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Matrices and Determinants question

2019 · 8 Apr · Shift 1 · Q39
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  5. /2019 · 8 Apr · Shift 1 · Q39

Matrices and Determinants question

2019 · 8 Apr · Shift 1 · Q39

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The greatest value of c ∈\in∈ R for which the system of linear equations x – cy – cz = 0 cx – y + cz = 0 cx + cy – z = 0 has a non-trivial solution, is :
  1. A
    -1
  2. B
    0
  3. C
    1/2
  4. D
    2
View written solutionFree

Correct answer: C

We need the system to have a non-trivial solution. For a homogeneous system of linear equations, this happens iff the determinant of the coefficient matrix is zero.

1. Write the coefficient matrix

The system is

{x−cy−cz=0cx−y+cz=0cx+cy−z=0\begin{cases} x-cy-cz=0\\ cx-y+cz=0\\ cx+cy-z=0 \end{cases}⎩⎨⎧​x−cy−cz=0cx−y+cz=0cx+cy−z=0​

So the coefficient matrix is

A=(1−c−cc−1ccc−1).A=\begin{pmatrix} 1 & -c & -c\\ c & -1 & c\\ c & c & -1 \end{pmatrix}.A=​1cc​−c−1c​−cc−1​​.

For a non-trivial solution,

det⁡(A)=0.\det(A)=0.det(A)=0.

2. Compute the determinant

Expand along the first row:

det⁡(A)=1∣−1cc−1∣−(−c)∣ccc−1∣+(−c)∣c−1cc∣.\det(A)=1\begin{vmatrix}-1 & c\\ c & -1\end{vmatrix}-(-c)\begin{vmatrix}c & c\\ c & -1\end{vmatrix}+(-c)\begin{vmatrix}c & -1\\ c & c\end{vmatrix}.det(A)=1​−1c​c−1​​−(−c)​cc​c−1​​+(−c)​cc​−1c​​.

Now compute each minor:

First minor

∣−1cc−1∣=(−1)(−1)−c2=1−c2.\begin{vmatrix}-1 & c\\ c & -1\end{vmatrix}=(-1)(-1)-c^2=1-c^2.​−1c​c−1​​=(−1)(−1)−c2=1−c2.

Second minor

∣ccc−1∣=c(−1)−c⋅c=−c−c2.\begin{vmatrix}c & c\\ c & -1\end{vmatrix}=c(-1)-c\cdot c=-c-c^2.​cc​c−1​​=c(−1)−c⋅c=−c−c2.

So its contribution is

−(−c)∣ccc−1∣=c(−c−c2)=−c2−c3.-(-c)\begin{vmatrix}c & c\\ c & -1\end{vmatrix} = c(-c-c^2)=-c^2-c^3.−(−c)​cc​c−1​​=c(−c−c2)=−c2−c3.

But note from cofactor expansion, since the second term is actually

−a12M12=−(−c)M12=cM12,- a_{12} M_{12} = -(-c)M_{12}=cM_{12},−a12​M12​=−(−c)M12​=cM12​,

this is correct.

Third minor

∣c−1cc∣=c⋅c−(−1)c=c2+c.\begin{vmatrix}c & -1\\ c & c\end{vmatrix}=c\cdot c-(-1)c=c^2+c.​cc​−1c​​=c⋅c−(−1)c=c2+c.

So its contribution is

(−c)(c2+c)=−c3−c2.(-c)(c^2+c)=-c^3-c^2.(−c)(c2+c)=−c3−c2.

Hence,

det⁡(A)=(1−c2)+(−c2−c3)+(−c3−c2).\det(A)=(1-c^2)+(-c^2-c^3)+(-c^3-c^2).det(A)=(1−c2)+(−c2−c3)+(−c3−c2).

Therefore,

det⁡(A)=1−3c2−2c3.\det(A)=1-3c^2-2c^3.det(A)=1−3c2−2c3.

So we need

1−3c2−2c3=0.1-3c^2-2c^3=0.1−3c2−2c3=0.

Equivalently,

2c3+3c2−1=0.2c^3+3c^2-1=0.2c3+3c2−1=0.

3. Factor the cubic

Try rational roots. For c=−1c=-1c=−1:

2(−1)3+3(−1)2−1=−2+3−1=0.2(-1)^3+3(-1)^2-1=-2+3-1=0.2(−1)3+3(−1)2−1=−2+3−1=0.

So (c+1)(c+1)(c+1) is a factor.

Divide:

2c3+3c2−1=(c+1)(2c2+c−1).2c^3+3c^2-1=(c+1)(2c^2+c-1).2c3+3c2−1=(c+1)(2c2+c−1).

Now factor the quadratic:

2c2+c−1=(2c−1)(c+1).2c^2+c-1=(2c-1)(c+1).2c2+c−1=(2c−1)(c+1).

Thus,

2c3+3c2−1=(c+1)2(2c−1).2c^3+3c^2-1=(c+1)^2(2c-1).2c3+3c2−1=(c+1)2(2c−1).

So the values of ccc are

c=−1,c=12.c=-1,\quad c=\frac12.c=−1,c=21​.

4. Find the greatest value

Among −1-1−1 and 12\frac1221​, the greatest is

12.\boxed{\frac12}.21​​.

5. Compare with stored answer

Stored correct answer: C

Option C is 12\frac1221​, which matches our result.

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