Matrices and Systems of Equations: Practice

Dimensions are always written rows first, then columns.

$3 \text{ rows} \times 2 \text{ columns} = 3 \times 2$

So it is a $3 \times 2$ matrix.

The entry $a_{12}$ sits in row $1$, column $2$. Read across the first row to the second column.

$a_{12} = 7$

The row number comes first, the column number second.

An augmented matrix puts each equation's coefficients in the left columns and its constant to the right of the vertical bar.

$\left[\begin{array}{cc|c}a & b & c\end{array}\right] \text{ means } ax + by = c$

So the constants live in the last column, past the bar.

Read the first equation $x + 2y = 7$. Its coefficients are $1$ (for $x$) and $2$ (for $y$), and its constant is $7$.

$x + 2y = 7 \;\Rightarrow\; \left[\begin{array}{cc|c}1 & 2 & 7\end{array}\right]$

The missing number in front of $x$ counts as $1$.

A variable written with no number in front of it has coefficient $1$, since $x$ means $1x$.

$x - y = 5 \;\Rightarrow\; 1x - 1y = 5$

So the coefficient of $x$ is $1$.

The variable $y$ does not appear, so its coefficient is $0$. That $0$ still holds the $y$ column open in the matrix.

$x + z = 4 \;\Rightarrow\; 1x + 0y + 1z = 4$

The row is $\left[\begin{array}{ccc|c}1 & 0 & 1 & 4\end{array}\right]$.

The three legal operations are swapping two rows, multiplying a row by a nonzero constant, and adding a multiple of one row to another.

$R_2 \to R_2 + 2R_1$

Adding a multiple of one row to another is exactly the third operation. Multiplying by $0$, deleting a row, or collapsing a row to one number are all disallowed.

The notation replaces row $2$ with row $2$ minus $3$ copies of row $1$, entry by entry.

$R_2 \to R_2 - 3R_1$

Each entry of the new row $2$ is (old row $2$ entry) minus $3$ times the matching row $1$ entry.

Translate the row into an equation using the column labels: $0x + 1y = 6$.

$0x + 1y = 6 \;\Rightarrow\; y = 6$

The bottom row reads $0x + 0y + 1z = 5$.

$0x + 0y + 1z = 5 \;\Rightarrow\; z = 5$

This single value is the starting point for back-substitution.

The bar marks where the left-hand coefficients end and the right-hand constants begin.

$\left[\begin{array}{cc|c}a & b & c\end{array}\right] \text{ means } ax + by = c$

It stands in for the equals signs of the system.

The equation $0 = 0$ is true for every choice of the variables, so it places no constraint on them.

$0 = 0 \text{ is always true}$

It often signals a dependent system with infinitely many solutions.