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7.6.1: Inverse Matrices (Exercises)

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    83744
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    SECTION 2.4 PROBLEM SET: INVERSE MATRICES

    In problems 1- 2, verify that the given matrices are inverses of each other.

    1. \(\left[\begin{array}{ll}
      7 & 3 \\
      2 & 1
      \end{array}\right]\left[\begin{array}{rr}
      1 & -3 \\
      -2 & 7
      \end{array}\right]\)
    1. \(\left[\begin{array}{ccc}
      1 & -1 & 0 \\
      1 & 0 & -1 \\
      2 & 3 & -4
      \end{array}\right]\left[\begin{array}{ccc}
      3 & -4 & 1 \\
      2 & -4 & 1 \\
      3 & -5 & 1
      \end{array}\right]\)

    In problems 3- 6, find the inverse of each matrix by the row-reduction method.

    1. \(\left[\begin{array}{rr}
      3 & -5 \\
      -1 & 2
      \end{array}\right]\)
    1. \(\left[\begin{array}{lll}
      1 & 0 & 2 \\
      0 & 1 & 4 \\
      0 & 0 & 1
      \end{array}\right]\)

    SECTION 2.4 PROBLEM SET: INVERSE MATRICES

    In problems 5 - 6, find the inverse of each matrix by the row-reduction method.

    1. \(\left[\begin{array}{ccc}
      1 & 1 & -1 \\
      1 & 0 & 1 \\
      2 & 1 & 1
      \end{array}\right]\)
    1. \(\left[\begin{array}{lll}
      1 & 1 & 1 \\
      3 & 1 & 0 \\
      1 & 1 & 2
      \end{array}\right]\)

    Problems 7 -10: Express the system as \(AX = B\); then solve using matrix inverses found in problems 3 - 6.

    1. \(\begin{array}{l}
      3 x-5 y=2 \\
      -x+2 y=0
      \end{array}\)
    1. \(\begin{aligned}
      x+\quad 2 z &=8 \\
      y+4 z &=8 \\
      z &=3
      \end{aligned}\)

    SECTION 2.4 PROBLEM SET: INVERSE MATRICES

    Problems 9 -10: Express the system as \(AX = B\); then solve using matrix inverses found in problems 3 - 6.

    1. \(\begin{aligned}
      x+y-z &=2 \\
      x+ z&=7 \\
      2 x+y+z &=13
      \end{aligned}\)
    1. \(\begin{array}{l}
      x+y+z=2 \\
      3 x+y=7 \\
      x+y+2 z=3
      \end{array}\)
    1. Why is it necessary that a matrix be a square matrix for its inverse to exist? Explain by relating the matrix to a system of equations.
    1. Suppose we are solving a system \(AX = B\) by the matrix inverse method, but discover \(A\) has no inverse. How else can we solve this system? What can be said about the solutions of this system?

    This page titled 7.6.1: Inverse Matrices (Exercises) is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Rupinder Sekhon and Roberta Bloom via source content that was edited to the style and standards of the LibreTexts platform.