1. (3 pts) For each of the matrices below, find its inverse. $$ A=\left[\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right], B=\left[\begin{array}{lll}1 & 1 & 1 \ 0 & 1 & 1 \ 1 & 0 & 1\end{array}\right], \quad C=\left[\begin{array}{llll}1 & 2 & 1 & 0 \ 2 & 1 & 0 & 1 \ 2 & 0 & 1 & 3 \ 0 & 1 & 3 & 1\end{array}\right] $$ Comment: This problem should be done by hand (but you can use Matlab to check your 2. (2 pts) Express the matrix $$ B $$ above as a product of elementary matrices. Hint: You can/should refer to your work on Problem 1 to help with this.

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Matrices by following steps:
- step0: Find the matrix inverse:
  $$\left[\begin{array}{rr}{1}&{2}\{3}&{4}\end{array}\right]$$
- step1: Evaluate the determinant:
  $$-2$$
- step2: Using the formula:
  $$\frac{1}{1\times 4-2\times 3}\times \left[\begin{array}{rr}{4}&{-2}\{-3}&{1}\end{array}\right]$$
- step3: Evaluate:
  $$-\frac{1}{2}\times \left[\begin{array}{rr}{4}&{-2}\{-3}&{1}\end{array}\right]$$
- step4: Multiply the terms:
  $$\left[\begin{array}{rr}{-\frac{1}{2}\times 4}&{-\frac{1}{2}\left(-2\right)}\{-\frac{1}{2}\left(-3\right)}&{-\frac{1}{2}\times 1}\end{array}\right]$$
- step5: Evaluate:
  $$\left[\begin{array}{rr}{-2}&{1}\{\frac{3}{2}}&{-\frac{1}{2}}\end{array}\right]$$
Find the inverse matrix of $$ \begin{bmatrix}1 & 1 & 1 \ 0 & 1 & 1 \ 1 & 0 & 1\end{bmatrix} $$.
Matrices by following steps:
- step0: Find the matrix inverse:
  $$\left[\begin{array}{rrr}{1}&{1}&{1}\{0}&{1}&{1}\{1}&{0}&{1}\end{array}\right]$$
- step1: Begin by adjoining the identity matrix to form the matrix:
  $$\left[\begin{array}{rrr|rrr}{1}&{1}&{1}&{1}&{0}&{0}\{0}&{1}&{1}&{0}&{1}&{0}\{1}&{0}&{1}&{0}&{0}&{1}\end{array}\right]$$
- step2: Simplify the row:
  $$\left[\begin{array}{rrr|rrr}{1}&{1}&{1}&{1}&{0}&{0}\{0}&{1}&{1}&{0}&{1}&{0}\{0}&{-1}&{0}&{-1}&{0}&{1}\end{array}\right]$$
- step3: Simplify the row:
  $$\left[\begin{array}{rrr|rrr}{1}&{1}&{1}&{1}&{0}&{0}\{0}&{1}&{1}&{0}&{1}&{0}\{0}&{0}&{1}&{-1}&{1}&{1}\end{array}\right]$$
- step4: Simplify the row:
  $$\left[\begin{array}{rrr|rrr}{1}&{1}&{1}&{1}&{0}&{0}\{0}&{1}&{0}&{1}&{0}&{-1}\{0}&{0}&{1}&{-1}&{1}&{1}\end{array}\right]$$
- step5: Simplify the row:
  $$\left[\begin{array}{rrr|rrr}{1}&{1}&{0}&{2}&{-1}&{-1}\{0}&{1}&{0}&{1}&{0}&{-1}\{0}&{0}&{1}&{-1}&{1}&{1}\end{array}\right]$$
- step6: Simplify the row:
  $$\left[\begin{array}{rrr|rrr}{1}&{0}&{0}&{1}&{-1}&{0}\{0}&{1}&{0}&{1}&{0}&{-1}\{0}&{0}&{1}&{-1}&{1}&{1}\end{array}\right]$$
- step7: Write the right half matrix:
  $$\left[\begin{array}{rrr}{1}&{-1}&{0}\{1}&{0}&{-1}\{-1}&{1}&{1}\end{array}\right]$$
Find the inverse matrix of $$ \begin{bmatrix}1 & 2 & 1 & 0 \ 2 & 1 & 0 & 1 \ 2 & 0 & 1 & 3 \ 0 & 1 & 3 & 1\end{bmatrix} $$.
Matrices by following steps:
- step0: Find the matrix inverse:
  $$\left[\begin{array}{rrrr}{1}&{2}&{1}&{0}\{2}&{1}&{0}&{1}\{2}&{0}&{1}&{3}\{0}&{1}&{3}&{1}\end{array}\right]$$
- step1: Begin by adjoining the identity matrix to form the matrix:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{2}&{1}&{0}&{1}&{0}&{1}&{0}&{0}\{2}&{0}&{1}&{3}&{0}&{0}&{1}&{0}\{0}&{1}&{3}&{1}&{0}&{0}&{0}&{1}\end{array}\right]$$
- step2: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{-3}&{-2}&{1}&{-2}&{1}&{0}&{0}\{2}&{0}&{1}&{3}&{0}&{0}&{1}&{0}\{0}&{1}&{3}&{1}&{0}&{0}&{0}&{1}\end{array}\right]$$
- step3: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{-3}&{-2}&{1}&{-2}&{1}&{0}&{0}\{0}&{-4}&{-1}&{3}&{-2}&{0}&{1}&{0}\{0}&{1}&{3}&{1}&{0}&{0}&{0}&{1}\end{array}\right]$$
- step4: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{1}&{\frac{2}{3}}&{-\frac{1}{3}}&{\frac{2}{3}}&{-\frac{1}{3}}&{0}&{0}\{0}&{-4}&{-1}&{3}&{-2}&{0}&{1}&{0}\{0}&{1}&{3}&{1}&{0}&{0}&{0}&{1}\end{array}\right]$$
- step5: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{1}&{\frac{2}{3}}&{-\frac{1}{3}}&{\frac{2}{3}}&{-\frac{1}{3}}&{0}&{0}\{0}&{0}&{\frac{5}{3}}&{\frac{5}{3}}&{\frac{2}{3}}&{-\frac{4}{3}}&{1}&{0}\{0}&{1}&{3}&{1}&{0}&{0}&{0}&{1}\end{array}\right]$$
- step6: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{1}&{\frac{2}{3}}&{-\frac{1}{3}}&{\frac{2}{3}}&{-\frac{1}{3}}&{0}&{0}\{0}&{0}&{\frac{5}{3}}&{\frac{5}{3}}&{\frac{2}{3}}&{-\frac{4}{3}}&{1}&{0}\{0}&{0}&{\frac{7}{3}}&{\frac{4}{3}}&{-\frac{2}{3}}&{\frac{1}{3}}&{0}&{1}\end{array}\right]$$
- step7: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{1}&{\frac{2}{3}}&{-\frac{1}{3}}&{\frac{2}{3}}&{-\frac{1}{3}}&{0}&{0}\{0}&{0}&{1}&{1}&{\frac{2}{5}}&{-\frac{4}{5}}&{\frac{3}{5}}&{0}\{0}&{0}&{\frac{7}{3}}&{\frac{4}{3}}&{-\frac{2}{3}}&{\frac{1}{3}}&{0}&{1}\end{array}\right]$$
- step8: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{1}&{\frac{2}{3}}&{-\frac{1}{3}}&{\frac{2}{3}}&{-\frac{1}{3}}&{0}&{0}\{0}&{0}&{1}&{1}&{\frac{2}{5}}&{-\frac{4}{5}}&{\frac{3}{5}}&{0}\{0}&{0}&{0}&{-1}&{-\frac{8}{5}}&{\frac{11}{5}}&{-\frac{7}{5}}&{1}\end{array}\right]$$
- step9: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{1}&{\frac{2}{3}}&{-\frac{1}{3}}&{\frac{2}{3}}&{-\frac{1}{3}}&{0}&{0}\{0}&{0}&{1}&{1}&{\frac{2}{5}}&{-\frac{4}{5}}&{\frac{3}{5}}&{0}\{0}&{0}&{0}&{1}&{\frac{8}{5}}&{-\frac{11}{5}}&{\frac{7}{5}}&{-1}\end{array}\right]$$
- step10: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{1}&{\frac{2}{3}}&{-\frac{1}{3}}&{\frac{2}{3}}&{-\frac{1}{3}}&{0}&{0}\{0}&{0}&{1}&{0}&{-\frac{6}{5}}&{\frac{7}{5}}&{-\frac{4}{5}}&{1}\{0}&{0}&{0}&{1}&{\frac{8}{5}}&{-\frac{11}{5}}&{\frac{7}{5}}&{-1}\end{array}\right]$$
- step11: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{1}&{\frac{2}{3}}&{0}&{\frac{6}{5}}&{-\frac{16}{15}}&{\frac{7}{15}}&{-\frac{1}{3}}\{0}&{0}&{1}&{0}&{-\frac{6}{5}}&{\frac{7}{5}}&{-\frac{4}{5}}&{1}\{0}&{0}&{0}&{1}&{\frac{8}{5}}&{-\frac{11}{5}}&{\frac{7}{5}}&{-1}\end{array}\right]$$
- step12: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{1}&{0}&{1}&{0}&{0}&{0}\{0}&{1}&{0}&{0}&{2}&{-2}&{1}&{-1}\{0}&{0}&{1}&{0}&{-\frac{6}{5}}&{\frac{7}{5}}&{-\frac{4}{5}}&{1}\{0}&{0}&{0}&{1}&{\frac{8}{5}}&{-\frac{11}{5}}&{\frac{7}{5}}&{-1}\end{array}\right]$$
- step13: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{2}&{0}&{0}&{\frac{11}{5}}&{-\frac{7}{5}}&{\frac{4}{5}}&{-1}\{0}&{1}&{0}&{0}&{2}&{-2}&{1}&{-1}\{0}&{0}&{1}&{0}&{-\frac{6}{5}}&{\frac{7}{5}}&{-\frac{4}{5}}&{1}\{0}&{0}&{0}&{1}&{\frac{8}{5}}&{-\frac{11}{5}}&{\frac{7}{5}}&{-1}\end{array}\right]$$
- step14: Simplify the row:
  $$\left[\begin{array}{rrrr|rrrr}{1}&{0}&{0}&{0}&{-\frac{9}{5}}&{\frac{13}{5}}&{-\frac{6}{5}}&{1}\{0}&{1}&{0}&{0}&{2}&{-2}&{1}&{-1}\{0}&{0}&{1}&{0}&{-\frac{6}{5}}&{\frac{7}{5}}&{-\frac{4}{5}}&{1}\{0}&{0}&{0}&{1}&{\frac{8}{5}}&{-\frac{11}{5}}&{\frac{7}{5}}&{-1}\end{array}\right]$$
- step15: Write the right half matrix:
  $$\left[\begin{array}{rrrr}{-\frac{9}{5}}&{\frac{13}{5}}&{-\frac{6}{5}}&{1}\{2}&{-2}&{1}&{-1}\{-\frac{6}{5}}&{\frac{7}{5}}&{-\frac{4}{5}}&{1}\{\frac{8}{5}}&{-\frac{11}{5}}&{\frac{7}{5}}&{-1}\end{array}\right]$$
### Problem 1: Finding Inverses of Matrices

1. **Matrix A:**
   $$
   A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}
   $$
   The inverse of matrix $$ A $$ is:
   $$
   A^{-1} = \begin{bmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}
   $$

2. **Matrix B:**
   $$
   B = \begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 1 & 0 & 1 \end{bmatrix}
   $$
   The inverse of matrix $$ B $$ is:
   $$
   B^{-1} = \begin{bmatrix} 1 & -1 & 0 \ 1 & 0 & -1 \ -1 & 1 & 1 \end{bmatrix}
   $$

3. **Matrix C:**
   $$
   C = \begin{bmatrix} 1 & 2 & 1 & 0 \ 2 & 1 & 0 & 1 \ 2 & 0 & 1 & 3 \ 0 & 1 & 3 & 1 \end{bmatrix}
   $$
   The inverse of matrix $$ C $$ is:
   $$
   C^{-1} = \begin{bmatrix} -\frac{9}{5} & \frac{13}{5} & -\frac{6}{5} & 1 \ 2 & -2 & 1 & -1 \ -\frac{6}{5} & \frac{7}{5} & -\frac{4}{5} & 1 \ \frac{8}{5} & -\frac{11}{5} & \frac{7}{5} & -1 \end{bmatrix}
   $$

### Problem 2: Expressing Matrix B as a Product of Elementary Matrices

To express matrix $$ B $$ as a product of elementary matrices, we can perform row operations to transform the identity matrix into $$ B $$. The elementary matrices correspond to these row operations.

1. Start with the identity matrix $$ I $$:
   $$
   I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}
   $$

2. Perform the following row operations to obtain $$ B $$:
   - **R1 + R2 → R2**: This operation adds the first row to the second row.
   - **R1 + R3 → R3**: This operation adds the first row to the third row.

The corresponding elementary matrices for these operations are:
- For $$ R1 + R2 $$:
  $$
  E_1 = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}
  $$
- For $$ R1 + R3 $$:
  $$
  E_2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix}
  $$

Thus, we can express $$ B $$ as:
$$
B = E_2 E_1 I
$$

In conclusion, the matrix $$ B $$ can be expressed as a product of elementary matrices corresponding to the row operations performed.
Matrix inverses and elementary matrix expressions for the given matrices are as follows: 1. **Matrix A:** $$ A^{-1} = \begin{bmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} $$ 2. **Matrix B:** $$ B^{-1} = \begin{bmatrix} 1 & -1 & 0 \ 1 & 0 & -1 \ -1 & 1 & 1 \end{bmatrix} $$ Matrix $$ B $$ can be expressed as: $$ B = E_2 E_1 I $$ where $$ E_1 $$ and $$ E_2 $$ are elementary matrices corresponding to the row operations: - $$ R1 + R2 $$ - $$ R1 + R3 $$ 3. **Matrix C:** $$ C^{-1} = \begin{bmatrix} -\frac{9}{5} & \frac{13}{5} & -\frac{6}{5} & 1 \ 2 & -2 & 1 & -1 \ -\frac{6}{5} & \frac{7}{5} & -\frac{4}{5} & 1 \ \frac{8}{5} & -\frac{11}{5} & \frac{7}{5} & -1 \end{bmatrix} $$

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