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11TH•Mathematics•Matrices And Determinants•Exercise 4.3

4.3 Properties of Matrix Operations

Practice set 4.3 (inverse matrices, rank, systems of linear equations, transformations, and coding/decoding).

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Question 1

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

Select Part:Part (i) of 3
(i)

[26−30−20−256]\begin{bmatrix}2&6&-3\\0&-2&0\\-2&5&6\end{bmatrix}

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Step-by-Step Solution

Question 1 • Part (i)

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Inversion Lab•Interactive

Matrix Inversion & 3D Space Un-Warping Visualizer

Step 0 of 6:Initial Augmented State$[A \mid I_3]$
Short Math Notes: Row Reduction Sequence
Geometric Transformation Effect:

Original deformed space. The columns of A form a skewed 3D parallelepiped with volume = 12.00.

Watch the vectors physically untangle and rotate toward standard coordinate axes.
Left Block: Current Matrix|Right Block: Accumulated Inverse Operations
2
6
-3
1
0
0
0
-2
0
0
1
0
-2
5
6
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1
Progress: 0% CompleteRow Operations Active

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

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

[26−31000−20010−256001]\left[\begin{array}{ccc|ccc} 2 & 6 & -3 & 1 & 0 & 0 \\ 0 & -2 & 0 & 0 & 1 & 0 \\ -2 & 5 & 6 & 0 & 0 & 1 \end{array}\right]

  1. R1→12R1R_1 \to \frac{1}{2} R_1: Leading 1 in Row 1

    ∼[13−3212000−20010−256001]\sim \left[\begin{array}{ccc|ccc} 1 & 3 & -\frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & -2 & 0 & 0 & 1 & 0 \\ -2 & 5 & 6 & 0 & 0 & 1 \end{array}\right]

  2. R3→R3+2R1R_3 \to R_3 + 2R_1: Clear Column 1 in Row 3

    ∼[13−3212000−200100113101]\sim \left[\begin{array}{ccc|ccc} 1 & 3 & -\frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & -2 & 0 & 0 & 1 & 0 \\ 0 & 11 & 3 & 1 & 0 & 1 \end{array}\right]

  3. R2→−12R2R_2 \to -\frac{1}{2} R_2: Leading 1 in Row 2

    ∼[13−3212000100−1200113101]\sim \left[\begin{array}{ccc|ccc} 1 & 3 & -\frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 11 & 3 & 1 & 0 & 1 \end{array}\right]

  4. R1→R1−3R2,R3→R3−11R2R_1 \to R_1 - 3R_2, \quad R_3 \to R_3 - 11R_2: Clear Column 2

    ∼[10−32123200100−12000311121]\sim \left[\begin{array}{ccc|ccc} 1 & 0 & -\frac{3}{2} & \frac{1}{2} & \frac{3}{2} & 0 \\ 0 & 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 3 & 1 & \frac{11}{2} & 1 \end{array}\right]

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

    ∼[10−32123200100−1200011311613]\sim \left[\begin{array}{ccc|ccc} 1 & 0 & -\frac{3}{2} & \frac{1}{2} & \frac{3}{2} & 0 \\ 0 & 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & \frac{1}{3} & \frac{11}{6} & \frac{1}{3} \end{array}\right]

  6. R1→R1+32R3R_1 \to R_1 + \frac{3}{2} R_3: Clear Column 3   ⟹  [I3∣A−1]\implies [I_3 \mid A^{-1}]

    ∼[1001174120100−1200011311613]\sim \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & \frac{17}{4} & \frac{1}{2} \\ 0 & 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & \frac{1}{3} & \frac{11}{6} & \frac{1}{3} \end{array}\right]
A−1=[1174120−1201311613]\boxed{A^{-1} = \begin{bmatrix} 1 & \frac{17}{4} & \frac{1}{2} \\ 0 & -\frac{1}{2} & 0 \\ \frac{1}{3} & \frac{11}{6} & \frac{1}{3} \end{bmatrix}}
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