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Worked Examples · Example 36

Q.Find the inverse of matrix AA using inverse of the coefficient matrix method when A=[23104−6569−20]A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}.

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With det⁡A=1200\det A=1200 and adj A\mathrm{adj}\,A from the cofactors, A−1=11200[7515075110−10030720−24]A^{-1}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}.

A−1=1det⁡A adj(A)A^{-1}=\dfrac{1}{\det A}\,\mathrm{adj}(A), where adj(A)\mathrm{adj}(A) is the transpose of the cofactor matrix, Cij=(−1)i+jMijC_{ij}=(-1)^{i+j}M_{ij}.

  1. A=[23104−6569−20].A=\begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}.

  2. Determinant (expand along R1R_1):

det⁡A=2(120−45)−3(−80−30)+10(36+36)=150+330+720=1200.\det A = 2(120-45) - 3(-80-30) + 10(36+36) = 150 + 330 + 720 = 1200.

  1. Cofactors:

C11=75, C12=110, C13=72,C_{11}=75,\ C_{12}=110,\ C_{13}=72,

C21=150, C22=−100, C23=0,C_{21}=150,\ C_{22}=-100,\ C_{23}=0,

C31=75, C32=30, C33=−24.C_{31}=75,\ C_{32}=30,\ C_{33}=-24.

  1. Adjoint = transpose of the cofactor matrix:

adj A=[7515075110−10030720−24].\mathrm{adj}\,A = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}.

  1. Inverse: …

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