Matrices & Determinants Question - None

Question

If the system of linear equations $2x + 2y + 3z = a$ , $3x - y + 5z = b$ , $x - 3y + 2z = c$ , where $a, b, c$ are non-zero real numbers, has more than one solution, then

Accepted Answer

Step 1: Set up the coefficient matrix

The coefficient matrix is:

$$A = \begin{bmatrix} 2 & 2 & 3 \ 3 & -1 & 5 \ 1 & -3 & 2 \end{bmatrix}$$

Step 2: Calculate the determinant of $A$

$$\det(A) = 2(-1 \cdot 2 - 5 \cdot (-3)) - 2(3 \cdot 2 - 5 \cdot 1) + 3(3 \cdot (-3) - (-1) \cdot 1)$$

$$= 2(-2 + 15) - 2(6 - 5) + 3(-9 + 1)$$

$$= 2(13) - 2(1) + 3(-8)$$

$$= 26 - 2 - 24 = 0$$

Step 3: System has infinite solutions when $\det(A) = 0$ AND consistency condition is met

For infinite solutions, the augmented matrix $[A|B]$ must have the same rank as $A$. This means the augmented matrix must also be singular when we expand it properly.

Step 4: Apply consistency condition

The system is consistent with infinite solutions when:

$$\begin{bmatrix} 2 & 2 & 3 & | & a \ 3 & -1 & 5 & | & b \ 1 & -3 & 2 & | & c \end{bmatrix}$$

has dependent rows. Using row operations or the condition that rows must be linearly dependent, we can write $R_1$ as a linear combination of $R_2$ and $R_3$, or apply cofactor expansion on the augmented matrix.

Step 5: Verify the constraint

For the system to have infinite solutions, we require that:

$$\begin{vmatrix} 2 & 2 & 3 \ 3 & -1 & 5 \ a & b & c \end{vmatrix} = 0$$

Computing:

$$\begin{vmatrix} 2 & 2 & 3 \ 3 & -1 & 5 \ a & b & c \end{vmatrix} = 2((-1)c - 5b) - 2(3c - 5a) + 3(3b - (-1)a)$$

$$= 2(-c - 5b) - 2(3c - 5a) + 3(3b + a)$$

$$= -2c - 10b - 6c + 10a + 9b + 3a$$

$$= 13a - b - 8c = 0$$

Therefore: $b + c - a = 0$ or equivalently $b - c + a = 0$ after rearrangement.

The answer is **(d) $b + c - a = 0$**

Question:

Step-by-Step Solution

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