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Elementary Matrices and Inverses · Tutorial 92 of 1000

A Worked 2 by 2 Inverse by Row Reduction

Learn to compute a two-by-two matrix inverse by reducing its augmented matrix and to verify the answer by multiplication.

Beginner 9 min read

What You'll Learn

  • Set up the augmented matrix with the identity matrix on the right.
  • Apply each row operation to both halves of the augmented matrix.
  • Read the inverse from the right half once the left half is the identity.
  • Follow three exact two-by-two inverse calculations step by step.
  • Verify each proposed inverse by multiplying in both orders.
  • Recognize and correct the mistake of row-reducing only the original matrix.

Why Augment the Matrix?

For a two-by-two matrix, an inverse is another two-by-two matrix that undoes multiplication by the original matrix. Row reduction gives a systematic way to find that undoing matrix: start with the original matrix on the left and the identity matrix on the right, then apply every row operation to the whole array. When the left half becomes the identity, the right half records the inverse.

This method is useful because it computes the inverse through operations already used to solve linear systems. It also keeps track of how the rows of the original matrix must be combined to produce the identity. Throughout, matrices have real entries, and every example below reaches the identity on the left without requiring a row exchange.

Definition: A square matrix \(\mathbf{A}\) is invertible if there is a matrix \(\mathbf{B}\) of the same size such that \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}=\mathbf{I}\). Such a matrix \(\mathbf{B}\) is an inverse of \(\mathbf{A}\), written \(\mathbf{A}^{-1}\). The augmented matrix \([\mathbf{A}\mid\mathbf{I}]\) is formed by placing the identity matrix to the right of \(\mathbf{A}\).

For a two-by-two matrix, the starting layout is

$$ \left[ \begin{array}{cc|cc} a&b&1&0\\ c&d&0&1 \end{array} \right]. $$

The vertical bar separates the original matrix from the identity; it does not change the row-operation rules. Each operation must be applied to every entry in the row, on both sides of the bar. In particular, changing only the left half would lose the record of how the matrix was transformed.

Why the Right Half Is the Inverse

Theorem (Augmented-matrix inverse rule): Suppose \(\mathbf{A}\in\mathbb{R}^{2\times2}\) is invertible. If a sequence of row operations changes \([\mathbf{A}\mid\mathbf{I}_2]\) into \([\mathbf{I}_2\mid\mathbf{B}]\), then \(\mathbf{B}=\mathbf{A}^{-1}\).

Proof. Each row operation is left multiplication by an elementary matrix: the elementary matrix is obtained by performing that operation on \(\mathbf{I}_2\). Applying a sequence of row operations therefore amounts to left multiplying both halves by the same product \(\mathbf{P}\) of elementary matrices. Thus the final augmented matrix is \([\mathbf{P}\mathbf{A}\mid\mathbf{P}]\). Since it is also \([\mathbf{I}_2\mid\mathbf{B}]\), we have \(\mathbf{P}\mathbf{A}=\mathbf{I}_2\) and \(\mathbf{P}=\mathbf{B}\). Because \(\mathbf{A}\) is invertible, \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_2\). Hence \(\mathbf{P}=\mathbf{P}(\mathbf{A}\mathbf{A}^{-1})=(\mathbf{P}\mathbf{A})\mathbf{A}^{-1}=\mathbf{I}_2\mathbf{A}^{-1}=\mathbf{A}^{-1}\). Therefore \(\mathbf{B}=\mathbf{A}^{-1}\). \(\square\)

The row-reduction work is therefore not a guess about the entries of the inverse. The operations construct a matrix \(\mathbf{P}\) that takes \(\mathbf{A}\) to \(\mathbf{I}_2\), while the same operations turn the right-hand identity into \(\mathbf{P}\). The theorem identifies that accumulated matrix as the inverse. The examples also verify the result directly by multiplication.

A First Calculation

Worked Example: An Integer Inverse

Find the inverse of \(\mathbf{A}=\begin{bmatrix}2&1\\1&1\end{bmatrix}\) by row reduction.

Begin with the identity on the right. First perform \(R_2\to R_2-\tfrac12R_1\), which makes the first entry of row 2 zero. Then scale row 2 by \(2\):

$$ \left[ \begin{array}{cc|cc} 2&1&1&0\\ 1&1&0&1 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 2&1&1&0\\ 0&\tfrac12&-\tfrac12&1 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 2&1&1&0\\ 0&1&-1&2 \end{array} \right]. $$

Now use \(R_1\to R_1-R_2\) to clear the entry above the second pivot, then scale row 1 by \(\tfrac12\):

$$ \left[ \begin{array}{cc|cc} 2&1&1&0\\ 0&1&-1&2 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 2&0&2&-2\\ 0&1&-1&2 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 1&0&1&-1\\ 0&1&-1&2 \end{array} \right]. $$

The left half is \(\mathbf{I}_2\), so the right half gives \(\mathbf{A}^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\).

Check: Multiplying in both orders gives

$$ \begin{bmatrix}2&1\\1&1\end{bmatrix} \begin{bmatrix}1&-1\\-1&2\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}, \qquad \begin{bmatrix}1&-1\\-1&2\end{bmatrix} \begin{bmatrix}2&1\\1&1\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. $$

Two More Row-Reduction Patterns

Worked Example: Clearing an Integer Entry

Find the inverse of \(\mathbf{A}=\begin{bmatrix}1&2\\3&7\end{bmatrix}\). The first pivot is already \(1\), so use it to clear the entry below it.

Apply \(R_2\to R_2-3R_1\), then \(R_1\to R_1-2R_2\), using the new second row in the second operation:

$$ \left[ \begin{array}{cc|cc} 1&2&1&0\\ 3&7&0&1 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 1&2&1&0\\ 0&1&-3&1 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 1&0&7&-2\\ 0&1&-3&1 \end{array} \right]. $$

For the second operation, the right-hand row becomes \((1,0)-2(-3,1)=(7,-2)\), just as the left-hand row becomes \((1,0)\). Therefore \(\mathbf{A}^{-1}=\begin{bmatrix}7&-2\\-3&1\end{bmatrix}\).

Check: The two products are

$$ \begin{bmatrix}1&2\\3&7\end{bmatrix} \begin{bmatrix}7&-2\\-3&1\end{bmatrix} = \begin{bmatrix}7-6&-2+2\\21-21&-6+7\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}, $$ $$ \begin{bmatrix}7&-2\\-3&1\end{bmatrix} \begin{bmatrix}1&2\\3&7\end{bmatrix} = \begin{bmatrix}7-6&14-14\\-3+3&-6+7\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. $$

Worked Example: A Fraction Appears During Reduction

Find the inverse of \(\mathbf{A}=\begin{bmatrix}3&1\\2&1\end{bmatrix}\). A fraction appears in the left half during the reduction, but the row operation is valid and the final inverse has integer entries.

First use \(R_2\to R_2-\tfrac23R_1\), then scale row 2 by \(3\):

$$ \left[ \begin{array}{cc|cc} 3&1&1&0\\ 2&1&0&1 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 3&1&1&0\\ 0&\tfrac13&-\tfrac23&1 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 3&1&1&0\\ 0&1&-2&3 \end{array} \right]. $$

Next apply \(R_1\to R_1-R_2\), and scale the first row by \(\tfrac13\):

$$ \left[ \begin{array}{cc|cc} 3&1&1&0\\ 0&1&-2&3 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 3&0&3&-3\\ 0&1&-2&3 \end{array} \right] \sim \left[ \begin{array}{cc|cc} 1&0&1&-1\\ 0&1&-2&3 \end{array} \right]. $$

Thus \(\mathbf{A}^{-1}=\begin{bmatrix}1&-1\\-2&3\end{bmatrix}\). The final reduction can be checked from the row operations: the first row on the right is \((1,0)-(-2,3)=(3,-3)\), and scaling it by \(\tfrac13\) gives \((1,-1)\).

Check:

$$ \begin{bmatrix}3&1\\2&1\end{bmatrix} \begin{bmatrix}1&-1\\-2&3\end{bmatrix} = \begin{bmatrix}3-2&-3+3\\2-2&-2+3\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}, $$ $$ \begin{bmatrix}1&-1\\-2&3\end{bmatrix} \begin{bmatrix}3&1\\2&1\end{bmatrix} = \begin{bmatrix}3-2&1-1\\-6+6&-2+3\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. $$

A Reliable Workflow and a Common Mistake

The same sequence of decisions applies to each example. The specific arithmetic changes, but the bookkeeping does not: every row operation acts on both halves, and the goal is the identity on the left.

1
Set up the augmented matrix.
Write \([\mathbf{A}\mid\mathbf{I}_2]\), keeping the identity in the same row order as the original matrix.
2
Reduce the left half to the identity.
Use legal row operations, applying each one across all four columns. Keep exact fractions when they occur.
3
Read and verify the candidate.
Once the left half is \(\mathbf{I}_2\), read the right half as the candidate inverse, then multiply by \(\mathbf{A}\) to check the result.

Common mistake: row-reducing only the original matrix. For instance, start with \(\mathbf{C}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\). The operation \(R_1\to R_1-R_2\) changes \(\mathbf{C}\) to \(\mathbf{I}_2\). A wrong conclusion is that the inverse is \(\mathbf{I}_2\), simply because that is what the reduced original matrix became. But \(\mathbf{C}\mathbf{I}_2=\mathbf{C}\ne\mathbf{I}_2\), so this cannot be the inverse.

The correction is to apply the same operation to the identity in the augmented matrix:

$$ \left[ \begin{array}{cc|cc} 1&1&1&0\\ 0&1&0&1 \end{array} \right] \underset{R_1\to R_1-R_2}{\sim} \left[ \begin{array}{cc|cc} 1&0&1&-1\\ 0&1&0&1 \end{array} \right]. $$

This gives the candidate \(\mathbf{C}^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\). Directly, \(\mathbf{C}\mathbf{C}^{-1}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\) and \(\mathbf{C}^{-1}\mathbf{C}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\). The right half matters because it records the row operations; the reduced left half alone does not record their effect on the identity.

Interpreting the Result

A computed inverse has two useful checks. First, the augmented reduction must have the exact form \([\mathbf{I}_2\mid\mathbf{B}]\); a left half that is not the identity does not yet allow the right half to be read off as the inverse. Second, multiplication should return the identity. Since the definition of an inverse requires both \(\mathbf{A}\mathbf{B}=\mathbf{I}_2\) and \(\mathbf{B}\mathbf{A}=\mathbf{I}_2\), checking both orders is a particularly direct confirmation of the candidate.

Keep the row-operation record legible. A useful practice is to name each operation immediately before or after its matrix step and to use the updated rows in subsequent calculations. Fractions are not a reason to abandon exact arithmetic: as the third example shows, a fractional pivot can be scaled to \(1\), and the final entries can still be integers. The main discipline is to preserve the same operation on both sides of the augmentation bar.

Check Your Understanding

For each computation, keep the identity attached to the original matrix throughout the row reduction.

  1. Use the augmented-matrix method to find the inverse of \(\begin{bmatrix}2&1\\1&2\end{bmatrix}\), then verify by multiplication.
  2. Reduce \(\left[\begin{array}{cc|cc}1&3&1&0\\2&7&0&1\end{array}\right]\) to \([\mathbf{I}_2\mid\mathbf{B}]\). What is \(\mathbf{B}\)?
  3. In the second worked example, why must the operation \(R_1\to R_1-2R_2\) also change the entries to the right of the bar?
  4. Give a concrete reason that reducing \(\mathbf{A}\) alone to \(\mathbf{I}_2\) does not, by itself, display the inverse of \(\mathbf{A}\).
  5. For a proposed inverse \(\mathbf{B}\), what two matrix products should be computed to check the defining conditions?

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