Question

Adjoint of the matrix

$N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$ is

• A. $\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$

Answer 1

Answer: (A)

$\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$

Answer 2

To find the adjoint of a matrix $N$, we follow these steps:

1. Calculate the cofactor of each element of the matrix.
2. Form the cofactor matrix.
3. Take the transpose of the cofactor matrix to get the adjoint matrix.

Given matrix: $N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$

Step 1: Calculate the cofactors

The cofactor $C_{ij}$ of an element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of $a_{ij}$.

$C_{11} = (-1)^{1+1} \begin{vmatrix} 0 & 1 \\ 4 & 3 \end{vmatrix} = (0 \times 3 - 1 \times 4) = 0 - 4 = -4$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 1 & 1 \\ 4 & 3 \end{vmatrix} = -(1 \times 3 - 1 \times 4) = -(3 - 4) = -(-1) = 1$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 1 & 0 \\ 4 & 4 \end{vmatrix} = (1 \times 4 - 0 \times 4) = 4 - 0 = 4$

$C_{21} = (-1)^{2+1} \begin{vmatrix} -3 & -3 \\ 4 & 3 \end{vmatrix} = -((-3) \times 3 - (-3) \times 4) = -(-9 - (-12)) = -(-9 + 12) = -(3) = -3$

$C_{22} = (-1)^{2+2} \begin{vmatrix} -4 & -3 \\ 4 & 3 \end{vmatrix} = (-4) \times 3 - (-3) \times 4 = -12 - (-12) = -12 + 12 = 0$

$C_{23} = (-1)^{2+3} \begin{vmatrix} -4 & -3 \\ 4 & 4 \end{vmatrix} = -((-4) \times 4 - (-3) \times 4) = -(-16 - (-12)) = -(-16 + 12) = -(-4) = 4$

$C_{31} = (-1)^{3+1} \begin{vmatrix} -3 & -3 \\ 0 & 1 \end{vmatrix} = (-3) \times 1 - (-3) \times 0 = -3 - 0 = -3$

$C_{32} = (-1)^{3+2} \begin{vmatrix} -4 & -3 \\ 1 & 1 \end{vmatrix} = -((-4) \times 1 - (-3) \times 1) = -(-4 - (-3)) = -(-4 + 3) = -(-1) = 1$

$C_{33} = (-1)^{3+3} \begin{vmatrix} -4 & -3 \\ 1 & 0 \end{vmatrix} = (-4) \times 0 - (-3) \times 1 = 0 - (-3) = 3$

Step 2: Form the cofactor matrix

The cofactor matrix $C$ is: $C = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} = \begin{bmatrix} -4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3 \end{bmatrix}$

Step 3: Take the transpose of the cofactor matrix

The adjoint of N, denoted as adj(N), is $C^T$: $adj(N) = C^T = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$

Notice that the calculated adjoint matrix is identical to the original matrix N. $adj(N) = N$

This implies that the matrix N is an involutory matrix (a matrix that is its own inverse, since $N^{-1} = \frac{1}{\det(N)} adj(N)$ and if $adj(N)=N$, then $N^{-1} = \frac{1}{\det(N)} N$. For $N^{-1}=N$, we must have $\det(N)=1$). Let's verify the determinant of N: $\det(N) = -4(0 \times 3 - 1 \times 4) - (-3)(1 \times 3 - 1 \times 4) + (-3)(1 \times 4 - 0 \times 4)$ $\det(N) = -4(-4) + 3(-1) - 3(4)$ $\det(N) = 16 - 3 - 12 = 1$ Since $\det(N) = 1$, our result $adj(N) = N$ is consistent with $N^{-1} = N$.

The final answer is $N$

Explanation of the solution:

1. Calculate all nine cofactors $C_{ij}$ for the given matrix $N$.
2. Arrange these cofactors into a matrix, called the cofactor matrix.
3. The adjoint of matrix $N$ is the transpose of this cofactor matrix.
4. After calculation, the adjoint matrix is found to be identical to the original matrix $N$.

Answer:

The adjoint of the matrix $N$ is: $adj(N) = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} = N$