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Adjoint of a Matrix – Determinants

  • Last Updated : 03 Jan, 2021
Geek Week

The knowledge of matrices is necessary for various branches of mathematics. Matrices are one of the most powerful tools in mathematics. From matrices there comes Determinants, Now we see one of the properties of the Determinant in this article. In this article, we see how to find Adjoint of a Matrix. To know about Adjoint of a Matrix we have to know about the Cofactor of a matrix.

Cofactor of a Matrix

The Cofactor is the number we get when we remove the column and row of a designated element in a matrix. It means to take one element from a matrix and delete the entire row and column of that element from the matrix, then which elements present in that matrix, that are called the cofactor. 

Example

Let’s take a matrix, A =\begin{bmatrix} 1 & 2 & 3\\ 7 & 4 & 5 \\ 6 & 8 & 9 \end{bmatrix}

Now find the cofactor of element 3

  • Step 1: Delete the entire row and column that contains element 3 
  • Step 2: Take the present elements as it is in the matrix after Step 1. 

So, the cofactor of 3 is  \begin{bmatrix} 7 & 4 \\ 6 & 8 \end{bmatrix}



Adjoint of a Matrix

To find the Adjoint of a Matrix, first, we have to find the Cofactor of each element, and then find 2 more steps. see below the steps,

  • Step 1: Find the Cofactor of each element present in the matrix.
  • Step 2: Create another matrix with the cofactors and expand the cofactors, then we get a matrix
  • Step 3: Now find the transpose of the matrix which comes from after Step 2.

Let’s understand the whole thing by taking some examples. 

Examples

Example 1: Find the Adjoint of the given matrix A =\begin{bmatrix} 1 & 2 & 3\\ 7 & 4 & 5 \\ 6 & 8 & 9 \end{bmatrix}?

Solution:

Step 1: To find the cofactor of each element

To find the cofactor of each element, we have to delete the row and column of each element one by one and take the present elements after deleting.

Cofactor of element at A[0,0] = 1 : +\begin{bmatrix} 4 & 5 \\ 8 & 9 \end{bmatrix}      = +(4×9 – 8×5) = -4

Cofactor of elements at A[0,1] = 2 : -\begin{bmatrix} 7 & 5 \\ 6 & 9 \end{bmatrix}      = -(7×9 – 6×5) = -33



Cofactor of elements at A[0,2] = 3 : +\begin{bmatrix} 7 & 4 \\ 6 & 8 \end{bmatrix}      = +(7×8 – 6×4) = 32

Cofactor of elements at A[2,0] = 7 : -\begin{bmatrix} 2 & 3 \\ 8 & 9 \end{bmatrix}      = -(2×9 – 8×3) = 6

Cofactor of elements at A[2,1] = 4 : +\begin{bmatrix} 1 & 3 \\ 6 & 9 \end{bmatrix}      = +(1×9 – 6×3) = -9

Cofactor of elements at A[2,2] = 5 : -\begin{bmatrix} 1 & 2 \\ 6 & 8 \end{bmatrix}      = -(1×8 – 6×2) = 4

Cofactor of elements at A[3,0] = 6 : +\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}      = +(2×5 – 4×3) = -2

Cofactor of elements at A[3,1] = 8 : -\begin{bmatrix} 1 & 3 \\ 7 & 5 \end{bmatrix}      = -(1×5 – 7×3) = 16

Cofactor of elements at A[3,2] = 9 : +\begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix}      = +(1×4 – 7×2) = -10

The matrix looks like with the cofactors:

A =\begin{bmatrix} +\begin{bmatrix} 4 & 5 \\ 8 & 9 \end{bmatrix} & -\begin{bmatrix} 7 & 5 \\ 6 & 9 \end{bmatrix} & +\begin{bmatrix} 7 & 4 \\ 6 & 8 \end{bmatrix}\\ \\ -\begin{bmatrix} 2 & 3 \\ 8 & 9 \end{bmatrix} & +\begin{bmatrix} 1 & 3 \\ 6 & 9 \end{bmatrix} & -\begin{bmatrix} 1 & 2 \\ 6 & 8 \end{bmatrix} \\ \\         +\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} & -\begin{bmatrix} 1 & 3 \\ 7 & 5 \end{bmatrix} & +\begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix} \end{bmatrix}

The final cofactor matrix:



A =\begin{bmatrix} -4 & -33 & 32\\ 6 & -9 & 4 \\ -2 & 16 & -10 \end{bmatrix}

Step 2: Find the transpose of the matrix obtained in step 1

adj(A) =\begin{bmatrix} -4 & 6 & -2\\ -33 & -9 & 16 \\ 32 & 4 & -10 \end{bmatrix}

This is the Adjoint of the matrix.

Example 2: Find the Adjoint of the given matrix A =\begin{bmatrix} -1 & -2 & -2\\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}?

Solution:

Step 1: To find the cofactor of each element

To find the cofactor of each element, we have to delete the row and column of each element one by one and take the present elements after deleting.

Cofactor of element at A[0,0] = -1 :  +\begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix}    = +(1×1 – (-2)x(-2)) = -3

Cofactor of elements at A[0,1] = -2 : -\begin{bmatrix} 2 & -2 \\ 2 & 1 \end{bmatrix}    = -(2×1 – 2x(-2)) = -6



Cofactor of elements at A[0,2] = -2 : +\begin{bmatrix} 2 & 1 \\ 2 & -2 \end{bmatrix}   = +(2x(-2) – 2×1) = -6

Cofactor of elements at A[2,0] = 2 : -\begin{bmatrix} -2 & -2 \\ -2 & 1 \end{bmatrix}    = -((-2)x1 – (-2)x(-2)) = 6

Cofactor of elements at A[2,1] = 1 :  +\begin{bmatrix} -1 & -2 \\ 2 & 1 \end{bmatrix}     = +((-1)x1 – 2x(-2)) = 3

Cofactor of elements at A[2,2] = -2 :  -\begin{bmatrix} -1 & -2 \\ 2 & -2 \end{bmatrix}   = -((-1)x(-2) – 2x(-2)) = -6

Cofactor of elements at A[3,0] = 2 : +\begin{bmatrix} -2 & -2 \\ 1 & -2 \end{bmatrix}    = +((-2)x(-2) – 1x(-2)) = 6

Cofactor of elements at A[3,1] = -2 : -\begin{bmatrix} -1 & -2 \\ 2 & -2 \end{bmatrix}    = -((-1)x(-2) – 2x(-2)) = -6

Cofactor of elements at A[3,2] = 1 : +\begin{bmatrix} -1 & -2 \\ 2 & 1 \end{bmatrix}  = +((-1)x(-1)- 2x(-2)) = 3

The final cofactor matrix:

A =\begin{bmatrix} -3 & -6 & -6\\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{bmatrix}

Step 2: Find the transpose of the matrix obtained in Step 1



adj(A) =\begin{bmatrix} -3 & 6 & 6\\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}

This is the Adjoint of the matrix.

Singular Matrix 

A matrix is said to be a singular matrix if the determinant of that matrix is ZERO. This singularity is achieved with only square matrices because only square matrices have determinant. Also, the inversion of singular matrices is not possible because to find the inverse of a matrix we need to divide the adjoint of a matrix with the determinant of the matrix, but for a singular matrix, the value of the determinant is ZERO. So the division is not possible here. To find a matrix is singular or not there is some rule, see below:

  • Rule 1: First check if the matrix square or not.
  • Rule 2: If square, then calculate its determinant and check if the value is ZERO or not. If ZERO then it is a singular matrix.

Examples

Example 1: Check if the given matrix is singular or not, A =\begin{bmatrix} 4 & 1\\ 8 & 2\end{bmatrix}?

Solution:

Here this matrix is a square matrix, so let’s find the determinant of this matrix,

det A = (4×2 – 8×1) = 0

Here determinant of A is ZERO, 

Also \frac{Adj(A)}{det (A)} = \frac{Adj(A)}{0} = \infty = inversion not possible

So we can say that matrix A is a singular matrix. 



Example 2: Check if the given matrix is singular or not, A =\begin{bmatrix} 2 & 4 & 6\\ 2 & 0 & 2 \\ 6 & 8 & 14 \end{bmatrix}?

Solution: 

det(A) = 2(0 – 16) – 4(28 – 12) + 6(16 – 0)  [using 1st row]

            = −2(16) + 2(16) = 0 

Here determinant of A is ZERO

Also  \frac{Adj(A)}{det (A)} = \frac{Adj(A)}{0} = \infty = inversion not possible

So we can say that matrix A is a singular matrix. 

Properties of Singular Matrix

  • All singular matrix has its determinant value ZERO.
  • All singular matrix must be of a square matrix.
  • All the singular matrix cannot be inverted.
  • There is no multiplicative inverse for a singular matrix.

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