How Do You Find the Inverse of a Matrix Using the Adjoint Method?

• MHB
• karush
In summary, to find the inverse of a matrix using the method of adjoint matrix, you must first find the determinant of the matrix. Then, you must replace each element in the matrix with the determinant of the smaller matrix created by deleting its row and column. This will give you the adjoint matrix. Next, you must supply the cofactors by multiplying each element in the adjoint matrix by -1 to the power of the sum of its row and column number. Finally, you must transpose the resulting matrix and multiply it by 1 over the determinant of the original matrix to get the inverse matrix. It is important to be careful and systematic in this process to avoid errors.
karush
Gold Member
MHB
$\textsf{Find the inverse of matrix}$
$$A=\left| \begin{array}{rrr} 1&0&2\\ 1&0&0 \\ 3&2&0 \end{array} \right|$$
$\textsf{by method of adjoint matrix }\\$
$\textsf{adj$A = |C_{ij}|^T$}\\$
$\textsf{det$A =4$}\\$
$\textsf{so then}\\$
$A^{-1}=\frac{1}{det A}(adj A)= \frac{1}{4}(adj A)$
$\textit{ok I don't know how to get (adj A)}$

It's a little tedious.

Pick each element in the 3x3 matrix, one at a time.

Delete the row and the column and calculate the determinant of the remaining 2x2 matrix.

Replace the selected element with the result.

Do this for each of the nine elements. Some of the results might be 0.

Don't forget the cofactor for each.

Transpose and you are done.

Pick each element in the 3x3 matrix, one at a time.

Delete the row and the column and calculate the determinant of the remaining 2x2 matrix.

like this?
$A=\left| \begin{array}{rrr} 1&0&2\\ 1&0&0 \\ 3&2&0 \end{array} \right|\\$
$=\begin{bmatrix}0& 0 \\ 2& 0 \\\end{bmatrix} +0 +2\begin{bmatrix}1& 0 \\ 3& 2 \\\end{bmatrix} +\begin{bmatrix}0& 2 \\ 2& 0 \\\end{bmatrix} +0+0 +3\begin{bmatrix}0& 2 \\ 0& 0 \\\end{bmatrix} +2\begin{bmatrix}1& 2 \\ 1& 0 \\\end{bmatrix}\\$
$=0+0+2(2)-4+0+0+0+2(-2)=-4$

ok W|A says it is 4 ?

Last edited:
karush said:
like this?

$A=\left| \begin{array}{rrr} 1&0&2\\ 1&0&0 \\ 3&2&0 \end{array} \right|\\$
$=\begin{bmatrix}0& 0 \\ 2& 0 \\\end{bmatrix} +0 +2\begin{bmatrix}1& 0 \\ 3& 2 \\\end{bmatrix} +\begin{bmatrix}0& 2 \\ 2& 0 \\\end{bmatrix} +0+0 +3\begin{bmatrix}0& 2 \\ 0& 0 \\\end{bmatrix} +2\begin{bmatrix}1& 2 \\ 1& 0 \\\end{bmatrix}\\$
$=0+0+2(-1)+0+0+0+0+2(-2)=-6$ok W|A says it is 4 ?
No. You calculated them all, but then you added all the pieces. Don't do that. Do one at a time and substitute them into a new matrix - replacing the element you used to produce the smaller matrix. You should get a new 3x3 matrix out of the process.

tkhunny said:
No. You calculated them all, but then you added all the pieces. Don't do that. Do one at a time and substitute them into a new matrix - replacing the element you used to produce the smaller matrix. You should get a new 3x3 matrix out of the process.

$$\begin{vmatrix}0& 0& -2 \\ -4& 0& 0\\ 0& -4& 0 \end{vmatrix}$$

not convinced this is correct
but couldn't find error

karush said:
$$\begin{vmatrix}0& 0& -2 \\ -4& 0& 0\\ 0& -4& 0 \end{vmatrix}$$

not convinced this is correct
but couldn't find error
Closer, but the Adjunct piece does NOT include the element itself. You still think you're trying to evaluate a determinant. That is not what you are doing. You must find all nine 2x2 matrices and evaluate their determinants. The element is replaced in each case. Don't multiply by it.

tkhunny said:
Closer, but the Adjunct piece does NOT include the element itself. You still think you're trying to evaluate a determinant. That is not what you are doing. You must find all nine 2x2 matrices and evaluate their determinants. The element is replaced in each case. Don't multiply by it.

$=\begin{bmatrix}0& 0 \\ 2& 0 \\\end{bmatrix} +0 +\begin{bmatrix}1& 0 \\ 3& 2 \\\end{bmatrix} +\begin{bmatrix}0& 2 \\ 2& 0 \\\end{bmatrix} +0+0 +\begin{bmatrix}0& 2 \\ 0& 0 \\\end{bmatrix} +\begin{bmatrix}1& 2 \\ 1& 0 \\\end{bmatrix}+0$$\begin{bmatrix}0& 0 &-2 \\ 4& 0 &0 \\ -2& 2& 0 \end{bmatrix} W|A gave this as inverse? View attachment 7985 Attachments • q2.gif 1.5 KB · Views: 99 karush said: =\begin{bmatrix}0& 0 \\ 2& 0 \\\end{bmatrix} +0 +\begin{bmatrix}1& 0 \\ 3& 2 \\\end{bmatrix} +\begin{bmatrix}0& 2 \\ 2& 0 \\\end{bmatrix} +0+0 +\begin{bmatrix}0& 2 \\ 0& 0 \\\end{bmatrix} +\begin{bmatrix}1& 2 \\ 1& 0 \\\end{bmatrix}+0$$\begin{bmatrix}0& 0 &-2 \\ 4& 0 &0 \\ -2& 2& 0 \end{bmatrix}$

W|A gave this as inverse?

$\begin{bmatrix}0& 0 &2 \\ -4& -6 &2 \\ 0& -2& 0 \end{bmatrix}$

After applying the cofactors

$\begin{bmatrix}0& 0 &2 \\ 4& -6 &-2 \\ 0& 2& 0 \end{bmatrix}$

Transpose and you should be done.

tkhunny said:

$\begin{bmatrix}0& 0 &2 \\ -4& -6 &2 \\ 0& -2& 0 \end{bmatrix}$

Transpose and you should be done.

Ok I don't see how you got the Adjunct Matrix

what do you mean "reproduce the results"

$\begin{bmatrix}\begin{vmatrix}0& 0 \\ 2& 0 \end{vmatrix}& \begin{vmatrix}1& 0 \\ 3& 0 \end{vmatrix} &\begin{vmatrix}1& 0 \\ 3& 2 \end{vmatrix} \\ \begin{vmatrix}0& 2 \\ 2& 0 \end{vmatrix}& \begin{vmatrix}1& 2 \\ 3& 0 \end{vmatrix} &\begin{vmatrix}1& 0 \\ 3& 2 \end{vmatrix} \\ \begin{vmatrix}0& 2 \\ 0& 0 \end{vmatrix}& \begin{vmatrix}1& 2 \\ 1& 0 \end{vmatrix}& \begin{vmatrix}1& 0 \\ 1& 0 \end{vmatrix} \end{bmatrix}$

Don't ever make me do that again. :-)

Every element is replaced by the determinant of the smaller matrix created be deleting its row and column. The original element is gone. We're creating a new animal.

Supply the cofactors.
Transpose.
You're done.

I presume this...
$\text{3. } A^{-1} =\displaystyle\frac{1}{det A}(adj A) =\frac{1}{4} \begin{bmatrix}0& 0 &2 \\ -4& -6 &2 \\ 0& -2& 0 \end{bmatrix}$

Seems there should be a Transpose in there.

Please check your results. Multiply by the original matrix. What should be the result?

thusly...

$\text{3. } A^{-1} =\displaystyle\frac{1}{det A}(adj A) =\frac{1}{4} \begin{bmatrix} 0& 0 &2 \\ -4& -6 &2 \\ 0& -2& 0 \end{bmatrix}^T =\frac{1}{4} \begin{bmatrix} 0& -4 &0\\ 0& -6 &-2 \\ 2& 2& 0 \end{bmatrix}$
$= \begin{bmatrix} 0& -1 &0\\ 0& -3/2 &-1/2 \\ 1/2& 1/2& 0 \end{bmatrix}$

Last edited:
Did you check it?

I know there are errors but couldn't find them

karush said:
I know there are errors but couldn't find them

Keep looking.

Be more careful and systematic.

You'll learn more. Plus, the process will annoy you a little and you'll be even more careful on the next problem.

1. How do you find the inverse of a matrix?

To find the inverse of a matrix, you need to use the Gaussian elimination method or the adjugate matrix method. The Gaussian elimination method involves row operations to reduce the matrix to its reduced row echelon form, while the adjugate matrix method involves finding the determinant and using it to calculate the inverse.

2. Can any matrix have an inverse?

No, not all matrices have an inverse. Only square matrices, which have the same number of rows and columns, can have an inverse. Additionally, the determinant of the matrix must not be equal to zero in order for the inverse to exist.

3. What is the purpose of finding the inverse of a matrix?

The inverse of a matrix is useful in solving systems of linear equations, as well as in performing operations such as division and finding solutions to certain types of problems in physics and engineering.

4. Are there any specific steps to follow when finding the inverse of a matrix?

Yes, there are specific steps to follow when finding the inverse of a matrix, depending on the method you choose to use. For the Gaussian elimination method, you need to use row operations to reduce the matrix to its reduced row echelon form. For the adjugate matrix method, you need to find the determinant and use it in the calculation of the inverse.

5. Can the inverse of a matrix be calculated by hand?

Yes, the inverse of a matrix can be calculated by hand using the Gaussian elimination method or the adjugate matrix method. However, for larger matrices, it may be more convenient to use a calculator or a computer program to find the inverse.

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