Row operations and linear models

What you will be able to do

• use the three elementary row operations;
• find an inverse by turning $[A \,|\, I]$ into $[I \,|\, A^{-1}]$;
• read the three cases of a system from the reduced augmented matrix;
• set up a real situation, such as traffic or an economy, as a system and solve it.

Elementary row operations

There are three elementary row operations:

1. Swap two rows: $R_i \leftrightarrow R_j$.
2. Scale a row by a non-zero number: $R_i \to kR_i$.
3. Add a multiple of one row to another: $R_i \to R_i + kR_j$.

Each operation can be undone by another one, so none of them changes the solutions of a system of equations. Row operations are the steps you use when you eliminate unknowns by hand.

The inverse by row operations

Write $A$ next to the identity matrix: $[A \,|\, I]$. Now use row operations until the left half becomes $I$. The same operations turn the right half into $A^{-1}$:

Why does it work? Each row operation is a multiplication on the left by some matrix. All the operations together multiply $A$ by one matrix $E$ so that $EA = I$. Then $E = A^{-1}$, and since $E$ also multiplied $I$, the right half is $EI = A^{-1}$.

Example with $2 \times 2$. Find the inverse of $A =

$.

So $A^{-1} =

$, the same as in the last lesson.

Example with $3 \times 3$. Take $A =

$. Work column by column and clear it.

1. Column 1. $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ give rows $(0, 1, 2 \,|\, -1, 1, 0)$ and $(0, 2, 5 \,|\, -1, 0, 1)$.
2. Column 2. $R_1 \to R_1 - R_2$ gives $(1, 0, -1 \,|\, 2, -1, 0)$. $R_3 \to R_3 - 2R_2$ gives $(0, 0, 1 \,|\, 1, -2, 1)$.
3. Column 3. $R_1 \to R_1 + R_3$ gives $(1, 0, 0 \,|\, 3, -3, 1)$. $R_2 \to R_2 - 2R_3$ gives $(0, 1, 0 \,|\, -3, 5, -2)$.

The result is $A^{-1} =

$. Multiply$A$by it to check that you get$I$.

If a row of zeros appears in the left half, $A$ is singular and has no inverse. Stop: the work has told you that $|A| = 0$.

Solving a system by row operations

Write the augmented matrix $[A \,|\, B]$ and reduce the left half as far as you can.

• If you can reach $[I \,|\, X]$, the system has the unique solution $X$.
• If a row becomes $(0\ 0\ 0 \,|\, c)$ with $c \ne 0$, the system is inconsistent: no solution.
• If a row becomes $(0\ 0\ 0 \,|\, 0)$ and there is no contradiction, there are infinitely many solutions.

For $x + y + z = 6$, $x + 2y + 3z = 14$, $2x + 3y + 4z = 21$, the operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - 2R_1$, then $R_3 \to R_3 - R_2$, give a last row $(0\ 0\ 0 \,|\, 1)$. The system has no solution. With $20$ in place of $21$, the last row is $(0\ 0\ 0 \,|\, 0)$ and there are infinitely many.

Linear models

An economy. Two industries, say steel and power, each need some of their own output and some of the other's. Let $C$ hold these amounts: $c_{ij}$ is the amount of industry $i$ used to make one unit of industry $j$. If the outside demand is $d$ and the industries produce $x$, then production minus the amount used up equals the demand:

For $C =

$and$d = (30, 60)$, the equations are$0.8x_1 - 0.3x_2 = 30$and$-0.4x_1 + 0.9x_2 = 60$. The determinant is$0.72 - 0.12 = 0.6$, and the solution is$x_1 = 75$and$x_2 = 100$. Check:$0.8 \cdot 75 - 0.3 \cdot 100 = 30$.

A traffic network. At a junction, the cars that arrive equal the cars that leave. Each junction gives one linear equation. The unknowns are the flows on the roads. A problem below sets up three junctions.

A worked example

Use row operations to find the inverse of $A =

$.

1. Start. $\left[\begin{array}{cc|cc} 1 & 3 & 1 & 0 \\ 2 & 7 & 0 & 1 \end{array}\right]$.
2. Clear column 1. $R_2 \to R_2 - 2R_1$ gives $\left[\begin{array}{cc|cc} 1 & 3 & 1 & 0 \\ 0 & 1 & -2 & 1 \end{array}\right]$.
3. Clear column 2. $R_1 \to R_1 - 3R_2$ gives $\left[\begin{array}{cc|cc} 1 & 0 & 7 & -3 \\ 0 & 1 & -2 & 1 \end{array}\right]$.
4. Read the inverse. $A^{-1} =

$. Check with the$2 \times 2$rule:$|A| = 7 - 6 = 1$, and the adjoint is$

$.

Practise with new numbers

Try it in Python

The program keeps the augmented matrix $[A \,|\, I]$ as exact fractions. It finds a pivot in each column, scales it to $1$, clears the rest of the column and prints the matrix after each column. The last print shows the inverse.

Common mistakes

• Applying a row operation to only one half. The operation acts on the whole row of $[A \,|\, I]$.
• Scaling a row by $0$. The scaling number must not be $0$.
• Clearing a column and spoiling the one before. Work column by column, using the pivot row.
• Reading the wrong half. The inverse is the right half once the left half is $I$.
• Ignoring a row of zeros. It means the matrix is singular, or the system has a free unknown.

Remember these

• The three row operations: swap, scale by a non-zero number, add a multiple of another row.
• $[A \,|\, I] \to [I \,|\, A^{-1}]$.
• On $[A \,|\, B]$: a row $(0\ 0\ 0 \,|\, c)$ with $c \ne 0$ means no solution.
• Models: junction flows give $Ax = b$, and an economy gives $x = (I - C)^{-1}d$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Apply $R_2 \to R_2 - 3R_1$ to $\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 3 & 7 & 0 & 1 \end{array}\right]$. Write the new row 2.

2. Continue the last exercise with $R_1 \to R_1 - 2R_2$ and read off the inverse.

$.

3. After row operations an augmented matrix has the last row $(0\ 0\ 0 \,|\, 5)$. What does this say?

4. After row operations an augmented matrix has the last row $(0\ 0\ 0 \,|\, 0)$ and no contradiction. What does this say?

5. Three junctions give $x + y = 5$, $y + z = 7$ and $x + z = 6$. Add all three equations and find $x + y + z$.

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 3 (Matrices), the section on elementary operations and the inverse of a matrix.
• OpenStax, Precalculus 2e, Section 9.6 (Solving Systems with Gaussian Elimination).