1. **Find the inverse of the matrix** \( M = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 5 & 7 \\ 1 & 2 & 4 \end{bmatrix} \) using elementary row operations. Step 1: Write the augmented matrix \( [M | I] \): $$\left[ \begin{array}{ccc|ccc} 2 & 3 & 4 & 1 & 0 & 0 \\ 3 & 5 & 7 & 0 & 1 & 0 \\ 1 & 2 & 4 & 0 & 0 & 1 \end{array} \right]$$ Step 2: Use row operations to convert the left side to the identity matrix. - Make \( R1 \) leading 1: \( R1 \to \frac{1}{2} R1 \) - Eliminate below and above leading 1 in column 1. Step 3: Continue row operations until left is identity and right is inverse. Final inverse matrix is: $$\begin{bmatrix} 6 & -4 & 1 \\ -5 & 3 & -1 \\ 1 & 0 & 0 \end{bmatrix}$$ --- 2. **Find eigenvalues and eigenvectors of matrix** \( A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 2 & 2 & 4 \end{bmatrix} \). Step 1: Find eigenvalues by solving \( \det(A - \lambda I) = 0 \). $$\det \begin{bmatrix} 1-\lambda & 0 & 0 \\ 1 & 2-\lambda & 0 \\ 2 & 2 & 4-\lambda \end{bmatrix} = 0$$ Step 2: Since matrix is upper-triangular, eigenvalues are diagonal entries: $$\lambda_1 = 1, \quad \lambda_2 = 2, \quad \lambda_3 = 4$$ Step 3: Find eigenvectors for each \( \lambda \): - For \( \lambda=1 \), solve \( (A - I)\mathbf{v} = 0 \). - For \( \lambda=2 \), solve \( (A - 2I)\mathbf{v} = 0 \). - For \( \lambda=4 \), solve \( (A - 4I)\mathbf{v} = 0 \). Eigenvectors: - \( \lambda=1 \): \( \mathbf{v} = t \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \) - \( \lambda=2 \): \( \mathbf{v} = s \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} \) - \( \lambda=4 \): \( \mathbf{v} = r \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \) --- 3. **Solve the system using Gauss-Jordan elimination:** \[ \begin{cases} 3x + y + z + w = 0 \\ 5x - y + z - w = 0 \end{cases} \] Step 1: Write augmented matrix: $$\left[ \begin{array}{cccc|c} 3 & 1 & 1 & 1 & 0 \\ 5 & -1 & 1 & -1 & 0 \end{array} \right]$$ Step 2: Use row operations to reduce to reduced row echelon form. - \( R2 \to R2 - \frac{5}{3} R1 \) Step 3: Solve for variables: - From second row, express \( y, z, w \) in terms of \( x \). General solution: $$x = t, \quad y = 2t, \quad z = -2t, \quad w = t$$ where \( t \) is a free parameter. **Final answers:** - Inverse of \( M \) is \( \begin{bmatrix} 6 & -4 & 1 \\ -5 & 3 & -1 \\ 1 & 0 & 0 \end{bmatrix} \). - Eigenvalues of \( A \) are \( 1, 2, 4 \) with eigenvectors as above. - Solution to system is \( (x,y,z,w) = t(1,2,-2,1) \).