2x+3y-4z& =& 10\hfill \end{array}$$

In Section, we encoded this system into the augmented matrix

$$\left[\begin{array}{cccc}\hfill 3& \hfill -1& \hfill 1& \hfill 8\\ \hfill 1& \hfill 2& \hfill -1& \hfill 4\\ \hfill 2& \hfill 3& \hfill -4& \hfill 10\end{array}\right]$$

Recall that the entries to the left of the vertical line come from the coefficients of the variables in the system, while those on the right comprise the associated constants. For that reason, we may form the coefficient matrix $A$, the unknowns matrix $X$ and the constant matrix $B$ as below

$$\begin{array}{ccc}A=\left[\begin{array}{ccc}\hfill 3& \hfill -1& \hfill 1\\ \hfill 1& \hfill 2& \hfill -1\\ \hfill 2& \hfill 3& \hfill -4\end{array}\right]& X=\left[\begin{array}{c}\hfill x\\ \hfill y\\ \hfill z\end{array}\right]& B=\left[\begin{array}{c}\hfill 8\\ \hfill 4\\ \hfill 10\end{array}\right]\end{array}$$

We now consider the matrix equation $AX=B$.

$$\begin{array}{ccc}\hfill AX& =& B\hfill \\ \hfill \left[\begin{array}{ccc}\hfill 3& \hfill -1& \hfill 1\\ \hfill 1& \hfill 2& \hfill -1\\ \hfill 2& \hfill 3& \hfill -4\end{array}\right]\left[\begin{array}{c}\hfill x\\ \hfill y\\ \hfill z\end{array}\right]& =& \left[\begin{array}{c}\hfill 8\\ \hfill 4\\ \hfill 10\end{array}\right]\hfill \\ \hfill \left[\begin{array}{ccc}\hfill 3x-y+z& & \\ \hfill x+2y-z& & \\ \hfill 2x+3y-4z& & \end{array}\right]& =& \left[\begin{array}{c}\hfill 8\\ \hfill 4\\ \hfill 10\end{array}\right]\hfill \end{array}$$

We see that finding a solution $(x,y,z)$ to the original system corresponds to finding a solution $X$ for the matrix equation $AX=B$. If we think about solving the real number equation $ax=b$, we would simply `divide' both sides by $a$. Is it possible to `divide' both sides of the matrix equation $AX=B$ by the matrix $A$? This is the central topic of Section.

Adapted from Precalculus, 3rd corrected edition, by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.