Determinant of a 3x3 matrix

Every $3\times3$ matrix carries a single hidden number that tells you whether the system of equations it represents has a unique solution, whether the matrix can be inverted, and how a linear transformation scales area or volume. That number is the determinant, and computing it for order 3 is the one calculation this whole chapter of CBSE Class 12 Mathematics is built around — minors, cofactors, the adjoint and the matrix inverse all start from expanding a $3\times3$ determinant correctly. The good news is that the method is completely mechanical: pick a row or column, attach alternating signs, and reduce three numbers to a sum of $2\times2$ determinants you already know how to evaluate. Get the sign pattern and the row/column choice right, and a $3\times3$ determinant takes under a minute.

The core insight

The determinant of a $3\times3$ matrix is found by expanding along any row or column into a signed sum of three $2\times2$ minors, and every valid choice of row or column gives the same numerical answer.

The core idea

A $3\times3$ matrix has the form

A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}

where: each entry $a_i, b_i, c_i$ sits in row $i$ and the column named by its letter. Its determinant $|A|$ is defined by expansion along a row or column, most commonly the first row:

|A| = a_1\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - b_1\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}

where: each $2\times2$ determinant is a minor — delete the row and column of the multiplying entry and evaluate what remains by the cross-multiplication rule $\begin{vmatrix} p & q \\ r & s \end{vmatrix} = ps - qr$ . Three points make this mechanical rather than mysterious.

The signs alternate, and follow a fixed checkerboard. Along the first row the signs are $+, -, +$ . In general the sign attached to the entry in row $i$ , column $j$ is $(-1)^{i+j}$ , which is $+$ when $i+j$ is even and $-$ when $i+j$ is odd:

\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}

Reading this grid off directly avoids the sign slips that cause most errors in $3\times3$ determinant work.

Any row or column gives the same value. Expanding along the first column, the second row, or any other row/column always produces the identical number $|A|$ , provided the correct checkerboard signs for that row/column are used. This is stated here as a working fact (its full proof belongs to a later, more theoretical treatment) and is what makes the method flexible.

Zeros are free wins. Because a $0$ entry multiplies a minor to give $0$ , expanding along the row or column with the most zeros eliminates that many minor calculations entirely — often turning a three-minor problem into a one-minor problem.

Figure

A three by three grid showing the alternating sign pattern plus minus plus, minus plus minus, plus minus plus for expanding a determinant

Derivation

Derivation AExpanding a 3x3 determinant along the first row

The goal is to show, entry by entry, how the definition $|A| = a_1\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - b_1\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix}$ reduces to a fully expanded polynomial in the nine entries of $A$ , so that the alternating-sign expansion rule and the "any row or column works" fact both become concrete and checkable rather than things to memorize blindly.

Step 1. Expand each $2\times2$ minor using $\begin{vmatrix} p & q \\ r & s \end{vmatrix} = ps - qr$ . The first minor gives $b_2c_3 - c_2b_3$ , the second gives $a_2c_3 - c_2a_3$ , and the third gives $a_2b_3 - b_2a_3$ .

Step 2. Substitute these into the first-row expansion. $|A| = a_1(b_2c_3 - c_2b_3) - b_1(a_2c_3 - c_2a_3) + c_1(a_2b_3 - b_2a_3)$ .

Step 3. Distribute each coefficient across its bracket to get the fully expanded six-term form. $|A| = a_1b_2c_3 - a_1c_2b_3 - b_1a_2c_3 + b_1c_2a_3 + c_1a_2b_3 - c_1b_2a_3$ .

Step 4. Group the six terms into three "positive diagonal" products and three "negative diagonal" products: $|A| = (a_1b_2c_3 + b_1c_2a_3 + c_1a_2b_3) - (a_1c_2b_3 + b_1a_2c_3 + c_1b_2a_3)$ . This six-term expansion is exactly what any other row or column expansion also produces after full distribution, which is why the choice of row or column never changes the final value — they are all different routes to the same six terms with the same net signs.

Worked example

Evaluate $\begin{vmatrix} 2 & 3 & 1 \\ 0 & 1 & -2 \\ 1 & -1 & 3 \end{vmatrix}$ by expanding along the first row, then confirm the same value along a row with a zero.

Step 1. Write out the first-row expansion with alternating signs $+, -, +$ .

|A| = 2\begin{vmatrix} 1 & -2 \\ -1 & 3 \end{vmatrix} - 3\begin{vmatrix} 0 & -2 \\ 1 & 3 \end{vmatrix} + 1\begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix}

Step 2. Evaluate each $2\times2$ minor.

\begin{vmatrix} 1 & -2 \\ -1 & 3 \end{vmatrix} = (1)(3) - (-2)(-1) = 3 - 2 = 1

\begin{vmatrix} 0 & -2 \\ 1 & 3 \end{vmatrix} = (0)(3) - (-2)(1) = 0 + 2 = 2

\begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} = (0)(-1) - (1)(1) = 0 - 1 = -1

Step 3. Substitute back and simplify.

|A| = 2(1) - 3(2) + 1(-1) = 2 - 6 - 1 = -5

So $|A| = -5$ .

Competitive stretch

Since row 2 of this matrix has a zero in its first entry, expanding along row 2 should give the same answer with less work. With signs $-, +, -$ for row 2: $|A| = -0(\ldots) + 1\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} - (-2)\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix}$ . The surviving minors are $\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = 6 - 1 = 5$ and $\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = -2 - 3 = -5$ , giving $|A| = 0 + 1(5) + 2(-5) = 5 - 10 = -5$ — the same $-5$ , confirming the row choice does not matter.

Common traps

Forgetting the alternating sign pattern, especially the minus sign on the middle term of a first-row or first-column expansion.
Deleting the wrong row/column when forming a minor — always remove the row and column that intersect at the multiplying entry, not any nearby row/column.
Not exploiting a zero-heavy row or column, wasting time computing minors that will be multiplied by zero anyway.
Sign errors inside the $2\times2$ minors themselves, particularly when the matrix has several negative entries.

Ask the tutor: Want to work through expanding this same matrix along its third column, or try a fresh $3\times3$ matrix of your own?

Why does it not matter which row or column I expand along?

Every expansion, whichever row or column you pick, reduces algebraically to the same six-term polynomial in the nine matrix entries once you fully distribute the minors — three positive products and three negative products. The derivation above shows this explicitly for the first row; any other row or column arrives at the identical six terms, just grouped differently on the way there. So the freedom to choose is genuine, not a shortcut that risks a wrong answer.

How do I decide which row or column to expand along?

Scan the matrix for the row or column with the most zero entries, since each zero entry contributes nothing and its minor never needs to be computed. If there are no zeros, any row or column works equally well and the first row is the conventional default because it matches how the determinant is usually written out. Picking wisely only saves arithmetic; it never changes the final numeric answer.

What is the difference between a minor and the determinant itself?

The determinant is the single number $|A|$ for the whole $3\times3$ matrix. A minor is the smaller $2\times2$ determinant left over after you delete the row and column of one particular entry — it is an ingredient used three times (once per entry of the expansion row or column) to build up $|A|$ , not the final answer on its own. Attaching the correct alternating sign to each minor before adding them up is what turns the three minors into the determinant.