2. The 3×3 Inverse

Video coming soon2 worked examples

Theory

For a $2\times 2$ matrix there is a formula for the inverse. For a $3\times 3$ there is no such shortcut worth memorising, so we use row reduction instead — the same elementary row operations used in Gaussian elimination.

The method rests on the definition $A^{-1}A = I$ . Write the matrix and the identity side by side in an augmented array, then apply row operations until the left block becomes the identity. Whatever those same operations do to the right block turns it into $A^{-1}$ :

\left(\begin{array}{ccc|ccc} & A & & & I & \end{array}\right) \quad \longrightarrow \quad \left(\begin{array}{ccc|ccc} & I & & & A^{-1} & \end{array}\right)

The three permitted operations are the familiar ones: interchange two rows, multiply a row by a non-zero constant, and add a multiple of one row to another. Every operation must be applied right across the array, both blocks together.

If at any stage a row of the left block becomes entirely zero, the matrix is singular — its determinant is zero and no inverse exists.

The Golden Rule: work down the leading diagonal one column at a time — first make the pivot entry $1$ , then clear every other entry in that column to $0$ , and only then move to the next column. Jumping around leads to undoing your own work.

⚠️ Common Examiner Traps

Forgetting the right-hand block: every row operation must be carried out on both halves. Missing one is the single most common error.
Clearing above as well as below: unlike Gaussian elimination, which stops at upper triangular form, here you must keep going until the left block is the full identity — zeros above the diagonal too.
Not stating the operations: write each one down ( $R_2 \to R_2 - 2R_1$ ). Marks are awarded for the operations, not just the answer.
Skipping the check: multiply $A$ by your answer. If you do not get $I$ exactly, there is an arithmetic slip to find.

Worked examples

Example 1

Find the inverse of $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 3 & 2 \\ 1 & 0 & 0 \end{pmatrix}$ using elementary row operations.

Step 1: Write $A$ alongside the identity matrix:

\left(\begin{array}{ccc|ccc} 2 & 1 & 1 & 1 & 0 & 0 \\ 1 & 3 & 2 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 \end{array}\right)

Step 2: Swap rows 1 and 3 to put a convenient $1$ in the top-left pivot position ( $R_1 \leftrightarrow R_3$ ):

\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 1 & 3 & 2 & 0 & 1 & 0 \\ 2 & 1 & 1 & 1 & 0 & 0 \end{array}\right)

Step 3: Clear the rest of column 1, using $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - 2R_1$ :

\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 3 & 2 & 0 & 1 & -1 \\ 0 & 1 & 1 & 1 & 0 & -2 \end{array}\right)

Step 4: Swap rows 2 and 3 so the next pivot is already $1$ ( $R_2 \leftrightarrow R_3$ ), then use $R_3 \to R_3 - 3R_2$ :

\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 & -2 \\ 0 & 0 & -1 & -3 & 1 & 5 \end{array}\right)

Step 5: Make the last pivot $1$ with $R_3 \to -R_3$ , then clear above it with $R_2 \to R_2 - R_3$ :

\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & -2 & 1 & 3 \\ 0 & 0 & 1 & 3 & -1 & -5 \end{array}\right)

Step 6: The left block is now the identity, so the right block is the inverse:

A^{-1} = \begin{pmatrix} 0 & 0 & 1 \\ -2 & 1 & 3 \\ 3 & -1 & -5 \end{pmatrix}

Check: the first row of $A$ times the first column of $A^{-1}$ gives $2(0) + 1(-2) + 1(3) = 1$ , and the same row times the second column gives $2(0) + 1(1) + 1(-1) = 0$ — as required for $AA^{-1} = I$ .

Example 2

Show that $B = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 0 & 5 \end{pmatrix}$ has no inverse.

Step 1: Begin the row reduction. Using $R_2 \to R_2 - 2R_1$ :

\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 0 & 5 \end{pmatrix} \longrightarrow \begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 0 & 5 \end{pmatrix}

Step 2: The second row has become entirely zero. This happened because row 2 of $B$ was exactly twice row 1 — the rows are not independent.

Step 3: A row of zeros in the left block means it can never be reduced to the identity, so $B$ is singular and has no inverse.

Confirming with the determinant: expanding along the first row,

\det B = 1(4 \times 5 - 6 \times 0) - 2(2 \times 5 - 6 \times 1) + 3(2 \times 0 - 4 \times 1)

= 1(20) - 2(4) + 3(-4) = 20 - 8 - 12 = 0

A zero determinant confirms the matrix is singular.