1. **Solve the system** given by the augmented matrix \(\begin{bmatrix}1 & 3 & 1 & 0 \\ 2 & 6 & 4 & 8 \\ 0 & 0 & 2 & 4\end{bmatrix} \begin{bmatrix}x \\ y \\ z \\ t\end{bmatrix} = \begin{bmatrix}1 \\ 3\end{bmatrix}\). Note: The system is inconsistent as the matrix dimensions do not match the vector on the right side (3x4 matrix times 4x1 vector should equal 3x1 vector, but right side is 2x1). Assuming the right side vector is \(\begin{bmatrix}1 \\ 3 \\ 0\end{bmatrix}\) to match dimensions. 2. Write the system of equations: \[ \begin{cases} 1x + 3y + 1z + 0t = 1 \\ 2x + 6y + 4z + 8t = 3 \\ 0x + 0y + 2z + 4t = 0 \end{cases} \] 3. From the third equation: \(2z + 4t = 0 \Rightarrow z = -2t\). 4. Substitute \(z = -2t\) into the first and second equations: \[ \begin{cases} x + 3y - 2t = 1 \\ 2x + 6y + 4(-2t) + 8t = 3 \end{cases} \] Simplify second equation: \[ 2x + 6y - 8t + 8t = 3 \Rightarrow 2x + 6y = 3 \] 5. From the first equation: \[ x + 3y = 1 + 2t \] 6. Multiply the first equation by 2: \[ 2x + 6y = 2 + 4t \] 7. Compare with the second equation \(2x + 6y = 3\), so: \[ 2 + 4t = 3 \Rightarrow 4t = 1 \Rightarrow t = \frac{1}{4} \] 8. Substitute \(t = \frac{1}{4}\) back into \(x + 3y = 1 + 2t\): \[ x + 3y = 1 + 2 \times \frac{1}{4} = 1 + \frac{1}{2} = \frac{3}{2} \] 9. Express \(x = \frac{3}{2} - 3y\). 10. Recall \(z = -2t = -2 \times \frac{1}{4} = -\frac{1}{2}\). **Solution:** \[ \begin{cases} x = \frac{3}{2} - 3y \\ z = -\frac{1}{2} \\ t = \frac{1}{4} \\ y \text{ is free} \end{cases} \] --- **Proof of equivalence of statements about invertible matrix \(A\):** 1. (a) \(\Rightarrow\) (b): If \(A\) is invertible, then \(\exists A^{-1}\) such that \(AA^{-1} = I_n\). Applying Gaussian elimination to \(A\) reduces it to \(I_n\), so the reduced row echelon form (RREF) of \(A\) is \(I_n\). 2. (b) \(\Rightarrow\) (c): If the RREF of \(A\) is \(I_n\), then \(A\) can be transformed into \(I_n\) by a sequence of elementary row operations. Each elementary row operation corresponds to multiplication by an elementary matrix. Hence, \(A\) is a product of elementary matrices. 3. (c) \(\Rightarrow\) (a): If \(A = E_1 E_2 \cdots E_k\) where each \(E_i\) is an elementary matrix, and since each \(E_i\) is invertible, their product \(A\) is invertible. Thus, (a), (b), and (c) are equivalent. --- **Proof that if \(A\) is invertible, then \(A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)\):** 1. Recall the definition of the adjugate matrix \(\operatorname{adj}(A)\) as the transpose of the cofactor matrix. 2. The matrix identity: \[ A \times \operatorname{adj}(A) = \operatorname{adj}(A) \times A = \det(A) I_n \] 3. Since \(A\) is invertible, \(\det(A) \neq 0\), so multiply both sides by \(\frac{1}{\det(A)}\): \[ A \times \frac{1}{\det(A)} \operatorname{adj}(A) = I_n \] 4. By definition of inverse, this means: \[ A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) \] --- **Summary:** - The system has infinitely many solutions parameterized by \(y\). - The three statements about invertibility are equivalent. - The formula for the inverse matrix in terms of adjugate and determinant holds for invertible matrices.