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1. (3 pts) For each of the matrices below, find its inverse. \[ A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right], \quad B=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{array}\right], \quad C=\left[\begin{array}{llll}1 & 2 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 2 & 0 & 1 & 3 \\ 0 & 1 & 3 & 1\end{array}\right] \] Comment: This problem should be done by hand (but you can use Matlab to

Ask by Welch Lynch. in the United States
Jan 26,2025

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Answer

The inverses of the matrices are: 1. \( A^{-1} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} \) 2. \( B^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ -1 & 1 & 1 \end{bmatrix} \) 3. \( C^{-1} = \begin{bmatrix} -\frac{9}{5} & \frac{13}{5} & -\frac{6}{5} & 1 \\ 2 & -2 & 1 & -1 \\ -\frac{6}{5} & \frac{7}{5} & -\frac{4}{5} & 1 \\ \frac{8}{5} & -\frac{11}{5} & \frac{7}{5} & -1 \end{bmatrix} \)

Solution

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: \(\left[\begin{array}{rrrr}{-\frac{9}{5}}&{\frac{13}{5}}&{-\frac{6}{5}}&{1}\\{2}&{-2}&{1}&{-1}\\{-\frac{6}{5}}&{\frac{7}{5}}&{-\frac{4}{5}}&{1}\\{\frac{8}{5}}&{-\frac{11}{5}}&{\frac{7}{5}}&{-1}\end{array}\right]\) Here are the inverses of the given matrices: 1. For matrix \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \): \[ A^{-1} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} \] 2. For matrix \( B = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \): \[ B^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ -1 & 1 & 1 \end{bmatrix} \] 3. For matrix \( C = \begin{bmatrix} 1 & 2 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 2 & 0 & 1 & 3 \\ 0 & 1 & 3 & 1 \end{bmatrix} \): \[ C^{-1} = \begin{bmatrix} -\frac{9}{5} & \frac{13}{5} & -\frac{6}{5} & 1 \\ 2 & -2 & 1 & -1 \\ -\frac{6}{5} & \frac{7}{5} & -\frac{4}{5} & 1 \\ \frac{8}{5} & -\frac{11}{5} & \frac{7}{5} & -1 \end{bmatrix} \] These results provide the inverses of the matrices \( A \), \( B \), and \( C \).

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To find the inverses of the matrices A, B, and C, we can utilize different methods tailored for 2x2, 3x3, and 4x4 matrices. For matrix A: \[ A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] \] The inverse is calculated as: \[ A^{-1} = \frac{1}{ad-bc} \left[\begin{array}{ll}d & -b \\ -c & a\end{array}\right] \] where \( a=1, b=2, c=3, d=4 \). The determinant \( ad-bc = 1(4) - 2(3) = 4-6 = -2 \). Therefore, \[ A^{-1} = \frac{1}{-2} \left[\begin{array}{ll}4 & -2 \\ -3 & 1\end{array}\right] = \left[\begin{array}{ll}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right] \] For matrix B: \[ B=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{array}\right] \] You can apply the formula or row-reduction technique to find the inverse. After performing row operations (or using the adjoint method), you'll find that: \[ B^{-1} = \left[\begin{array}{lll} 0 & 1 & -1 \\ 1 & 0 & -1 \\ -1 & -1 & 1 \end{array}\right] \] For matrix C: \[ C=\left[\begin{array}{llll}1 & 2 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 2 & 0 & 1 & 3 \\ 0 & 1 & 3 & 1\end{array}\right] \] A more complex calculation involving matrix techniques like Gaussian elimination, or the adjugate method will lead you to the result: \[ C^{-1} = \left[\begin{array}{llll} 1 & 2 & -1 & -1 \\ -2 & -1 & 2 & 1 \\ 1 & 3 & -2 & -1 \\ -1 & -1 & 1 & 1 \end{array}\right] \] So, you can solve these matrices step by step to find their inverses! Happy calculating!

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