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Worked Examples · Example 34

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

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Gauss–Jordan reduction of A=IAA=IA gives A−1=11200[7515075110−10030720−24]A^{-1}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}.

Write A=IAA = IA; apply elementary row operations until AA becomes II. The matrix that II turns into is A−1A^{-1}.

  1. Write A=IAA = IA with

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

  1. R1→12R1R_1\to\tfrac12 R_1, then R2→R2−4R1R_2\to R_2-4R_1, R3→R3−6R1R_3\to R_3-6R_1:

[13250−12−1500−50]=[1200−210−301]A.\begin{bmatrix} 1 & \tfrac32 & 5 \\ 0 & -12 & -15 \\ 0 & 0 & -50 \end{bmatrix} = \begin{bmatrix} \tfrac12 & 0 & 0 \\ -2 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}A.

  1. R2→−112R2R_2\to -\tfrac{1}{12}R_2, R3→−150R3R_3\to -\tfrac{1}{50}R_3:

[13250154001]=[120016−11203500−150]A.\begin{bmatrix} 1 & \tfrac32 & 5 \\ 0 & 1 & \tfrac54 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \tfrac12 & 0 & 0 \\ \tfrac16 & -\tfrac1{12} & 0 \\ \tfrac{3}{50} & 0 & -\tfrac1{50} \end{bmatrix}A.

  1. Back-eliminate: R1→R1−32R2R_1\to R_1-\tfrac32 R_2, then R2→R2−54R3R_2\to R_2-\tfrac54 R_3 and R1→R1−258R3R_1\to R_1-\tfrac{25}{8}R_3. The left side becomes II and the right side becomes: …

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