Let's solve this matrix problem step-by-step.

The matrix $A$ given is:

$A = \begin{bmatrix} 2 & 0 & 3 \\ 2 & 1 & -2 \\ -3 & 0 & 3 \end{bmatrix}$

We need to solve four parts:

Part (a): Find the Determinant of $A$

To find $\det(A)$ of a 3x3 matrix, use the formula:

$\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$

Where:

$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$

For matrix $A$, we have:

$a = 2, \; b = 0, \; c = 3$ $d = 2, \; e = 1, \; f = -2$ $g = -3, \; h = 0, \; i = 3$

Substitute these values:

$\det(A) = 2(1 \cdot 3 - (-2) \cdot 0) - 0(2 \cdot 3 - (-2) \cdot (-3)) + 3(2 \cdot 0 - 1 \cdot (-3))$ $= 2(3) + 3(3) = 6 + 9 = 15$

So, $\det(A) = 15$.

Part (b): Find the Matrix of Cofactors of $A$

To find the cofactor matrix $C$, calculate the cofactor for each element by removing the row and column of each element and finding the determinant of the resulting 2x2 matrix.

Let's calculate each cofactor:

1. $C_{11} = \det \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} = 1 \cdot 3 - (-2) \cdot 0 = 3$
2. $C_{12} = -\det \begin{bmatrix} 2 & -2 \\ -3 & 3 \end{bmatrix} = -(2 \cdot 3 - (-2) \cdot (-3)) = -(6 - 6) = 0$
3. $C_{13} = \det \begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix} = 2 \cdot 0 - 1 \cdot (-3) = 3$
4. $C_{21} = -\det \begin{bmatrix} 0 & 3 \\ 0 & 3 \end{bmatrix} = 0$
5. $C_{22} = \det \begin{bmatrix} 2 & 3 \\ -3 & 3 \end{bmatrix} = 2 \cdot 3 - 3 \cdot (-3) = 15$
6. $C_{23} = -\det \begin{bmatrix} 2 & 0 \\ -3 & 0 \end{bmatrix} = 0$
7. $C_{31} = \det \begin{bmatrix} 0 & 3 \\ 1 & -2 \end{bmatrix} = 0 \cdot (-2) - 3 \cdot 1 = -3$
8. $C_{32} = -\det \begin{bmatrix} 2 & 3 \\ 2 & -2 \end{bmatrix} = -(2 \cdot (-2) - 3 \cdot 2) = 10$
9. $C_{33} = \det \begin{bmatrix} 2 & 0 \\ 2 & 1 \end{bmatrix} = 2 \cdot 1 - 0 = 2$

Thus, the cofactor matrix $C$ is:

$C = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 15 & 0 \\ -3 & 10 & 2 \end{bmatrix}$

Part (c): Find the Adjoint of $A$

The adjoint of $A$, $\text{adj}(A)$, is the transpose of the cofactor matrix $C$:

$\text{adj}(A) = C^T = \begin{bmatrix} 3 & 0 & -3 \\ 0 & 15 & 10 \\ 3 & 0 & 2 \end{bmatrix}$

Part (d): Find the Inverse of $A$

The inverse of $A$ is given by:

$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$

Since $\det(A) = 15$:

$A^{-1} = \frac{1}{15} \begin{bmatrix} 3 & 0 & -3 \\ 0 & 15 & 10 \\ 3 & 0 & 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{15} & 0 & \frac{-3}{15} \\ 0 & 1 & \frac{10}{15} \\ \frac{3}{15} & 0 & \frac{2}{15} \end{bmatrix}$

Simplify each entry:

$A^{-1} = \begin{bmatrix} \frac{1}{5} & 0 & -\frac{1}{5} \\ 0 & 1 & \frac{2}{3} \\ \frac{1}{5} & 0 & \frac{2}{15} \end{bmatrix}$

Let me know if you would like more details on any step, or if you have questions!

Related Questions

1. What is the significance of the determinant when finding the inverse of a matrix?
2. How can you determine if a matrix is invertible just by looking at its determinant?
3. What are some applications of the adjoint matrix in solving linear systems?
4. How does cofactor expansion help in finding determinants for larger matrices?
5. Can a matrix with zero entries have an inverse? Why or why not?

Tip

For 3x3 matrices, always simplify each minor determinant as you go to keep calculations manageable.