Solveeit Logo
Login

Question

Question: Adjoint of the matrix $N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$...

Adjoint of the matrix N=[−4−3−3101443]N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}

Answer

[−4−3−3101443]\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}

Explanation

Solution

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

  1. Calculate the matrix of cofactors: For each element nijn_{ij} in the matrix NN, its cofactor CijC_{ij} is given by Cij=(−1)i+jMijC_{ij} = (-1)^{i+j} M_{ij}, where MijM_{ij} is the minor of the element nijn_{ij} (the determinant of the submatrix formed by deleting the ii-th row and jj-th column).

  2. Transpose the matrix of cofactors: The adjoint of NN, denoted as adj(N)adj(N), is the transpose of the matrix of cofactors.

Given matrix: N=[−4−3−3101443]N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}

Let's calculate each cofactor:

C11=(−1)1+1∣0143∣=(0×3)−(1×4)=0−4=−4C_{11} = (-1)^{1+1} \begin{vmatrix} 0 & 1 \\ 4 & 3 \end{vmatrix} = (0 \times 3) - (1 \times 4) = 0 - 4 = -4

C12=(−1)1+2∣1143∣=−((1×3)−(1×4))=−(3−4)=−(−1)=1C_{12} = (-1)^{1+2} \begin{vmatrix} 1 & 1 \\ 4 & 3 \end{vmatrix} = -((1 \times 3) - (1 \times 4)) = -(3 - 4) = -(-1) = 1

C13=(−1)1+3∣1044∣=(1×4)−(0×4)=4−0=4C_{13} = (-1)^{1+3} \begin{vmatrix} 1 & 0 \\ 4 & 4 \end{vmatrix} = (1 \times 4) - (0 \times 4) = 4 - 0 = 4

C21=(−1)2+1∣−3−343∣=−((−3×3)−(−3×4))=−(−9−(−12))=−(−9+12)=−(3)=−3C_{21} = (-1)^{2+1} \begin{vmatrix} -3 & -3 \\ 4 & 3 \end{vmatrix} = -((-3 \times 3) - (-3 \times 4)) = -(-9 - (-12)) = -(-9 + 12) = -(3) = -3

C22=(−1)2+2∣−4−343∣=(−4×3)−(−3×4)=−12−(−12)=−12+12=0C_{22} = (-1)^{2+2} \begin{vmatrix} -4 & -3 \\ 4 & 3 \end{vmatrix} = (-4 \times 3) - (-3 \times 4) = -12 - (-12) = -12 + 12 = 0

C23=(−1)2+3∣−4−344∣=−((−4×4)−(−3×4))=−(−16−(−12))=−(−16+12)=−(−4)=4C_{23} = (-1)^{2+3} \begin{vmatrix} -4 & -3 \\ 4 & 4 \end{vmatrix} = -((-4 \times 4) - (-3 \times 4)) = -(-16 - (-12)) = -(-16 + 12) = -(-4) = 4

C31=(−1)3+1∣−3−301∣=(−3×1)−(−3×0)=−3−0=−3C_{31} = (-1)^{3+1} \begin{vmatrix} -3 & -3 \\ 0 & 1 \end{vmatrix} = (-3 \times 1) - (-3 \times 0) = -3 - 0 = -3

C32=(−1)3+2∣−4−311∣=−((−4×1)−(−3×1))=−(−4−(−3))=−(−4+3)=−(−1)=1C_{32} = (-1)^{3+2} \begin{vmatrix} -4 & -3 \\ 1 & 1 \end{vmatrix} = -((-4 \times 1) - (-3 \times 1)) = -(-4 - (-3)) = -(-4 + 3) = -(-1) = 1

C33=(−1)3+3∣−4−310∣=(−4×0)−(−3×1)=0−(−3)=3C_{33} = (-1)^{3+3} \begin{vmatrix} -4 & -3 \\ 1 & 0 \end{vmatrix} = (-4 \times 0) - (-3 \times 1) = 0 - (-3) = 3

Now, form the matrix of cofactors, let's call it CC: C=[C11C12C13C21C22C23C31C32C33]=[−414−304−313]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}

Finally, the adjoint of NN is the transpose of the matrix of cofactors CC: adj(N)=CT=[−4−3−3101443]adj(N) = C^T = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}

Notice that the adjoint of the matrix NN is the matrix NN itself. This implies that NN is an involutory matrix and its determinant is 1.