1. **State the problem:** We have a system of linear equations: $$\begin{cases} 5x_1 + 2x_2 + 5x_3 + 2x_4 = 2 \\ -20x_1 - 3x_2 - 13x_3 - 8x_4 = -8 \\ -15x_1 - 6x_2 - 15x_3 - 6x_4 = -6 \end{cases}$$ We want to analyze the system using Kronecker's Rank Theorem. 2. **Part (a): Find the rank of coefficient matrix $C$.** The coefficient matrix $C$ is: $$C = \begin{bmatrix} 5 & 2 & 5 & 2 \\ -20 & -3 & -13 & -8 \\ -15 & -6 & -15 & -6 \end{bmatrix}$$ We need to find a maximal square submatrix $M$ of $C$ with nonzero determinant. - Try $3 \times 3$ submatrices (since $C$ is $3 \times 4$, we can only take $3 \times 3$ submatrices by removing one column). - For example, remove the 4th column: $$M = \begin{bmatrix} 5 & 2 & 5 \\ -20 & -3 & -13 \\ -15 & -6 & -15 \end{bmatrix}$$ Calculate $\det(M)$: $$\det(M) = 5 \begin{vmatrix} -3 & -13 \\ -6 & -15 \end{vmatrix} - 2 \begin{vmatrix} -20 & -13 \\ -15 & -15 \end{vmatrix} + 5 \begin{vmatrix} -20 & -3 \\ -15 & -6 \end{vmatrix}$$ Calculate each minor: - $\begin{vmatrix} -3 & -13 \\ -6 & -15 \end{vmatrix} = (-3)(-15) - (-13)(-6) = 45 - 78 = -33$ - $\begin{vmatrix} -20 & -13 \\ -15 & -15 \end{vmatrix} = (-20)(-15) - (-13)(-15) = 300 - 195 = 105$ - $\begin{vmatrix} -20 & -3 \\ -15 & -6 \end{vmatrix} = (-20)(-6) - (-3)(-15) = 120 - 45 = 75$ So, $$\det(M) = 5(-33) - 2(105) + 5(75) = -165 - 210 + 375 = 0$$ Try removing the 3rd column: $$M = \begin{bmatrix} 5 & 2 & 2 \\ -20 & -3 & -8 \\ -15 & -6 & -6 \end{bmatrix}$$ Calculate $\det(M)$: $$\det(M) = 5 \begin{vmatrix} -3 & -8 \\ -6 & -6 \end{vmatrix} - 2 \begin{vmatrix} -20 & -8 \\ -15 & -6 \end{vmatrix} + 2 \begin{vmatrix} -20 & -3 \\ -15 & -6 \end{vmatrix}$$ Calculate minors: - $\begin{vmatrix} -3 & -8 \\ -6 & -6 \end{vmatrix} = (-3)(-6) - (-8)(-6) = 18 - 48 = -30$ - $\begin{vmatrix} -20 & -8 \\ -15 & -6 \end{vmatrix} = (-20)(-6) - (-8)(-15) = 120 - 120 = 0$ - $\begin{vmatrix} -20 & -3 \\ -15 & -6 \end{vmatrix} = 120 - 45 = 75$ So, $$\det(M) = 5(-30) - 2(0) + 2(75) = -150 + 0 + 150 = 0$$ Try removing the 2nd column: $$M = \begin{bmatrix} 5 & 5 & 2 \\ -20 & -13 & -8 \\ -15 & -15 & -6 \end{bmatrix}$$ Calculate $\det(M)$: $$\det(M) = 5 \begin{vmatrix} -13 & -8 \\ -15 & -6 \end{vmatrix} - 5 \begin{vmatrix} -20 & -8 \\ -15 & -6 \end{vmatrix} + 2 \begin{vmatrix} -20 & -13 \\ -15 & -15 \end{vmatrix}$$ Calculate minors: - $\begin{vmatrix} -13 & -8 \\ -15 & -6 \end{vmatrix} = (-13)(-6) - (-8)(-15) = 78 - 120 = -42$ - $\begin{vmatrix} -20 & -8 \\ -15 & -6 \end{vmatrix} = 120 - 120 = 0$ - $\begin{vmatrix} -20 & -13 \\ -15 & -15 \end{vmatrix} = 300 - 195 = 105$ So, $$\det(M) = 5(-42) - 5(0) + 2(105) = -210 + 0 + 210 = 0$$ Try removing the 1st column: $$M = \begin{bmatrix} 2 & 5 & 2 \\ -3 & -13 & -8 \\ -6 & -15 & -6 \end{bmatrix}$$ Calculate $\det(M)$: $$\det(M) = 2 \begin{vmatrix} -13 & -8 \\ -15 & -6 \end{vmatrix} - 5 \begin{vmatrix} -3 & -8 \\ -6 & -6 \end{vmatrix} + 2 \begin{vmatrix} -3 & -13 \\ -6 & -15 \end{vmatrix}$$ Calculate minors: - $\begin{vmatrix} -13 & -8 \\ -15 & -6 \end{vmatrix} = -42$ - $\begin{vmatrix} -3 & -8 \\ -6 & -6 \end{vmatrix} = -30$ - $\begin{vmatrix} -3 & -13 \\ -6 & -15 \end{vmatrix} = (-3)(-15) - (-13)(-6) = 45 - 78 = -33$ So, $$\det(M) = 2(-42) - 5(-30) + 2(-33) = -84 + 150 - 66 = 0$$ All $3 \times 3$ submatrices have zero determinant, so rank of $C$ is less than 3. Check $2 \times 2$ submatrices for rank 2: For example, submatrix formed by first two rows and first two columns: $$\begin{vmatrix} 5 & 2 \\ -20 & -3 \end{vmatrix} = 5(-3) - 2(-20) = -15 + 40 = 25 \neq 0$$ So rank of $C$ is at least 2. Check if rank is 3? No, as all $3 \times 3$ determinants are zero. Therefore, \textbf{rank of $C$ is 2}. 3. **Part (b): Find the rank of augmented matrix $A$.** Augmented matrix $A$ is: $$A = \begin{bmatrix} 5 & 2 & 5 & 2 \\ -20 & -3 & -13 & -8 \\ -15 & -6 & -15 & -6 \end{bmatrix}$$ We check $3 \times 3$ submatrices including the augmented column. Try submatrix formed by first two rows and last three columns: $$N = \begin{bmatrix} 2 & 5 & 2 \\ -3 & -13 & -8 \\ -6 & -15 & -6 \end{bmatrix}$$ Calculate $\det(N)$: $$\det(N) = 2 \begin{vmatrix} -13 & -8 \\ -15 & -6 \end{vmatrix} - 5 \begin{vmatrix} -3 & -8 \\ -6 & -6 \end{vmatrix} + 2 \begin{vmatrix} -3 & -13 \\ -6 & -15 \end{vmatrix}$$ From previous calculations, this determinant is 0. Try other $3 \times 3$ submatrices including the augmented column, all determinants are zero (similar to above). Check $2 \times 2$ submatrices including augmented column: For example, submatrix: $$\begin{vmatrix} 5 & 2 \\ -20 & -8 \end{vmatrix} = 5(-8) - 2(-20) = -40 + 40 = 0$$ Try another: $$\begin{vmatrix} 2 & 2 \\ -3 & -8 \end{vmatrix} = 2(-8) - 2(-3) = -16 + 6 = -10 \neq 0$$ So rank of $A$ is at least 2. Since rank of $C$ is 2 and rank of $A$ is also 2, the ranks are equal. 4. **Part (c): Is the system consistent or inconsistent?** By Kronecker's Rank Theorem, the system is consistent if and only if rank($C$) = rank($A$). Here, rank($C$) = rank($A$) = 2, so the system is \textbf{consistent}. 5. **Part (d): Which statement is true?** Number of variables $n = 4$. Since rank($C$) = 2 < 4, the system has infinitely many solutions (because the number of free variables is $4 - 2 = 2$). **Summary:** - Rank of $C$ is 2. - Rank of $A$ is 2. - The system is consistent. - The system has infinitely many solutions. **Final answers:** - a) Rank of $C$ = 2 - b) Rank of $A$ = 2 - c) The system is consistent. - d) The system has infinitely many solutions because rank $<$ number of variables.