1

Use cofactor expansion to compute: \(\begin{vmatrix} 1 \amp 0 \amp 3 \\ 2 \amp 0 \amp 6 \\ 1 \amp 1 \amp 1 \end{vmatrix}\)

2

Use cofactor expansion to compute: \(\begin{vmatrix} 1 \amp 99 \amp 17 \\ 0 \amp 2 \amp 50 \\ 0 \amp 0 \amp 3 \end{vmatrix}\)

3

Use cofactor expansion to compute: \(\begin{vmatrix} 1 \amp 3 \amp 2 \\ 3 \amp 2 \amp 0 \\ 0 \amp 1 \amp -3 \end{vmatrix}\)

4

Use cofactor expansion to compute: \(\begin{vmatrix} 1 \amp 2 \amp 3 \\ 4 \amp 5 \amp 6 \\ 7 \amp 8 \amp 9 \end{vmatrix}\)

5

Use cofactor expansion to compute: \(\begin{vmatrix} 2 \amp 1 \amp 0 \amp 3 \\ 0 \amp 2 \amp 0 \amp 0 \\ -1 \amp 3 \amp 0 \amp 1 \\ 0 \amp 0 \amp 3 \amp 2 \end{vmatrix}\)

6

Use Cramer's rule (see pages 211—213 in your textbook) to solve the following linear system of equations. \begin{equation*} \left\{ \begin{alignedat}{5} 2x \amp {}-{} \amp 3y \amp {}={} \amp 7 \\ 3x \amp {}-{} \amp 5y \amp {}={} \amp 12 \end{alignedat} \right. \end{equation*}

7

Use Cramer's rule (see pages 211—213 in your textbook) to solve the following linear system of equations. \begin{equation*} \left\{ \begin{alignedat}{7} x \amp {}+{} \amp y \amp {}-{} \amp z \amp {}={} \amp 1 \\ 3x \amp {}-{} \amp y \amp {}+{} \amp 2z \amp {}={} \amp 10 \\ x \amp {}-{} \amp y \amp {}+{} \amp 2z \amp {}={} \amp 6 \end{alignedat} \right. \end{equation*}

8

Let \(A = \begin{bmatrix}a \amp b \\ c \amp d\end{bmatrix}, \quad B = \begin{bmatrix}e \amp f \\ g \amp h\end{bmatrix}.\)

Use these two matrices to prove: \(\abs{AB} = \abs{A} \cdot \abs{B}\)

9

Use the result from the previous exercise to prove: \begin{equation*} \abs{A_1 A_2 A_3} = \abs{A_1} \abs{A_2} \abs{A_3} \end{equation*}

10

Consider the following: \begin{align*} A = \begin{bmatrix}1 \amp 2 \\ 3 \amp 4 \end{bmatrix} \quad \abs{A} = -2\\ A^{-1} = \begin{bmatrix} -2 \amp 1 \\ 3/2 \amp -1/2 \end{bmatrix} \quad \abs{A^{-1}} = -\frac{1}{2} \end{align*} Thus, at least in this particular case we have \begin{equation} \abs{A^{-1}} = \abs{A}^{-1}. \label{men-29}\tag{4.5.1} \end{equation}

Convince yourself that the above equality is always true by doing the following:

1. Encode the three elementary row operations which transform \(A\) to \(I\) as matrices, \(E_1, E_2, E_3\text{,}\) i.e. \begin{equation*} \underbrace{E_3 E_2 E_1}_{A^{-1}} A = I \end{equation*} Hint: The three elementary row operations are:

   • \(E_1: R_2 = R_2 + (-3)R_1\)

   • \(E_2: R_1 = R_1 + R_2\)

   • \(E_3: R_2 = (-1/2)R_2\)

2. Compute \(\abs{E_3}\text{,}\) \(\abs{E_2}\text{,}\) \(\abs{E_1}\text{.}\)

3. Show: \(\abs{A^{-1}} = \abs{E_3} \abs{E_2} \abs{E_1} = \dfrac{\abs{I}}{\abs{A}} = \dfrac{1}{\abs{A}} = \abs{A}^{-1}.\)