1. **Problem Statement:** Find the inverse of matrix 21 and verify that $M^{-1}M=I$. Matrix 21 is: $$M=\begin{bmatrix}0 & 1 & -3 \\ 2 & -1 & 3 \\ 1 & 0 & 7\end{bmatrix}$$ 2. **Formula for Inverse of a 3x3 Matrix:** The inverse $M^{-1}$ of a matrix $M$ exists if $\det(M) \neq 0$ and is given by: $$M^{-1} = \frac{1}{\det(M)} \mathrm{adj}(M)$$ where $\mathrm{adj}(M)$ is the adjugate matrix (transpose of the cofactor matrix). 3. **Step 1: Calculate $\det(M)$** $$\det(M) = 0 \cdot \begin{vmatrix}-1 & 3 \\ 0 & 7\end{vmatrix} - 1 \cdot \begin{vmatrix}2 & 3 \\ 1 & 7\end{vmatrix} + (-3) \cdot \begin{vmatrix}2 & -1 \\ 1 & 0\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}-1 & 3 \\ 0 & 7\end{vmatrix} = (-1)(7) - (3)(0) = -7$$ $$\begin{vmatrix}2 & 3 \\ 1 & 7\end{vmatrix} = (2)(7) - (3)(1) = 14 - 3 = 11$$ $$\begin{vmatrix}2 & -1 \\ 1 & 0\end{vmatrix} = (2)(0) - (-1)(1) = 0 + 1 = 1$$ So, $$\det(M) = 0 \cdot (-7) - 1 \cdot 11 + (-3) \cdot 1 = 0 - 11 - 3 = -14$$ Since $\det(M) \neq 0$, the inverse exists. 4. **Step 2: Find the Cofactor Matrix $C$** Calculate each cofactor $C_{ij} = (-1)^{i+j} M_{ij}$ where $M_{ij}$ is the minor of element in row $i$, column $j$. - $C_{11} = (+1) \cdot \begin{vmatrix}-1 & 3 \\ 0 & 7\end{vmatrix} = -7$ - $C_{12} = (-1) \cdot \begin{vmatrix}2 & 3 \\ 1 & 7\end{vmatrix} = -11$ - $C_{13} = (+1) \cdot \begin{vmatrix}2 & -1 \\ 1 & 0\end{vmatrix} = 1$ - $C_{21} = (-1) \cdot \begin{vmatrix}1 & -3 \\ 0 & 7\end{vmatrix} = - (1 \cdot 7 - (-3) \cdot 0) = -7$ - $C_{22} = (+1) \cdot \begin{vmatrix}0 & -3 \\ 1 & 7\end{vmatrix} = 0 \cdot 7 - (-3) \cdot 1 = 3$ - $C_{23} = (-1) \cdot \begin{vmatrix}0 & 1 \\ 1 & 0\end{vmatrix} = - (0 \cdot 0 - 1 \cdot 1) = 1$ - $C_{31} = (+1) \cdot \begin{vmatrix}1 & -3 \\ -1 & 3\end{vmatrix} = 1 \cdot 3 - (-3)(-1) = 3 - 3 = 0$ - $C_{32} = (-1) \cdot \begin{vmatrix}0 & -3 \\ 2 & 3\end{vmatrix} = - (0 \cdot 3 - (-3) \cdot 2) = -6$ - $C_{33} = (+1) \cdot \begin{vmatrix}0 & 1 \\ 2 & -1\end{vmatrix} = 0 \cdot (-1) - 1 \cdot 2 = -2$ So, $$C = \begin{bmatrix}-7 & -11 & 1 \\ -7 & 3 & 1 \\ 0 & -6 & -2\end{bmatrix}$$ 5. **Step 3: Find the Adjugate Matrix $\mathrm{adj}(M)$** Transpose the cofactor matrix: $$\mathrm{adj}(M) = C^T = \begin{bmatrix}-7 & -7 & 0 \\ -11 & 3 & -6 \\ 1 & 1 & -2\end{bmatrix}$$ 6. **Step 4: Calculate the Inverse** $$M^{-1} = \frac{1}{-14} \begin{bmatrix}-7 & -7 & 0 \\ -11 & 3 & -6 \\ 1 & 1 & -2\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & \frac{1}{2} & 0 \\ \frac{11}{14} & -\frac{3}{14} & \frac{3}{7} \\ -\frac{1}{14} & -\frac{1}{14} & \frac{1}{7}\end{bmatrix}$$ 7. **Step 5: Verify $M^{-1}M = I$** Multiply $M^{-1}$ by $M$: $$\begin{bmatrix}\frac{1}{2} & \frac{1}{2} & 0 \\ \frac{11}{14} & -\frac{3}{14} & \frac{3}{7} \\ -\frac{1}{14} & -\frac{1}{14} & \frac{1}{7}\end{bmatrix} \begin{bmatrix}0 & 1 & -3 \\ 2 & -1 & 3 \\ 1 & 0 & 7\end{bmatrix} = I$$ Calculations confirm the product is the identity matrix. --- **Summary:** $$M^{-1} = \begin{bmatrix}\frac{1}{2} & \frac{1}{2} & 0 \\ \frac{11}{14} & -\frac{3}{14} & \frac{3}{7} \\ -\frac{1}{14} & -\frac{1}{14} & \frac{1}{7}\end{bmatrix}$$ and $$M^{-1}M = I$$ This completes the solution for problem 21. --- **Note:** The user message contains multiple problems, but per instructions, only the first problem (21) is solved here.