Inverse of a 3 by 3 Matrix by Adjoint: Signs, Steps, and Checks

Finding the inverse of a 3 by 3 matrix can feel like a question with too many moving parts. There is a determinant, nine minors, nine signs, a transpose, a division, and then a multiplication check.

That sounds like a lot. But the method becomes calm and predictable when you give every step its own job.

Think of the matrix as a nine-tile lock. The determinant first tells you whether the lock can open at all. The cofactors cut the nine key pieces. The transpose turns those pieces into the correct arrangement. Dividing by the determinant finishes the key. Verification proves that it really opens the lock.

The complete route is:

determinant -> minors -> cofactors -> transpose -> divide -> verify

This guide will take you through the adjoint method slowly, show exactly where the signs come from, and finish with checks that catch almost every common mistake.

The Formula You Are Building Towards

For a square matrix A:

A inverse = (1 / det(A)) adj(A)

This formula can be used only when:

det(A) != 0

If det(A) = 0, the matrix is singular and its inverse does not exist.

If det(A) != 0, the matrix is non-singular and its inverse exists.

If that distinction needs a quick revision, read this guide to singular and non-singular matrices before continuing.

The Six-Step Adjoint Method

Suppose A is a 3 by 3 matrix.

Use this order every time:

1. Find det(A).
2. If det(A) = 0, stop. There is no inverse.
3. Find all nine cofactors and arrange them in the cofactor matrix.
4. Transpose the cofactor matrix to get adj(A).
5. Multiply adj(A) by 1 / det(A).
6. Verify that A x A inverse = I.

The order matters. Most errors happen when two of these steps are mixed together.

First Building Block: A Minor

The minor of an element is the 2 by 2 determinant left after deleting the row and column containing that element.

For a general matrix:

A = [ a11  a12  a13 ]
    [ a21  a22  a23 ]
    [ a31  a32  a33 ]

To find the minor M12, delete row 1 and column 2.

The entries left are:

[ a21  a23 ]
[ a31  a33 ]

Therefore:

M12 = a21(a33) - a23(a31)

The instruction is always the same:

Delete the element's row.
Delete the element's column.
Find the determinant of what remains.

Second Building Block: A Cofactor

A cofactor is a minor with the correct sign attached.

The rule is:

Cij = (-1)^(i+j) Mij

You do not need to calculate the power each time. Memorise the sign map:

[ +  -  + ]
[ -  +  - ]
[ +  -  + ]

The top-left position is positive, and the signs alternate like a checkerboard.

This means:

C11 = +M11    C12 = -M12    C13 = +M13
C21 = -M21    C22 = +M22    C23 = -M23
C31 = +M31    C32 = -M32    C33 = +M33

Here is the important part: a minus sign in the sign map does not guarantee a negative cofactor.

Suppose:

M12 = -5

Then:

C12 = -M12
    = -(-5)
    = 5

The position contributes one sign, while the 2 by 2 determinant contributes its own value. Keep them separate until the last line.

Third Building Block: The Adjoint

Once all nine cofactors are ready, arrange them in their original positions:

Cofactor matrix = [ C11  C12  C13 ]
                  [ C21  C22  C23 ]
                  [ C31  C32  C33 ]

The adjoint is the transpose of this matrix:

adj(A) = [ C11  C21  C31 ]
         [ C12  C22  C32 ]
         [ C13  C23  C33 ]

Transposing changes rows into columns.

So the first row of the cofactor matrix becomes the first column of the adjoint. The second row becomes the second column. The third row becomes the third column.

This is the most frequently forgotten step in the entire method.

A Complete Example From Start to Finish

Find the inverse of:

A = [ 2  1  3 ]
    [ 0  1  2 ]
    [ 1  0  1 ]

We will keep every stage separate.

Step 1: Find the Determinant

Expand along the first row:

det(A) = 2[(1)(1) - (2)(0)]
       - 1[(0)(1) - (2)(1)]
       + 3[(0)(0) - (1)(1)]

       = 2(1) - 1(-2) + 3(-1)
       = 2 + 2 - 3
       = 1

Since:

det(A) = 1 != 0

the inverse exists.

Step 2: Find the First Row of Cofactors

For C11, delete row 1 and column 1:

C11 = +[(1)(1) - (2)(0)]
    = 1

For C12, delete row 1 and column 2, then apply the negative sign belonging to position (1, 2):

C12 = -[(0)(1) - (2)(1)]
    = -[-2]
    = 2

For C13, delete row 1 and column 3:

C13 = +[(0)(0) - (1)(1)]
    = -1

So the first cofactor row is:

[ 1  2  -1 ]

Step 3: Find the Second Row of Cofactors

For C21, delete row 2 and column 1. Position (2, 1) carries a negative sign:

C21 = -[(1)(1) - (3)(0)]
    = -1

For C22:

C22 = +[(2)(1) - (3)(1)]
    = 2 - 3
    = -1

For C23, position (2, 3) carries a negative sign:

C23 = -[(2)(0) - (1)(1)]
    = -[-1]
    = 1

So the second cofactor row is:

[ -1  -1  1 ]

Step 4: Find the Third Row of Cofactors

For C31:

C31 = +[(1)(2) - (3)(1)]
    = 2 - 3
    = -1

For C32, position (3, 2) carries a negative sign:

C32 = -[(2)(2) - (3)(0)]
    = -4

For C33:

C33 = +[(2)(1) - (1)(0)]
    = 2

So the third cofactor row is:

[ -1  -4  2 ]

Step 5: Assemble the Cofactor Matrix

Put each cofactor back in the position whose row and column numbers it carries:

C = [  1   2  -1 ]
    [ -1  -1   1 ]
    [ -1  -4   2 ]

Pause here and compare this with the sign map. The values do not have to display the same visible pattern because some minors were already negative. The sign map tells you the operation to apply, not the final sign of every answer.

Step 6: Transpose to Obtain the Adjoint

Turn the rows of C into columns:

adj(A) = C transpose

       = [  1  -1  -1 ]
         [  2  -1  -4 ]
         [ -1   1   2 ]

Notice two entries that make the transpose easy to check:

C23 = 1 moves to position (3, 2) in adj(A)
C32 = -4 moves to position (2, 3) in adj(A)

If those entries stay in their old positions, the matrix has not been transposed.

Step 7: Divide by the Determinant

Now use:

A inverse = (1 / det(A)) adj(A)

Here det(A) = 1, so:

A inverse = [  1  -1  -1 ]
            [  2  -1  -4 ]
            [ -1   1   2 ]

That is the required inverse.

How to Verify the Answer

An inverse must reverse the action of the original matrix. Therefore:

A x A inverse = I

where I is the 3 by 3 identity matrix:

I = [ 1  0  0 ]
    [ 0  1  0 ]
    [ 0  0  1 ]

For our example:

A x A inverse

= [ 2  1  3 ] [  1  -1  -1 ]
  [ 0  1  2 ] [  2  -1  -4 ]
  [ 1  0  1 ] [ -1   1   2 ]

= [ 2(1)+1(2)+3(-1)   2(-1)+1(-1)+3(1)   2(-1)+1(-4)+3(2) ]
  [ 0(1)+1(2)+2(-1)   0(-1)+1(-1)+2(1)   0(-1)+1(-4)+2(2) ]
  [ 1(1)+0(2)+1(-1)   1(-1)+0(-1)+1(1)   1(-1)+0(-4)+1(2) ]

= [ 1  0  0 ]
  [ 0  1  0 ]
  [ 0  0  1 ]

= I

The answer is verified.

If row-column multiplication needs refreshing, use this guide to matrix multiplication, compatibility, and order.

A Faster Check Before Fractions Appear

Sometimes det(A) is not 1. The inverse may contain several fractions, and multiplying with them can become untidy.

There is a cleaner identity:

A x adj(A) = det(A) I

You can use this before dividing by the determinant.

Suppose det(A) = 5. Then a correct adjoint must satisfy:

A x adj(A) = [ 5  0  0 ]
             [ 0  5  0 ]
             [ 0  0  5 ]

This check catches cofactor and transpose errors without fractions. Once it passes, divide the adjoint by 5 to obtain the inverse.

Why the Adjoint Formula Works

The central identity is:

A x adj(A) = det(A) I

Why do the diagonal entries become det(A)?

When a row of A is multiplied by the matching column of adj(A), you are multiplying the elements of that row by their corresponding cofactors. Their sum is precisely the expansion of the determinant.

Why do the other entries become zero?

An off-diagonal entry combines one row with cofactors belonging to a different row. This behaves like a determinant with two identical rows, so its value is zero.

If det(A) != 0, divide both sides by det(A):

A x [(1 / det(A)) adj(A)] = I

The matrix in square brackets must therefore be A inverse.

This is why the adjoint method is more than a collection of steps. The cofactor structure is designed to produce the identity matrix.

What If the Determinant Is Zero?

Consider:

B = [ 1  2  3 ]
    [ 2  4  6 ]
    [ 0  1  1 ]

Row 2 is obtained by multiplying row 1 by 2. Therefore, the rows are dependent and:

det(B) = 0

So B inverse does not exist.

There is no reason to calculate nine cofactors for an inverse question after reaching this result. Even if adj(B) can be calculated, the inverse formula would require division by zero.

The correct conclusion is simply:

Since det(B) = 0, B is singular.
Therefore, B inverse does not exist.

How to Choose the Best Row or Column for the Determinant

A 3 by 3 determinant can be expanded along any row or column. The answer will be the same.

Choose the row or column containing the most zeros. Each zero removes one full 2 by 2 calculation.

In our main example:

A = [ 2  1  3 ]
    [ 0  1  2 ]
    [ 1  0  1 ]

the second row and second column each contain a zero. Either can shorten the determinant calculation.

This shortcut helps with det(A), but remember that finding the full adjoint still requires all nine cofactors.

A Parameter Question: When Does the Inverse Exist?

Consider:

P = [ 1  2  0 ]
    [ 0  k  1 ]
    [ 1  0  1 ]

Find the values of k for which the inverse exists.

Expand the determinant along the first row:

det(P) = 1[(k)(1) - (1)(0)]
       - 2[(0)(1) - (1)(1)]
       + 0

       = k - 2(-1)
       = k + 2

The inverse exists when the determinant is not zero:

k + 2 != 0
k != -2

At k = -2, the determinant becomes zero, so the inverse does not exist.

Using the Inverse to Solve Three Equations

The adjoint method also connects directly to a system of equations.

Suppose:

2x + y + 3z = 12
     y + 2z = 5
 x       + z = 5

Write it as:

AX = B

where:

A = [ 2  1  3 ]      X = [ x ]      B = [ 12 ]
    [ 0  1  2 ]          [ y ]          [  5 ]
    [ 1  0  1 ]          [ z ]          [  5 ]

We already found:

A inverse = [  1  -1  -1 ]
            [  2  -1  -4 ]
            [ -1   1   2 ]

Since X = A inverse B:

X = [  1  -1  -1 ] [ 12 ]
    [  2  -1  -4 ] [  5 ]
    [ -1   1   2 ] [  5 ]

  = [ 12 - 5 - 5 ]
    [ 24 - 5 - 20 ]
    [ -12 + 5 + 10 ]

  = [  2 ]
    [ -1 ]
    [  3 ]

Therefore:

x = 2, y = -1, z = 3

For a fuller decision guide about the number of solutions, read matrix equations: unique, infinite, or no solution.

The Seven Most Common Mistakes

1. Skipping the Determinant Test

Calculating nine cofactors before finding that det(A) = 0 wastes time. Always test the determinant first.

2. Deleting the Wrong Row or Column

For M23, delete row 2 and column 3. Say the position aloud before crossing anything out.

3. Using the Minor as the Cofactor

At negative positions, the minor must have its sign changed. Keep the checkerboard beside your work.

4. Assuming Every Negative Position Gives a Negative Answer

If the minor is negative, a negative cofactor position can make the final cofactor positive.

5. Forgetting the Transpose

The cofactor matrix must be transposed to create the adjoint. Label both matrices separately.

6. Dividing Only Some Entries

The factor 1 / det(A) multiplies every entry of adj(A). Keeping it outside the matrix until the end is often neater.

7. Treating Verification as Optional

One wrong sign can create a complete-looking but incorrect inverse. A product check exposes the error.

A Reliable Page Layout for Your Solution

Long calculations are easier to check when the page is organised.

Use this layout:

1. Write A and find det(A).
2. State whether the inverse exists.
3. Write the cofactor sign map.
4. Calculate C11 to C33 in three rows.
5. Display the cofactor matrix C.
6. Write adj(A) = C transpose.
7. Apply A inverse = (1 / det(A)) adj(A).
8. Verify the result.

Do not squeeze a 3 by 3 inverse into one crowded block. Clear working is not just presentation. It also makes self-correction possible.

A Two-Minute Self-Check

Before moving to the next question, ask:

• Did I prove that det(A) != 0?
• Did I use the sign pattern + - + / - + - / + - +?
• Did I calculate each 2 by 2 determinant in the order ad - bc?
• Did I transpose the cofactor matrix?
• Did I divide every adjoint entry by the determinant?
• Did the product give the identity matrix?

If all six answers are yes, the solution is usually secure.

Practice Questions

Have a go at these before opening the answer section.

Question 1

For the main example matrix, find M12 and C12 separately.

Question 2

State the signs of C21, C22, and C23 before finding any minors.

Question 3

Does this matrix have an inverse?

Q = [ 1  0  2 ]
    [ 2  0  4 ]
    [ 3  1  5 ]

Question 4

The cofactor matrix of R is:

C = [ 1  2  3 ]
    [ 4  5  6 ]
    [ 7  8  9 ]

Write adj(R).

Question 5

For what value of t does the following matrix fail to have an inverse?

S = [ 1  0  1 ]
    [ 0  t  2 ]
    [ 1  1  0 ]

Answers to Practice Questions

Answer 1

M12 = (0)(1) - (2)(1) = -2
C12 = -M12 = 2

Answer 2

The signs are:

C21: negative
C22: positive
C23: negative

These are position signs, not guaranteed final signs.

Answer 3

No. Row 2 is two times row 1, so det(Q) = 0. The matrix is singular and has no inverse.

Answer 4

Transpose the cofactor matrix:

adj(R) = [ 1  4  7 ]
         [ 2  5  8 ]
         [ 3  6  9 ]

Answer 5

Expanding along the first row gives:

det(S) = -2 - t

The inverse fails to exist when:

-2 - t = 0
t = -2

A Final Memory Map

When you see “find the inverse by the adjoint method,” remember this sentence:

Test, sign, turn, divide, check.

Test the determinant.

Sign the minors to form cofactors.

Turn the cofactor matrix by transposing it.

Divide the adjoint by the determinant.

Check that the product is the identity matrix.

The formal ideas and identities used here are explained in the NCERT Mathematics Part I chapter on determinants. The MIT notes on matrix inverses independently show the same minors, checkerboard signs, transpose, and division sequence.

Frequently Asked Questions

1. What is the formula for the inverse of a 3 by 3 matrix by the adjoint method?

If det(A) != 0, then:

A inverse = (1 / det(A)) adj(A)

2. Why must I find the determinant first?

The determinant tells you whether an inverse exists. If it is zero, the inverse does not exist and there is no need to continue with nine cofactors.

3. What is the sign pattern for cofactors of a 3 by 3 matrix?

The pattern is:

[ +  -  + ]
[ -  +  - ]
[ +  -  + ]

4. What is the difference between a minor and a cofactor?

A minor is the 2 by 2 determinant obtained after deleting an element’s row and column. A cofactor is that minor multiplied by the sign belonging to its position.

5. Is the cofactor matrix the same as the adjoint?

Not generally. The adjoint is the transpose of the cofactor matrix. The two happen to match only in special cases where the cofactor matrix is symmetric.

6. Why do we transpose the cofactor matrix?

The transpose puts each cofactor in the position needed for the identity A x adj(A) = det(A) I to work. Without that rearrangement, the inverse check will usually fail.

7. Can I expand the determinant along any row or column?

Yes. Choose the row or column with the most zeros because it reduces the amount of arithmetic.

8. Do I have to calculate all nine cofactors?

For the full adjoint of a general 3 by 3 matrix, yes. Each entry of the adjoint comes from one of those cofactors.

9. What is the best way to verify the inverse?

Multiply A x A inverse. The answer should be the identity matrix. You may also verify A x adj(A) = det(A) I before introducing fractions.

10. What if the determinant is negative?

A negative determinant is perfectly acceptable. The inverse exists as long as the determinant is not zero. The negative value will simply affect the signs of the inverse entries.

11. Can an inverse contain fractions?

Yes. Fractions are common when the determinant does not divide every entry of the adjoint exactly. Keep 1 / det(A) outside the matrix until the final step to make the work neater.

12. What should I do if my product check is not the identity matrix?

Check the cofactor signs first, then confirm that you transposed the cofactor matrix, and finally inspect the row-column multiplication. Those three places cause most errors.