Theory of Intelligence
11TH•Mathematics•Matrices And Determinants•Question 1

(iii) Row-Reduction Steps to Find $A^{-1}$

Find the inverses of the following matrices by using row operations:

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(iii) Row-Reduction Steps to Find A−1A^{-1}

Form the augmented matrix [A∣I3][A \mid I_3]:

[16210021300100−11001]\left[\begin{array}{ccc|ccc} 1 & 6 & 2 & 1 & 0 & 0 \\ 2 & 13 & 0 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array}\right]

  1. R2→R2−2R1R_2 \to R_2 - 2R_1: Eliminate Column 1 in Row 2

    ∼[16210001−4−2100−11001]\sim \left[\begin{array}{ccc|ccc} 1 & 6 & 2 & 1 & 0 & 0 \\ 0 & 1 & -4 & -2 & 1 & 0 \\ 0 & -1 & 1 & 0 & 0 & 1 \end{array}\right]

  2. R1→R1−6R2,R3→R3+R2R_1 \to R_1 - 6R_2, \quad R_3 \to R_3 + R_2: Clear Column 2

    ∼[102613−6001−4−21000−3−211]\sim \left[\begin{array}{ccc|ccc} 1 & 0 & 26 & 13 & -6 & 0 \\ 0 & 1 & -4 & -2 & 1 & 0 \\ 0 & 0 & -3 & -2 & 1 & 1 \end{array}\right]

  3. R3→−13R3R_3 \to -\frac{1}{3} R_3: Leading 1 in Row 3

    ∼[102613−6001−4−21000123−13−13]\sim \left[\begin{array}{ccc|ccc} 1 & 0 & 26 & 13 & -6 & 0 \\ 0 & 1 & -4 & -2 & 1 & 0 \\ 0 & 0 & 1 & \frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \end{array}\right]

  4. R1→R1−26R3,R2→R2+4R3R_1 \to R_1 - 26R_3, \quad R_2 \to R_2 + 4R_3: Clear Column 3   ⟹  [I3∣A−1]\implies [I_3 \mid A^{-1}]

    ∼[100−1338326301023−13−4300123−13−13]\sim \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -\frac{13}{3} & \frac{8}{3} & \frac{26}{3} \\ 0 & 1 & 0 & \frac{2}{3} & -\frac{1}{3} & -\frac{4}{3} \\ 0 & 0 & 1 & \frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \end{array}\right]
A−1=[−1338326323−13−4323−13−13]\boxed{A^{-1} = \begin{bmatrix} -\frac{13}{3} & \frac{8}{3} & \frac{26}{3} \\ \frac{2}{3} & -\frac{1}{3} & -\frac{4}{3} \\ \frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \end{bmatrix}}

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