Question

consider the matrix $a = \

$$\begin{bmatrix} 4 & -16 \\\\ 0 & -1 \\end{bmatrix}$$

$. what is the determinant of $a$? select the inverse of $a$ can be written as $a^{-1} = \

$$\begin{bmatrix} x_1 & x_2 \\\\ x_3 & x_4 \\end{bmatrix}$$

$. what are the values of $x_1$ and $x_2$? $x_1 =$ select $x_2 =$ select

Explanation:

Step1: Calculate determinant of A

For a 2x2 matrix \(

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

\), determinant is \( ad - bc \).
For \( A =

$$\begin{bmatrix} 4 & -16 \\ 0 & -1 \end{bmatrix}$$

\), determinant \( = (4)(-1) - (-16)(0) = -4 - 0 = -4 \).

Step2: Find inverse of A

The inverse of a 2x2 matrix \(

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

\) is \( \frac{1}{\text{det}(A)}

$$\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

\).
We know \( \text{det}(A) = -4 \), \( a = 4, b = -16, c = 0, d = -1 \).
So \( A^{-1} = \frac{1}{-4}

$$\begin{bmatrix} -1 & 16 \\ 0 & 4 \end{bmatrix}$$

=

$$\begin{bmatrix} \frac{-1}{-4} & \frac{16}{-4} \\ \frac{0}{-4} & \frac{4}{-4} \end{bmatrix}$$

=

$$\begin{bmatrix} \frac{1}{4} & -4 \\ 0 & -1 \end{bmatrix}$$

\).
Thus, \( x_1 = \frac{1}{4} \), \( x_2 = -4 \).

Answer:

Determinant of \( A \): \(-4\)
\( x_1 = \frac{1}{4} \), \( x_2 = -4 \)