Question

Consider the system of linear equations

$x + 2y + z = 1$, $2x + y + z = a$, $4x + 5y + 3z = a^2$

Then the system has

• A. Infinitely many solutions when a = -1 or 2
• B. Infinitely many solutions when a = -2 or 1
• C. Unique solution $\forall$ a $\in$ R - {-1, 1, 2}
• D. Unique solution when a $\in$ {-1, 1, 2}

Answer 1

Answer: (A)

Infinitely many solutions when a = -1 or 2

Answer 2

The given system of linear equations is:

1. $x + 2y + z = 1$
2. $2x + y + z = a$
3. $4x + 5y + 3z = a^2$

We can represent this system in an augmented matrix form and use Gaussian elimination to determine the nature of its solutions.

The augmented matrix is: $M = \begin{bmatrix} 1 & 2 & 1 & | & 1 \\ 2 & 1 & 1 & | & a \\ 4 & 5 & 3 & | & a^2 \end{bmatrix}$

Step 1: Calculate the determinant of the coefficient matrix (D). The coefficient matrix is $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 4 & 5 & 3 \end{bmatrix}$. $D = \det(A) = 1(1 \times 3 - 1 \times 5) - 2(2 \times 3 - 1 \times 4) + 1(2 \times 5 - 1 \times 4)$ $D = 1(3 - 5) - 2(6 - 4) + 1(10 - 4)$ $D = 1(-2) - 2(2) + 1(6)$ $D = -2 - 4 + 6 = 0$

Since the determinant $D=0$, the system does not have a unique solution. This eliminates options C and D. The system will either have infinitely many solutions or no solution.

Step 2: Apply Gaussian elimination to the augmented matrix. Perform row operations to transform the matrix into row echelon form:

$M = \begin{bmatrix} 1 & 2 & 1 & | & 1 \\ 2 & 1 & 1 & | & a \\ 4 & 5 & 3 & | & a^2 \end{bmatrix}$

$R_2 \rightarrow R_2 - 2R_1$: $\begin{bmatrix} 1 & 2 & 1 & | & 1 \\ 2 - 2(1) & 1 - 2(2) & 1 - 2(1) & | & a - 2(1) \\ 4 & 5 & 3 & | & a^2 \end{bmatrix}$ $\sim \begin{bmatrix} 1 & 2 & 1 & | & 1 \\ 0 & -3 & -1 & | & a - 2 \\ 4 & 5 & 3 & | & a^2 \end{bmatrix}$

$R_3 \rightarrow R_3 - 4R_1$: $\begin{bmatrix} 1 & 2 & 1 & | & 1 \\ 0 & -3 & -1 & | & a - 2 \\ 4 - 4(1) & 5 - 4(2) & 3 - 4(1) & | & a^2 - 4(1) \end{bmatrix}$ $\sim \begin{bmatrix} 1 & 2 & 1 & | & 1 \\ 0 & -3 & -1 & | & a - 2 \\ 0 & -3 & -1 & | & a^2 - 4 \end{bmatrix}$

$R_3 \rightarrow R_3 - R_2$: $\begin{bmatrix} 1 & 2 & 1 & | & 1 \\ 0 & -3 & -1 & | & a - 2 \\ 0 - 0 & -3 - (-3) & -1 - (-1) & | & (a^2 - 4) - (a - 2) \end{bmatrix}$ $\sim \begin{bmatrix} 1 & 2 & 1 & | & 1 \\ 0 & -3 & -1 & | & a - 2 \\ 0 & 0 & 0 & | & a^2 - a - 2 \end{bmatrix}$

Step 3: Analyze the condition for infinitely many solutions. For the system to have infinitely many solutions, the last row of the augmented matrix must be entirely zeros (i.e., $0 = 0$). This means the expression in the last entry of the last row must be zero: $a^2 - a - 2 = 0$

Solve this quadratic equation for $a$: $(a - 2)(a + 1) = 0$

This yields two possible values for $a$: $a = 2$ or $a = -1$.

If $a = 2$ or $a = -1$, the last row becomes $0 \ 0 \ 0 \ | \ 0$, indicating that the system is consistent and has infinitely many solutions (since the rank of the coefficient matrix is 2, and the rank of the augmented matrix is also 2, which is less than the number of variables, 3).

If $a \neq 2$ and $a \neq -1$, then $a^2 - a - 2 \neq 0$. In this case, the last row would be $0 \ 0 \ 0 \ | \ \text{non-zero constant}$, which implies $0 = \text{non-zero constant}$. This is a contradiction, meaning the system has no solution.

Therefore, the system has infinitely many solutions when $a = -1$ or $a = 2$.