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Matrices and Determinants question

2003 · Shift 0 · Q81
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  5. /2003 · Shift 0 · Q81

Matrices and Determinants question

2003 · Shift 0 · Q81

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of linear equations x+2ay+az=0;x+3by+bz=0;  x+4cy+cz=0;x + 2ay + az = 0;x + 3by + bz = 0;\,\,x + 4cy + cz = 0;x+2ay+az=0;x+3by+bz=0;x+4cy+cz=0; has a non - zero solution, then a,b,ca, b, ca,b,c.
  1. A
    satisfy a+2b+3c=0a+2b+3c=0a+2b+3c=0
  2. B
    are in A.P
  3. C
    are in G.P
  4. D
    are in H.P.
View written solutionFree

Correct answer: D

  1. Write the homogeneous system in matrix form

The system is

{x+2ay+az=0x+3by+bz=0x+4cy+cz=0\begin{cases} x+2ay+az=0 \\ x+3by+bz=0 \\ x+4cy+cz=0 \end{cases}⎩⎨⎧​x+2ay+az=0x+3by+bz=0x+4cy+cz=0​

This can be written as

[12aa13bb14cc][xyz]=[000].\begin{bmatrix} 1 & 2a & a\\ 1 & 3b & b\\ 1 & 4c & c \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}.​111​2a3b4c​abc​​​xyz​​=​000​​.
  1. Condition for a non-zero solution

Since this is a homogeneous system, it has a non-zero solution iff the determinant of the coefficient matrix is zero:

∣12aa13bb14cc∣=0.\begin{vmatrix} 1 & 2a & a\\ 1 & 3b & b\\ 1 & 4c & c \end{vmatrix}=0.​111​2a3b4c​abc​​=0.
  1. Evaluate the determinant

Take column operation:

C2→C2−2C3.C_2 \to C_2-2C_3.C2​→C2​−2C3​.

Then the determinant becomes

∣10a1bb12cc∣.\begin{vmatrix} 1 & 0 & a\\ 1 & b & b\\ 1 & 2c & c \end{vmatrix}.​111​0b2c​abc​​.

Now expand along the first row:

Δ=1⋅∣bb2cc∣+a⋅∣1b12c∣.\Delta=1\cdot \begin{vmatrix} b & b\\ 2c & c \end{vmatrix} +a\cdot \begin{vmatrix} 1 & b\\ 1 & 2c \end{vmatrix}.Δ=1⋅​b2c​bc​​+a⋅​11​b2c​​.

Compute each minor:

∣bb2cc∣=bc−2bc=−bc,\begin{vmatrix} b & b\\ 2c & c \end{vmatrix}=bc-2bc=-bc,​b2c​bc​​=bc−2bc=−bc,

and

∣1b12c∣=2c−b.\begin{vmatrix} 1 & b\\ 1 & 2c \end{vmatrix}=2c-b.​11​b2c​​=2c−b.

Hence,

Δ=−bc+a(2c−b).\Delta=-bc+a(2c-b).Δ=−bc+a(2c−b).

So the condition is

a(2c−b)−bc=0.a(2c-b)-bc=0.a(2c−b)−bc=0.

That is,

2ac−ab−bc=0.2ac-ab-bc=0.2ac−ab−bc=0.
  1. Rewrite the relation

Divide by abcabcabc (assuming nonzero parameters; the relation obtained is the intended one):

2acabc−ababc−bcabc=0\frac{2ac}{abc}-\frac{ab}{abc}-\frac{bc}{abc}=0abc2ac​−abcab​−abcbc​=0 2b−1c−1a=0.\frac{2}{b}-\frac{1}{c}-\frac{1}{a}=0.b2​−c1​−a1​=0.

So,

1a+1c=2b.\frac{1}{a}+\frac{1}{c}=\frac{2}{b}.a1​+c1​=b2​.

This means that 1a,1b,1c\dfrac1a,\dfrac1b,\dfrac1ca1​,b1​,c1​ are in A.P.

Therefore a,b,ca,b,ca,b,c are in H.P.

  1. Check options
  • A: a+2b+3c=0a+2b+3c=0a+2b+3c=0 — not implied.
  • B: a,b,ca,b,ca,b,c are in A.P. — not necessary.
  • C: a,b,ca,b,ca,b,c are in G.P. — not necessary.
  • D: a,b,ca,b,ca,b,c are in H.P. — correct.

Therefore, the correct option is

D\boxed{\text{D}}D​
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