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3.3E: Exercises for Section 3.3

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    197418

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    Exercise \(\PageIndex{1}\)

    Determine whether the following matrices are invertible by computing their determinants.

    1. \( A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\)
    2. \( B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \)
    Answer
    1. \(\det(A)=5\). Invertible.
    2. \(\det(B)=0\). Not invertible.
    Exercise \(\PageIndex{2}\)

    Find the adjugate of the matrix:

    \[ A = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} \nonumber\]

    Answer

    The cofactor matrix is:

    \[ \text{cof}(A) = \begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix} \nonumber \]

    The adjugate (adjoint) matrix is the transpose of the cofactor matrix:

    \[ \text{adj}(A) = \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} \nonumber \]

    Exercise \(\PageIndex{3}\)

    Find the inverse of the matrix

    \[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \nonumber \]

    using the determinant and the adjugate matrix.

    Answer

    First, compute the determinant: \[ \det(A) = (1)(4) - (2)(3) = 4 - 6 = -2. \nonumber\]

    Next, compute the cofactor matrix:

    \[ \text{cof}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \nonumber \]

    Now, compute the adjugate (transpose of the cofactor matrix):

    \[ \text{adj}(A) = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix} \nonumber \]

    Using the formula for the inverse:

    \[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A), \nonumber\]

    we substitute the values:

    \[ A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix} \nonumber\]

    Thus, the inverse is: \[ A^{-1} = \begin{bmatrix} -2 & \frac{3}{2} \\ 1 & -\frac{1}{2} \end{bmatrix}. \nonumber\]

    Exercise \(\PageIndex{4}\)

    Find the inverse of the matrix \[ A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix} \nonumber \]

    using the determinant and the adjugate matrix

    Exercise \(\PageIndex{5}\)

    Find the cofactor matrix for:

    \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 6 & 7 & 8 \end{bmatrix} \nonumber \]

    Exercise \(\PageIndex{6}\)

    Find the inverse of

    \[ A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \nonumber \]

    by computing the cofactor matrix and adjugate.

    Exercise \(\PageIndex{7}\)

    Solve the system

    \[ \begin{cases} 2x + 3y = 5 \\ 4x + y = 6 \end{cases} \nonumber\]

    using Cramer’s Rule.

    Answer

    \(x=\frac{\det(A_x)}{det(A​)}​,y=\frac{\det(A_y)}{det(A​)}​\), yielding \(x = 1, y = 1\)

    Exercise \(\PageIndex{8}\)

    Solve the system

    \[ \begin{cases} x + 2y + 3z = 6 \\ 2x - y + z = 3 \\ 3x + y - 2z = 4 \end{cases} \nonumber\]

    using Cramer’s Rule

    Exercise \(\PageIndex{9}\)

    Find a \( 3 \times 3 \) system where Cramer’s Rule does not apply, and explain why.

    Hint

    What must be true about determinant of the coefficient matrix for Cramer's Rule to work?

    Exercise \(\PageIndex{10}\)

    Show that for any invertible \( 2 \times 2 \) matrix \( A \),

    \[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A). \nonumber\]

    Exercise \(\PageIndex{11}\)

    If \( A \) is an upper triangular \( 3 \times 3 \) matrix, explain how Cramer’s Rule simplifies solving \( A x = b \).

    Exercise \(\PageIndex{12}\)

    In an electrical circuit, Kirchhoff’s Laws lead to the system: \[ \begin{cases} 3I_1 + I_2 - I_3 = 5 \\ 2I_1 + 4I_2 + I_3 = 6 \\ -I_1 + 2I_2 + 5I_3 = 2 \end{cases} \nonumber \] Solve for the currents \( I_1, I_2, I_3 \) using Cramer’s Rule.

    Answer

    Using determinants, \(I_1 = 1, I_2 = 2, I_3 = -1\).

    Exercise \(\PageIndex{13}\)

    If the determinant of a matrix is very small but nonzero, what does this tell us about the matrix?

    Answer

    There are at least two reasons:

    1. While the matrix technically has an inverse, numerical computations involving the inverse may be unstable due to round-off errors.
    2. Even small changes in input (such as rounding errors) can cause large errors in the output when solving systems of equations.
    Exercise \(\PageIndex{14}\)

    Solve the system of equations using Cramer’s Rule:

    \[ \begin{cases} x + y + z + w = 4 \\ 2x - y + 3z + w = 5 \\ x + 2y - z - w = 2 \\ 3x + y + 2z + 4w = 8 \end{cases} \nonumber\]

    Answer

    \(x=2, y=-1, z=3, w=4 \)

    Exercise \(\PageIndex{15}\)

    Consider the system:

    \[ \begin{cases} ax + y = 3 \\ 2x + (a+1)y = 5 \end{cases} \nonumber\]

    1. For what values of \( a \) does the system have no unique solution?
    2. For what values of \( a \) does the system have a unique solution, and what is it in terms of \( a \)?

    This page titled 3.3E: Exercises for Section 3.3 is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Doli Bambhania, Fatemeh Yarahmadi, and Bill Wilson.