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    2.5 Notes and Examples For Matrix Equations: Example 1

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    七海未来
    This document discusses solving matrix equations by manipulating matrices and their inverses. It provides 5 examples of solving equations of the form AX=B, XA+B=C, and AX+B=C. The key steps are checking for invertibility, calculating matrix inverses, and multiplying equations by the inverse matrices to isolate the variable matrix X.

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    2.5 Notes and Examples For Matrix Equations: Example 1

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    七海未来
    13 views4 pages
    This document discusses solving matrix equations by manipulating matrices and their inverses. It provides 5 examples of solving equations of the form AX=B, XA+B=C, and AX+B=C. The key steps are checking for invertibility, calculating matrix inverses, and multiplying equations by the inverse matrices to isolate the variable matrix X.

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    2.5 Matrix Equations

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    This document discusses solving matrix equations by manipulating matrices and their inverses. It provides 5 examples of solving equations of the form AX=B, XA+B=C, and AX+B=C. The key steps are checking for invertibility, calculating matrix inverses, and multiplying equations by the inverse matrices to isolate the variable matrix X.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    13 views4 pages

    2.5 Notes and Examples For Matrix Equations: Example 1

    Uploaded by

    七海未来
    This document discusses solving matrix equations by manipulating matrices and their inverses. It provides 5 examples of solving equations of the form AX=B, XA+B=C, and AX+B=C. The key steps are checking for invertibility, calculating matrix inverses, and multiplying equations by the inverse matrices to isolate the variable matrix X.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 4

    2.

    5 Notes and Examples for Matrix Equations


    Throughout the previous sections, we introduced matrices and basic operations for adding,
    subtracting and multiplication of matrices. We also discussed some techniques how to
    evaluate the inverse of a squared matrix if it exists. Remember that these operations are not
    defined for all matrices and there are conditions that must be met in order to apply these
    operations. For instance, addition and subtraction make only sense if the two matrices have
    the same dimension and the product is only defined if the number of columns of the first
    matrix is equal to the number of rows of the second. The inverse of a matrix exists only for
    squared matrices with nonzero determinants. Further, we saw that the division of a matrix
    with another matrix is not defend at all!
    In this section we will use these operations to solve equations of matrices where the
    unknowns (variables) are matrices rather than numbers as we know them from real numbers
    analysis.
    For a matrix equation of the form 𝐴. 𝑋 = 𝐵, the variable 𝑋 is a matrix that has a dimension
    that allows the product with the matrix 𝐴.
    Let us see some examples:
    Example 1:
    Solve for 𝑋
    3 2 6 3
    𝑋+[ ]=[ ]
    1 0 7 −1
    3 2 3 2 6 3 3 2
    𝑋+[ ]−[ ]=[ ]−[ ]
    1 0 1 0 7 −1 1 0
    0 0 6−3 3−2 3 1
    𝑋+[ ]=[ ]=[ ]
    0 0 7 − 1 −1 − 0 6 −1
    𝟑 𝟏
    𝑿=[ ]
    𝟔 −𝟏
    Example 2:
    Solve for 𝑋
    −9 −3 4 0
    𝑋−[ ]=[ ]
    6 0 12 −10
    −9 −3 −9 −3 4 0 −9 −3
    𝑋−[ ]+[ ]=[ ]+[ ]
    6 0 6 0 12 −10 6 0
    0 0 4 − 9 0 + (−3) −5 −3
    𝑋−[ ]=[ ]=[ ]
    0 0 12 + 6 −10 + 0 18 −10

    −𝟓 −𝟑
    𝑿=[ ]
    𝟏𝟖 −𝟏𝟎
    For solving matrix equations involving multiplication, since there is no division defined for
    matrices, we will use the fact that 𝐼 = 𝐴−1 . 𝐴 = 𝐴. 𝐴−1 as shown in the next example:

    1
    Example 3:
    Given the matrices
    2 3 3 1
    𝐴=[ ] and 𝐵 = [ ]
    1 2 2 −5
    Solve the equation:
    𝐴∙𝑋 =𝐵
    Here, we expect 𝑋 to be a matrix with dimension 2 × 2. As you know, for solving such type of
    equations involving real numbers, we usually divide by 𝐴 to solve for 𝑋. Since we do not have
    division for matrices -as mentioned earlier- we will multiply be inverse of the matrix 𝐴 if it
    exists. Remember that 𝐴−1 exists if and only if 𝐴 is a squared matrix with det⁡(𝐴) ≠ 0. The
    latest we need to check first.
    2 3
    det ([ ]) = 2.2 − 1.3 = 4 − 3 = 1 ≠ 0
    1 2
    For 2 × 2 matrices, the inverse could be evaluated as follows:
    𝑎 𝑏 1 𝑑 −𝑏
    For 𝐴 = [ ],⁡then⁡𝐴−1 = det⁡(𝐴) [ ], if det⁡(𝐴) ≠ 0
    𝑐 𝑑 −𝑐 𝑎
    2 3 1 2 −3 2 −3
    So the inverse of [ ] is = 1 [ ]=[ ]
    1 2 −1 2 −1 2
    Now multiply be 𝐴−1:
    𝐴∙𝑋 =𝐵
    𝐴−1 ∙ 𝐴 ∙ 𝑋 = 𝐴−1 ∙ 𝐵
    𝐼 ∙ 𝑋 = 𝐴−1 ∙ 𝐵
    𝑋 = 𝐴−1 ∙ 𝐵

    We plug in 𝐴−1 ⁡and⁡𝐵 in 𝑋 = 𝐴−1 ∙ 𝐵 to solve the equation:


    2 −3 3 1 6−6 2 + 15 0 17
    𝑋=[ ]∙[ ]=[ ]=[ ]
    −1 2 2 −5 −3 + 4 −1 − 10 1 −11
    𝟎 𝟏𝟕
    𝑿=[ ]
    𝟏 −𝟏𝟏

    One more example involving both addition and multiplication:


    Example 4:
    Given the three matrices
    1 1 1 0 0 1
    𝐴=[ ], 𝐵 =[ ] and 𝐶 = [ ]
    1 2 0 2 1 0
    Solve the equation:

    2
    𝑋∙𝐴+𝐵 =𝐶
    We will use the techniques of the previous examples combined to solve this example.
    𝑋∙𝐴+𝐵 =𝐶
    𝑋∙𝐴+𝐵−𝐵 =𝐶−𝐵
    𝑋∙𝐴+𝟎=𝐶−𝐵
    𝑋∙𝐴=𝐶−𝐵
    Now, multiply both sides of the equation by 𝐴−1 ⁡if it exists to solve for 𝑋 as follows:
    𝑋 ∙ 𝐴 ∙ 𝐴−1 = (𝐶 − 𝐵) ∙ 𝐴−1
    𝑋 ∙ 𝐼 = (𝐶 − 𝐵) ∙ 𝐴−1
    𝑋 = (𝐶 − 𝐵) ∙ 𝐴−1
    We check the determinant of⁡det⁡(𝐴) ≠ 0
    1 1
    det ([ ]) = 2.1 − 1.1 = 2 − 1 = 1 ≠ 0, so the inverse exists. We use the method from
    1 2
    the previous example to evaluate 𝐴−1
    1 2 −1 2 −1
    𝐴−1 = [ ]=[ ]
    1 −1 1⁡ −1 1⁡
    Finally, we substitute to solve for 𝑋 in
    𝑋 = (𝐶 − 𝐵) ∙ 𝐴−1
    0 1 1 0 2 −1
    𝑋 = ([ ]−[ ]) ∙ [ ]
    1 0 0 2 −1 1⁡
    0−1 1−0 2 −1
    𝑋 = ([ ]) ∙ [ ]
    1−0 0−2 −1 1⁡
    −1 1 2 −1 −3 2
    𝑋 = ([ ]) ∙ [ ]=[ ]
    1 −2 −1 1⁡ 4 −3⁡
    −𝟑 𝟐
    𝑿=[ ]
    𝟒 −𝟑⁡
    Notice that in this example we multiplied the equation on both sides with⁡𝐴−1 from the left
    since 𝐴⁡was on the right. This is not the same as multiplying from the right since, in general,
    the matrix multiplication is noncommutative!
    To wrap-up this section, we will turn to an example involving a 3 × 3 matrix equation:
    Example 5:
    1 0 0 0 1 1 1 0 0
    Given 𝐴 = [1 1 0], ⁡𝐵 = [1 0 0] and 𝐶 = [0 1 0]
    1 1 1 0 0 1 1 0 1

    Solve the equation:

    3
    𝐴 ∙ 𝑋 + 2𝐵 = 3𝐶
    𝐴 ∙ 𝑋 + 2𝐵 − 2𝐵 = 3𝐶 − 2𝐵
    𝐴 ∙ 𝑋 + 𝟎 = 3𝐶 − 2𝐵
    𝐴 ∙ 𝑋 = 3𝐶 − 2𝐵
    If det⁡(𝐴) ≠ 0, then multiply by 𝐴−1 :
    𝐴−1 ∙ 𝐴 ∙ 𝑋 = 𝐴−1 ∙ (3𝐶 − 2𝐵)
    𝐼 ∙ 𝑋 = 𝐴−1 ∙ (3𝐶 − 2𝐵)
    𝑋 = 𝐴−1 ∙ (3𝐶 − 2𝐵)
    So we need to show that If det⁡(𝐴) ≠ 0 and evaluate 𝐴−1 :
    1 0 0
    1 0 1 0 1 1
    det(𝐴) = |1 1 0| = 1 ∙ | |−0∙| |+0∙| |= 1−0+0=1 ≠0
    1 1 1 1 1 1
    1 1 1
    I will leave it for you as an exercise to show that
    1 0 0
    𝐴−1 = [−1 1 0]
    0 −1 1
    You may use the Gaussian-elimination method to reduce the augmented matrix ⟨𝐴|𝐼⟩ into
    ⟨𝐼|𝐴−1 ⟩.
    Now plug in the value for 𝐵⁡, 𝐶⁡and⁡𝐴−1 in the simplified equation
    𝑋 = 𝐴−1 ∙ (3𝐶 − 2𝐵)
    1 0 0 1 0 0 0 1 1
    𝑋 = [−1 1 0] ∙ (3 [0 1 0] − 2 [1 0 0])
    0 −1 1 1 0 1 0 0 1
    1 0 0 3 0 0 0 2 2
    𝑋 = [−1 1 0] ∙ ([0 3 0] − [2 0 0])
    0 −1 1 3 0 3 0 0 2
    1 0 0 3 −2 −2
    𝑋 = [−1 1 0] ∙ ([−2 3 0 ])
    0 −1 1 3 0 1
    𝟑 −𝟐 −𝟐
    𝑿 = [−𝟓 𝟓 𝟐]
    𝟓 −𝟑 𝟏

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