Question

If the system of linear equations: \[ x + 2ay + az = 0 \] \[ x + 3by + bz = 0 \] \[ x + 4cy + cz = 0 \] has a non-zero solution, then $a, b, c$ are:

• A. satisfy $a + 2b + 3c = 0$
• B. in A.P.
• C. in G.P.
• D. in H.P.

Hint:

If $\frac1a, \frac1b, \frac1c$ are in A.P., then $a, b, c$ are in H.P. — use determinants to check for non-trivial solutions.

Answer 1

The Correct Option is D

Step 1: Write the system in matrix form: \[ \beginbmatrix 1 & 2a & a
1 & 3b & b
1 & 4c & c \endbmatrix \beginbmatrix x
y
z \endbmatrix = \beginbmatrix 0
0
0 \endbmatrix \] Step 2: For a non-trivial (non-zero) solution, the determinant of the coefficient matrix must be zero: \[ \beginvmatrix 1 & 2a & a
1 & 3b & b
1 & 4c & c \endvmatrix = 0 \] Step 3: Evaluate determinant: Expanding along the first column: \[ 1 \times \beginvmatrix 3b & b
4c & c \endvmatrix - 1 \times \beginvmatrix 2a & a
4c & c \endvmatrix + 1 \times \beginvmatrix 2a & a
3b & b \endvmatrix = 0 \] Step 4: Calculate each: First: $(3b)(c) - (b)(4c) = 3bc - 4bc = -bc$
Second: $(2a)(c) - (a)(4c) = 2ac - 4ac = -2ac$
Third: $(2a)(b) - (a)(3b) = 2ab - 3ab = -ab$
Step 5: Substituting into expansion: \[ (-bc) - (-2ac) + (-ab) = -bc + 2ac - ab = 0 \] Step 6: Rearrange: \[ 2ac - ab - bc = 0 \] Divide through by $abc$: \[ \frac2b - \frac1c - \frac1a = 0 \] Step 7: Rearranged: \[ \frac1a , \frac1b , \frac1c \ \textare in A.P. \] Which means $a, b, c$ are in $\mathbfH.P.$ (Harmonic Progression).