Expert Maths Tutoring in the UK

How to Prove that a Matrix is Invertible
How to Prove that a Matrix is Invertible

with tutors mapped to your child’s learning needs.

Inverse of 3×3 Matrix

Before going to see how to find the inverse of a 3×3 matrix, let us recall the what the inverse mean. The inverse of a number is a number which when multiplied by the given number results in the multiplicative identity, 1. In the same way, the product of a matrix A and its inverse A-1 gives the identity matrix, I. i.e., AA-1 = A-1A = I. Let us see how to find the inverse of 3×3 matrix.

Let us see the formula for finding the inverse of 3×3 matrix along with some other ways of finding it. Also, we will see some examples of finding the inverse of a 3×3 matrix.

What is the Inverse of 3×3 Matrix?

The inverse of a 3×3 matrix, say A, is a matrix of the same order denoted by A-1 where AA-1 = A-1A = I, where I is the identity matrix of order 3×3. i.e., I = \(\left[\begin{array}{rr}1 & 0 & 0 \\ 0&1&0 \\ 0 & 1&0 \end{array}\right]\). For example, if A = \(\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]\) then A-1 = \(\left[\begin{array}{rr}3 / 16 & 1 / 4 & -5 / 16 \\
1 / 4 & 0 & 1 / 4 \\
-5 / 16 & 1 / 4 & 3 / 16 \\ \end{array}\right]\). One can easily multiply these matrices and verify whether AA-1 = A-1A = I. We will see how to find the inverse of a 3×3 matrix in the upcoming section.

Elements Used to Find Inverse of 3×3 Matrix

Before going to know how to find the inverse of 3×3 matrix, let us see how to find the determinant and adjoint of 3×3 matrix. Let us use this same example (as in the previous section) in each explanation.

Adjoint of a 3×3 Matrix

The adjoint of a matrix A is obtained by finding the transpose of the cofactor matrix of A. To know how to find the adjoint of a matrix in detail click here. The cofactor of any element of a 3×3 matrix is the determinant of 2×2 matrix that is obtained by removing the row and the column containing the element. Also, we write alternate + and – signs while finding the cofactors. Here is an example.

Let A = \(\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]\).

Then its cofactor matrix is:

\(\left[\begin{array}{rr}\left|\begin{array}{ll} 1 & 2 \\
2 & 1
\end{array}\right| & -\left|\begin{array}{cc}
2 & 2 \\
-1 & 1
\end{array}\right| & -\left|\begin{array}{cc}
2 & 1 \\
-1 & 2
\end{array}\right|\\
-\left|\begin{array}{cc}
2 & -1 \\
2 & 1
\end{array}\right| & \left|\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right|&-\left|\begin{array}{cc}
1 & 2 \\
-1 & 2
\end{array}\right| \\ \left|\begin{array}{cc}
2 & -1 \\
1 & 2
\end{array}\right|& -\left|\begin{array}{rr}
1 & -1 \\
2 & 2
\end{array}\right|&\left|\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right| \end{array}\right]\)

Each 2×2 determinant is obtained by multiplying diagonals and subtracting the products (from left to right).

So the cofactor matrix = \(\left[\begin{array}{ccc}
1-4 & -(2+2) & 4+1 \\
-(2+2) & 1-1 & -(2+2) \\
4+1 & -(2+2) & 1-4
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
-3 & -4 & 5 \\
-4 & 0 & -4 \\
5 & -4 & -3
\end{array}\right]\)

By transposing the cofactor matrix, we get the adjoint matrix.

So adj A = \(\left[\begin{array}{ccc}
-3 & -4 & 5 \\
-4 & 0 & -4 \\
5 & -4 & -3
\end{array}\right]\).

(Of course, we have got both the cofactor matrix and adjoint matrix to be the same in this case. But it may not happen always).

Determinant of a 3×3 Matrix

To find the determinant of a 3×3 matrix, find the sum of the product of the elements of any of its row/column and their corresponding cofactors. Here is an example.

A = \(\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]\). Let us use the first row to find the determinant.

det A = 1 (cofactor of 1) + 2 (cofactor of 2) + (-1) cofactor of (-1)
= 1(-3) + 2(-4) + (-1)5
= -3 – 8 – 5
= -16

But here is a trick to find the determinant of any 3×3 A = \(\left[\begin{array}{ccc}a & b & c \\ p & q & r \\ x & y & z\end{array}\right]\) matrix faster. Here, we just write the same matrix twice next to each other and then apply the trick.

The inverse of a 3×3 matrix formula uses the determinant of the matrix.

Inverse of 3×3 Matrix Formula

The inverse of a 3×3 matrix A is calculated using the formula A-1 = (adj A)/(det A), where

  • adj A = The adjoint matrix of A
  • det A = determinant of A

det A is in the denominator in the formula of A-1. Thus, for A-1 to exist det A should not be 0. i.e.,

  • A-1 exists when det A ≠ 0 (i.e., when A is nonsingular)
  • A-1 does not exist when det A = 0 (i.e., when A is singular)

Thus, here are the steps to find the inverse of 3×3 matrix. The steps are explained through the same example A = \(\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]\). Let us find A-1.

  • Step – 1: Find adj A.
    We have already seen that adj A = \(\left[\begin{array}{ccc}
    -3 & -4 & 5 \\
    -4 & 0 & -4 \\
    5 & -4 & -3
    \end{array}\right]\).
  • Step – 2: Find det A.
    We have already seen that det A = -16
  • Step – 3: Apply the inverse of 3×3 matrix formula A-1 = (adj A)/(det A). i.e., divide every element of adj A by det A.
    Then A-1 = \(\left[\begin{array}{ccc}
    -3/-16 & -4/-16 & 5/-16 \\
    -4/-16 & 0/-16 & -4/-16 \\
    5/-16 & -4/-16 & -3/-16
    \end{array}\right]\)
    = \(\left[\begin{array}{rr}3 / 16 & 1 / 4 & -5 / 16 \\
    1 / 4 & 0 & 1 / 4 \\
    -5 / 16 & 1 / 4 & 3 / 16 \\ \end{array}\right]\).

Finding Inverse of 3×3 Matrix Using Row Operations

Like any other square matrix, we can use the elementary row operations to find the inverse of a 3×3 matrix as well. The process is explained below with an example.

  • We first write the given 3×3 matrix A and the identity matrix I of order 3×3 as an augmented matrix separated by a line where A is on the left side and I is on the right side.
  • Apply row operations so as to make the left side matrix to become the identity matrix I.
  • Then the matrix on the right side is A-1.

We can an see an example for this in the upcoming sections.

Solving System of 3×3 Equations Using Inverse

We can solve the system of 3×3 equations using the inverse of a matrix. The steps for this are explained here with an example where we are going to solve the system of 3×3 equations x + 2y – z = 10, 2x + y + 2z = 5, and -x + 2y + z = 6.

  • Step – 1: Write the given system of equations as AX = B.
    \(\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]\) \(\left[\begin{array}{rr}x \\y \\ z \end{array}\right]\) = \(\left[\begin{array}{rr}10 \\ 5 \\ 6 \end{array}\right]\)
    Here, A = \(\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]\), X = \(\left[\begin{array}{rr}x \\y \\ z\end{array}\right]\), and B = \(\left[\begin{array}{rr}10 \\ 5\\ 6 \end{array}\right]\).
  • Step – 2: Find the inverse of the 3×3 matrix. i.e., find A-1.
    In one of the previous sections, we found that A-1 = \(\left[\begin{array}{rr}3 / 16 & 1 / 4 & -5 / 16 \\
    1 / 4 & 0 & 1 / 4 \\
    -5 / 16 & 1 / 4 & 3 / 16 \\ \end{array}\right]\).
  • Step – 3: Find the solution matrix X using the formula X = A-1B.
    X = \(\left[\begin{array}{rr}3 / 16 & 1 / 4 & -5 / 16 \\
    1 / 4 & 0 & 1 / 4 \\
    -5 / 16 & 1 / 4 & 3 / 16 \\ \end{array}\right]\) \(\left[\begin{array}{rr}10 \\ 5 \\ 6 \end{array}\right]\)
    = \(\left[\begin{array}{rr}5/4 \\4 \\ -3/4 \end{array}\right]\)

Therefore, x = 5/4, y = 4, and z = -3/4 is the solution of the given system of equations.

Important Notes on Inverse of 3×3 Matrix:

  • A matrix A is invertible (inverse of A exists) only when det A ≠ 0.
  • If A and A-1 are the inverses of each other, then AA-1 = A-1A = I.
  • The inverse of a 3×3 identity matrix is itself. i.e., I-1 = I.
  • The inverse of 3×3 matrix is used to solve a system of 3×3 equations in 3 variables.

☛ Related Topics:

Inverse of 3×3 Matrix Examples

  1. Example 1: Determine which of the following 3×3 matrices have an inverse. (a) A = \(\left[\begin{array}{rr}2 & 6 & 3 \\ 4 & -1 & 3 \\ 1 & 3 & 2 \end{array}\right]\) (b) B = \(\left[\begin{array}{rr}-3 & 2 & 1\\ 4 & 8 & -1 \\ 6 & -4 & -2 \end{array}\right]\).

    Solution:

    A 3×3 matrix A is invertible only if det A ≠ 0. So Let us find the determinant of each of the given matrices.

    (a) det A = 2 (-2 – 9) – 6 (8 – 3) + 3 (12 + 1)

    = 2(-11) -6(5) + 3(13)
    = -22 – 30 + 39
    = -13
    ≠ 0

    Thus, A-1 exists. i.e., A is invertible.

    (b) det B = -3 (-16 – 4) – 2 ( -8 + 6) + 1 (-16 -48)

    = -3 (-20) – 2 (-2) + 1 (-64)
    = 60 + 4 – 64
    = 0

    Therefore, B-1 does NOT exist. i.e., B is NOT invertible.

    Answer: A has inverse but B does not.

  2. Example 2: Find the inverse of A = \(\left[\begin{array}{rr}4 & 2 & -3 \\ 1 & -1 &2\\ 5 & 3 & 0 \end{array}\right]\).

    Solution:

    We will find det A and adj A for the given matrix.

    Finding det A:

    det A = 4 (0 – 6) – 2 (0 – 10) – 3 (3 + 5) = -28

    Finding adj A:

    The co-factor matrix of A = \(\left[\begin{array}{rr}0-6 & -(0-10) & 3+5 \\ -(0+9) & 0+15 & -(12-10) \\ 4-3 & -(8+3) & -4-2 \end{array}\right]\) = \(\left[\begin{array}{rr}-6 &10 & 8\\ -9 & 15 & -2 \\ 1 & -11 & -6 \end{array}\right]\)

    Thus,adj A = \(\left[\begin{array}{rr}-6 &-9 & 1\\ 10 & 15 & -11 \\ 8 & -2 & -6 \end{array}\right]\)

    Finding A-1:

    Substitute the values of adj A and det Ain the formula A-1 = (adj A) / (det A),

    A-1 = \(\left[\begin{array}{rr}-6/-28 &-9/-28 & 1/-28\\ 10/-28 & 15/-28 & -11/-28 \\ 8/-28 & -2/-28 & -6/-28 \end{array}\right]\)

    = \(\left[\begin{array}{rr}3/14 &9/28 & -1/28\\ -5/14 & -15/28 & 11/28 \\ -2/7 & 1/14 & 3/14 \end{array}\right]\)

    Answer: The inverse of the given 3×3 matrix is A-1 = \(\left[\begin{array}{rr}3/14 &9/28 & -1/28\\ -5/14 & -15/28 & 11/28 \\ -2/7 & 1/14 & 3/14 \end{array}\right]\).

  3. Example 3: Find the inverse of the 3×3 matrix A from Example 2 using elementary row operations. Verify whether the answer you get is the same as that of Example 2.

    Solution:

    Given A = \(\left[\begin{array}{rr}4 & 2 & -3 \\ 1 & -1 &2\\ 5 & 3 & 0 \end{array}\right]\).

    Step 1: Write A and I as in a single matrix separated by a dotted line (as an augmented matrix).

    \(\left[\begin{array}{ccccccc}

    4 & 2 & -3 & | & 1 & 0 & 0 \\
    1 & -1 & 2 & | & 0 & 1 & 0 \\
    5 & 3 & 0 & | & 0 & 0 & 1
    \end{array}\right]\)

    Step 2: Apply elementary row operations to make the left side matrix converted to an identity matrix.

    Applying R\(_2\) → 4R\(_2\) – R\(_1\) and R\(_3\) → 4R\(_3\) – 5R\(_1\),

    \(\left[\begin{array}{ccccccc}

    4 & 2 & -3 & | & 1 & 0 & 0 \\
    0 & -6 & 11 & | & -1 & 4 & 0 \\
    0 & 2 & 15 & | & -5 & 0 & 4
    \end{array}\right]\)

    Applying R\(_1\) → 3R\(_1\) + R\(_2\) and R\(_3\) → 3R\(_3\) + R\(_2\),

    \(\left[\begin{array}{ccccccc}

    12 & 0 & 2 & |&2 & 4 & 0 \\
    0 & -6 & 11 & | & -1 & 4 & 0 \\
    0 & 0 & 56 &|& -16 & 4 & 12
    \end{array}\right]\)

    Applying R\(_1\) → 28R\(_1\) – R\(_3\) and R\(_2\) → 56R\(_2\) -11R\(_3\),

    \(\left[\begin{array}{cccccc}

    336 & 0 & 0 &|& 72 & 108 & -12 \\
    0 & -336 & 0 & | & 120 & 180 & -132 \\
    0 & 0 & 56 &|& -16 & 4 & 12
    \end{array}\right]\)

    Divide R\(_1\) by 336, R\(_2\) by -336, and R\(_3\) by 56,

    \(\left[\begin{array}{cccccc}

    1 & 0 & 0 & | & 3 / 14 & 9 / 28 & -1 / 28 \\
    0 & 1 & 0 & | & -5 / 14 & -15 / 28 & 11 / 28 \\
    0 & 0 &1 & | & -2/7 & 1 / 14 & 3 / 14
    \end{array}\right]\)

    Step 3: The right side matrix is our inverse matrix. i.e.,

    A-1 = \(\left[\begin{array}{rr}3/14 &9/28 & -1/28\\ -5/14 & -15/28 & 11/28 \\ -2/7 & 1/14 & 3/14 \end{array}\right]\)

    Answer: The answer we got here is matching with the answer from Example 2.

Practice Questions on Inverse of 3×3 Matrix

FAQs on Inverse of 3×3 Matrix

What is Meant by Inverse of a 3×3 Matrix?

The inverse of a 3×3 matrix A is denoted by A-1. Here, AA-1 = A-1A = I, where I is the identity matrix of order 3×3.

How to Find Inverse of a 3×3 Matrix?

Here are the steps to find the inverse of a 3×3 matrix A:

  • Find det A.
  • Find adj A.
  • Apply the formula A-1 = (adj A)/(det A).

What is an Example of a 3×3 Matrix with No Inverse?

A matrix cannot have inverse if its determinant is 0. A = \(\left[\begin{array}{rr}1 & 2 & 1 \\ 2&4&2 \\2 & 4 &5 \end{array}\right]\) has no inverse as det A = 0 in this case.

Are All 3×3 Matrices Invertible?

No, all 3×3 matrices are not invertible as a matrix cannot have its inverse when its determinant is 0. For example, A = \(\left[\begin{array}{rr}0 & 0 & 0 \\ -1&3&2 \\5 & 7 &5 \end{array}\right]\) is not invertible as det A = 0 in this case.

What is the Inverse of 3×3 Matrix Formula?

If A is a 3×3 matrix, its inverse formula is A-1 = (adj A)/(det A). Here,

  • det A = Determinant of the matrix A
  • adj A = Adjoint of the matrix A

Does a 3×3 Matrix have an Inverse?

A 3×3 matrix has inverse only if its determinant is not zero. If the determinant is zero, then the matrix has is not invertible (does not have inverse) and in that case, it is called a singular matrix.

How to Find Inverse of a 3×3 Matrix Using Elementary Row Operations?

For finding the inverse of a 3×3 matrix (A ) by elementary row operations,

  • Write A and I (identity matrix of same order) in a single matrix separating them by a vertical dotted line.
  • Apply elementary row operations so that the left side matrix becomes I.
  • The matrix that comes on the right side is A-1.

visual curriculum

You are watching: Expert Maths Tutoring in the UK. Info created by THVinhTuy selection and synthesis along with other related topics.

Rate this post

Related Posts