1. Key Concepts and Formulas

For a system of linear equations in the form $AX=B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix:

• Let $D$ be the determinant of the coefficient matrix $A$.
• Let $D_x, D_y, D_z$ be the determinants obtained by replacing the first, second, and third columns of $A$ respectively with the constant terms from matrix $B$.

The system of linear equations has infinitely many solutions if and only if:

1. $D = 0$
2. $D_x = 0$
3. $D_y = 0$
4. $D_z = 0$

Important Note: If $D=0$ but at least one of $D_x, D_y, D_z$ is non-zero, the system has no solution. If $D \neq 0$, the system has a unique solution.

2. Step-by-Step Solution

Step 1: Set up the Coefficient Matrix and its Determinant (D) The given system of equations is: $\begin{aligned} & (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\ & \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\ & (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9 \end{aligned}$ The coefficient matrix $A$ is: $A = \begin{pmatrix} \lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2) \end{pmatrix}$ For infinitely many solutions, the determinant $D$ must be zero. $D = \left|\begin{array}{ccc} \lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2) \end{array}\right|$

Step 2: Calculate D and Find Potential Values of $\lambda$ To simplify the determinant calculation, we use row operations. These operations do not change the value of the determinant. First, apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ to introduce zeros: $D = \left|\begin{array}{ccc} \lambda-1 & \lambda-4 & \lambda \\ \lambda - (\lambda-1) & (\lambda-1) - (\lambda-4) & (\lambda-4) - \lambda \\ (\lambda+1) - (\lambda-1) & (\lambda+2) - (\lambda-4) & -(\lambda+2) - \lambda \end{array}\right|$ $D = \left|\begin{array}{ccc} \lambda-1 & \lambda-4 & \lambda \\ 1 & 3 & -4 \\ 2 & 6 & -2\lambda-2 \end{array}\right|$ Next, apply $R_3 \to R_3 - 2R_2$ to introduce more zeros in the third row: $D = \left|\begin{array}{ccc} \lambda-1 & \lambda-4 & \lambda \\ 1 & 3 & -4 \\ 2 - 2(1) & 6 - 2(3) & (-2\lambda-2) - 2(-4) \end{array}\right|$ $D = \left|\begin{array}{ccc} \lambda-1 & \lambda-4 & \lambda \\ 1 & 3 & -4 \\ 0 & 0 & -2\lambda+6 \end{array}\right|$ Now, expand the determinant along the third row: $D = (-2\lambda+6) \cdot \left|\begin{array}{cc} \lambda-1 & \lambda-4 \\ 1 & 3 \end{array}\right|$ $D = -2(\lambda-3) [3(\lambda-1) - 1(\lambda-4)]$ $D = -2(\lambda-3) [3\lambda - 3 - \lambda + 4]$ $D = -2(\lambda-3) [2\lambda + 1]$ For infinitely many solutions, $D=0$: $-2(\lambda-3)(2\lambda+1) = 0$ This gives two possible values for $\lambda$: $\lambda = 3$ or $\lambda = -1/2$.

Step 3: Calculate $D_x$ and Verify $\lambda$ $D_x$ is obtained by replacing the first column of $A$ with the constant terms $(5, 7, 9)^T$. $D_x = \left|\begin{array}{ccc} 5 & \lambda-4 & \lambda \\ 7 & \lambda-1 & \lambda-4 \\ 9 & \lambda+2 & -(\lambda+2) \end{array}\right|$ Apply the same row operations: $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $D_x = \left|\begin{array}{ccc} 5 & \lambda-4 & \lambda \\ 2 & 3 & -4 \\ 4 & 6 & -2\lambda-2 \end{array}\right|$ Apply $R_3 \to R_3 - 2R_2$: $D_x = \left|\begin{array}{ccc} 5 & \lambda-4 & \lambda \\ 2 & 3 & -4 \\ 0 & 0 & -2\lambda+6 \end{array}\right|$ Expand along the third row: $D_x = (-2\lambda+6) \cdot \left|\begin{array}{cc} 5 & \lambda-4 \\ 2 & 3 \end{array}\right|$ $D_x = -2(\lambda-3) [5(3) - 2(\lambda-4)]$ $D_x = -2(\lambda-3) [15 - 2\lambda + 8]$ $D_x = -2(\lambda-3) [23 - 2\lambda]$ For infinitely many solutions, $D_x=0$: $-2(\lambda-3)(23-2\lambda) = 0$ This gives $\lambda = 3$ or $\lambda = 23/2$. Comparing the values for $\lambda$ from $D=0$ ($\lambda=3, -1/2$) and $D_x=0$ ($\lambda=3, 23/2$), the only common value is $\lambda=3$. This means $\lambda=3$ is the only candidate for infinitely many solutions.

Step 4: Verify $D_y=0$ for $\lambda=3$ $D_y$ is obtained by replacing the second column of $A$ with $(5, 7, 9)^T$. $D_y = \left|\begin{array}{ccc} \lambda-1 & 5 & \lambda \\ \lambda & 7 & \lambda-4 \\ \lambda+1 & 9 & -(\lambda+2) \end{array}\right|$ Substitute $\lambda=3$: $D_y = \left|\begin{array}{ccc} 3-1 & 5 & 3 \\ 3 & 7 & 3-4 \\ 3+1 & 9 & -(3+2) \end{array}\right| = \left|\begin{array}{ccc} 2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & 9 & -5 \end{array}\right|$ Expand the determinant: $D_y = 2(7(-5) - (-1)(9)) - 5(3(-5) - (-1)(4)) + 3(3(9) - 7(4))$ $D_y = 2(-35 + 9) - 5(-15 + 4) + 3(27 - 28)$ $D_y = 2(-26) - 5(-11) + 3(-1)$ $D_y = -52 + 55 - 3 = 0$ So, $D_y=0$ for $\lambda=3$.

Step 5: Verify $D_z=0$ for $\lambda=3$ $D_z$ is obtained by replacing the third column of $A$ with $(5, 7, 9)^T$. $D_z = \left|\begin{array}{ccc} \lambda-1 & \lambda-4 & 5 \\ \lambda & \lambda-1 & 7 \\ \lambda+1 & \lambda+2 & 9 \end{array}\right|$ Substitute $\lambda=3$: $D_z = \left|\begin{array}{ccc} 3-1 & 3-4 & 5 \\ 3 & 3-1 & 7 \\ 3+1 & 3+2 & 9 \end{array}\right| = \left|\begin{array}{ccc} 2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{array}\right|$ Expand the determinant: $D_z = 2(2(9) - 7(5)) - (-1)(3(9) - 7(4)) + 5(3(5) - 2(4))$ $D_z = 2(18 - 35) + 1(27 - 28) + 5(15 - 8)$ $D_z = 2(-17) + 1(-1) + 5(7)$ $D_z = -34 - 1 + 35 = 0$ So, $D_z=0$ for $\lambda=3$.

Step 6: Calculate the Final Expression Since $D=D_x=D_y=D_z=0$ for $\lambda=3$, the system has infinitely many solutions when $\lambda=3$. The question asks for the value of $\lambda^2+\lambda$. Substitute $\lambda=3$: $\lambda^2+\lambda = (3)^2 + 3 = 9 + 3 = 12$

3. Common Mistakes & Tips

• Determinant Calculation Errors: Be extremely careful with arithmetic and signs, especially when expanding determinants or performing row/column operations. Double-check calculations.
• Incomplete Conditions: Remember that for infinitely many solutions in a non-homogeneous system, all determinants ($D, D_x, D_y, D_z$) must be zero. Checking only $D=0$ is insufficient; it could lead to no solution.
• Systematic Approach: Using row/column operations to simplify determinants before expansion is often less error-prone than direct expansion for larger matrices. Aim to create rows or columns with many zeros.

4. Summary

We began by outlining the conditions for a system of linear equations to have infinitely many solutions, which require all determinants ($D, D_x, D_y, D_z$) to be zero. We then calculated the determinant of the coefficient matrix, $D$, and found the possible values of $\lambda$ for which $D=0$. Next, we calculated $D_x$ and identified the common value of $\lambda$ that made both $D=0$ and $D_x=0$. This unique candidate was $\lambda=3$. Finally, we verified that for $\lambda=3$, both $D_y$ and $D_z$ were also zero, confirming that $\lambda=3$ is indeed the value for which the system has infinitely many solutions. Substituting $\lambda=3$ into the expression $\lambda^2+\lambda$ yielded the final answer.

The final answer is $\boxed{\text{12}}$, which corresponds to option (D).