Differentiation (Part-38) | S N De

Determinants MCQ Solutions (Questions 69 to 79)

Q 69. If \( \Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \), then the value of \( \Delta^2 \) will be—

(a)

\( \begin{vmatrix} ab+bc+ca & a^2+b^2+c^2 & a^2+b^2+c^2 \\ a^2+b^2+c^2 & ab+bc+ca & a^2+b^2+c^2 \\ a^2+b^2+c^2 & a^2+b^2+c^2 & ab+bc+ca \end{vmatrix} \)

(b)

\( \begin{vmatrix} a^2+b^2+c^2 & ab+bc+ca & ab+bc+ca \\ ab+bc+ca & a^2+b^2+c^2 & ab+bc+ca \\ ab+bc+ca & ab+bc+ca & a^2+b^2+c^2 \end{vmatrix} \)

(c)

\( \begin{vmatrix} 2bc-(b^2+c^2) & c^2 & b^2 \\ c^2 & 2ac-(a^2+c^2) & a^2 \\ b^2 & a^2 & 2ab-(a^2+b^2) \end{vmatrix} \)

(d)

None of these

Ans 69.

The square of a determinant can be evaluated by multiplying the determinant by itself (row-by-row multiplication).

$$ \Delta^2 = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \times \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} $$

Evaluating the dot product for each row combination (\( R_i \cdot R_j \)):

\( R_1 \cdot R_1 = a^2 + b^2 + c^2 \)

\( R_1 \cdot R_2 = ab + bc + ca \)

\( R_1 \cdot R_3 = ac + ab + bc = ab + bc + ca \)

Proceeding similarly for all rows, we get:

$$ \Delta^2 = \begin{vmatrix} a^2+b^2+c^2 & ab+bc+ca & ab+bc+ca \\ ab+bc+ca & a^2+b^2+c^2 & ab+bc+ca \\ ab+bc+ca & ab+bc+ca & a^2+b^2+c^2 \end{vmatrix} $$

Correct Option: (b)

Q 70. If \( S_n = a^n + b^n + c^n \), then the value of \( \begin{vmatrix} s_0 & s_1 & s_2 \\ s_1 & s_2 & s_3 \\ s_2 & s_3 & s_4 \end{vmatrix} \) is—

(a)

\( abc(a-b)(b-c)(c-a) \)

(b)

\( abc(a-b)^2(b-c)^2(c-a)^2 \)

(c)

\( (a-b)^2(b-c)^2(c-a)^2 \)

(d)

None of these

Ans 70.

This matrix is the product of a Vandermonde matrix and its transpose. Let \( V = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \). The given determinant equals \( |V \cdot V^T| = |V|^2 \). The value of the Vandermonde determinant \( |V| \) is well-known:

$$ |V| = (a-b)(b-c)(c-a) $$

Therefore, the required determinant is:

$$ |V|^2 = (a-b)^2(b-c)^2(c-a)^2 $$

Correct Option: (c)

💡 Recommended Read for Students

Do It Today: Overcome Procrastination, Improve Productivity and Achieve More Meaningful Things

Struggling to stay focused during your math prep? This life-changing guide will help you beat procrastination, study smarter, and hit your exam goals without the burnout.

View on Amazon 📚

Q 71. Eliminating \( x, y, z \) from the equations \( \frac{1}{x} = \frac{b+c}{bz+cy} \), \( \frac{1}{y} = \frac{c+a}{cx+az} \) and \( \frac{1}{z} = \frac{a+b}{ax+by} \), we get—

(a)

\( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \)

(b)

\( \begin{vmatrix} b+c & c & b \\ c & a+c & a \\ a & b & a+b \end{vmatrix} = 0 \)

(c)

\( \begin{vmatrix} b+c & -c & -b \\ -c & a+c & -a \\ -a & -b & a+b \end{vmatrix} = 0 \)

(d)

None of these

Ans 71.

Cross-multiplying and rearranging the given equations into standard linear forms:

1) \( bz + cy = (b+c)x \implies -(b+c)x + cy + bz = 0 \)

2) \( cx + az = (c+a)y \implies cx - (c+a)y + az = 0 \)

3) \( ax + by = (a+b)z \implies bx + ay - (a+b)z = 0 \)

For non-trivial solutions (to successfully eliminate \( x, y, z \)), the determinant of the coefficients must be zero:

$$ \begin{vmatrix} -(b+c) & c & b \\ c & -(c+a) & a \\ b & a & -(a+b) \end{vmatrix} = 0 $$

Factoring out \(-1\) from each row (which multiplies the determinant by \((-1)^3 = -1\)):

$$ -1 \begin{vmatrix} b+c & -c & -b \\ -c & a+c & -a \\ -b & -a & a+b \end{vmatrix} = 0 \implies \begin{vmatrix} b+c & -c & -b \\ -c & a+c & -a \\ -b & -a & a+b \end{vmatrix} = 0 $$

Correct Option: (c)

Q 72. The value of \( \begin{vmatrix} 3 & 1975 & 1978 \\ 4 & 1982 & 1986 \\ 5 & 1995 & 2000 \end{vmatrix} \) is—

(a)

1

(b)

0

(c)

2

(d)

3

Ans 72.

Apply the column operation \( C_3 \to C_3 - C_2 \):

$$ \begin{vmatrix} 3 & 1975 & (1978 - 1975) \\ 4 & 1982 & (1986 - 1982) \\ 5 & 1995 & (2000 - 1995) \end{vmatrix} = \begin{vmatrix} 3 & 1975 & 3 \\ 4 & 1982 & 4 \\ 5 & 1995 & 5 \end{vmatrix} $$

Since Column 1 (\( C_1 \)) and Column 3 (\( C_3 \)) are now identical, the value of the determinant is \( 0 \).
Correct Option: (b) 0

Q 73. The value of \( \begin{vmatrix} 1 & \omega^2 & \omega \\ \omega & 1 & \omega^2 \\ \omega & \omega & 1 \end{vmatrix} \), [where \( \omega^3 = 1 \)] is—

(a)

-1

(b)

\( \omega^3 \)

(c)

0

(d)

\( -\omega^2 \)

Ans 73.

Expanding the determinant directly along the first row:

$$ \Delta = 1(1 \cdot 1 - \omega^2 \cdot \omega) - \omega^2(\omega \cdot 1 - \omega^2 \cdot \omega) + \omega(\omega \cdot \omega - 1 \cdot \omega) $$

$$ \Delta = 1(1 - \omega^3) - \omega^2(\omega - \omega^3) + \omega(\omega^2 - \omega) $$

Since \( \omega^3 = 1 \):

$$ \Delta = 1(1 - 1) - \omega^2(\omega - 1) + \omega^3 - \omega^2 $$

$$ \Delta = 0 - \omega^3 + \omega^2 + 1 - \omega^2 $$

$$ \Delta = -1 + \omega^2 + 1 - \omega^2 = 0 $$

Correct Option: (c) 0

Q 74. If \( \begin{vmatrix} -5 & 5 & 10 \\ 5 & -5 & x \\ 0 & 10 & 5 \end{vmatrix} = 0 \), then the value of \( x \) will be—

(a)

0

(b)

10

(c)

-10

(d)

1

Ans 74.

Expanding the determinant along the first column (\( C_1 \)):

$$ -5((-5)(5) - 10x) - 5(5(5) - 10(10)) + 0 = 0 $$

$$ -5(-25 - 10x) - 5(25 - 100) = 0 $$

$$ 125 + 50x - 5(-75) = 0 $$

$$ 125 + 50x + 375 = 0 $$

$$ 50x + 500 = 0 \implies 50x = -500 \implies x = -10 $$

Correct Option: (c) -10

Q 75. If \( a+b+c = 0 \), then the value of \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \) will be—

(a)

1

(b)

\( a \)

(c)

0

(d)

-1

Ans 75.

The standard expansion for this circulant determinant is:

$$ -(a^3 + b^3 + c^3 - 3abc) $$

Using the algebraic identity:

$$ a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) $$

Given that \( a+b+c = 0 \), the entire product becomes \( 0 \).

$$ \Delta = -(0) = 0 $$

Correct Option: (c) 0

Q 76. \( A = \begin{pmatrix} \alpha & 2 \\ 2 & \alpha \end{pmatrix} \) is a \( 2 \times 2 \) matrix and \( |A^3| = 125 \), then the value of \( \alpha \) will be—

(a)

1

(b)

3, -3

(c)

3

(d)

-3

Ans 76.

By properties of determinants, \( |A^n| = |A|^n \).

$$ |A^3| = 125 \implies |A|^3 = 5^3 \implies |A| = 5 $$

Calculate the determinant of matrix \( A \):

$$ |A| = (\alpha)(\alpha) - (2)(2) = \alpha^2 - 4 $$

Equating the values:

$$ \alpha^2 - 4 = 5 \implies \alpha^2 = 9 \implies \alpha = \pm 3 $$

Correct Option: (b) 3, -3

Q 77. For the given system of simultaneous equations \( x+y+3z=1, 2x+y+2z=3, 3x+2y+5z=3 \), which of the following is true?

(a)

Unique solution

(b)

Infinite solutions

(c)

Undefined and inconsistent (No solution)

(d)

None of these

Ans 77.

Calculate the determinant of the coefficient matrix (\( \Delta \)):

$$ \Delta = \begin{vmatrix} 1 & 1 & 3 \\ 2 & 1 & 2 \\ 3 & 2 & 5 \end{vmatrix} = 1(5 - 4) - 1(10 - 6) + 3(4 - 3) $$

$$ = 1(1) - 1(4) + 3(1) = 1 - 4 + 3 = 0 $$

Now, calculate \( \Delta_x \) (replacing the first column with constants):

$$ \Delta_x = \begin{vmatrix} 1 & 1 & 3 \\ 3 & 1 & 2 \\ 3 & 2 & 5 \end{vmatrix} = 1(5 - 4) - 1(15 - 6) + 3(6 - 3) $$

$$ = 1(1) - 9 + 3(3) = 1 - 9 + 9 = 1 $$

Since \( \Delta = 0 \) but \( \Delta_x \neq 0 \), by Cramer's Rule, the system is inconsistent and has no solution.
Correct Option: (c) Undefined and inconsistent

Q 78. The value of \( \begin{vmatrix} 9 & 9 & 12 \\ 1 & -3 & -4 \\ 1 & 9 & 12 \end{vmatrix} \) is—

(a)

0

(b)

1

(c)

-1

(d)

2

Ans 78.

Apply row operation \( R_1 \to R_1 - R_3 \):

$$ \Delta = \begin{vmatrix} 9 - 1 & 9 - 9 & 12 - 12 \\ 1 & -3 & -4 \\ 1 & 9 & 12 \end{vmatrix} = \begin{vmatrix} 8 & 0 & 0 \\ 1 & -3 & -4 \\ 1 & 9 & 12 \end{vmatrix} $$

Expanding along the first row:

$$ \Delta = 8((-3)(12) - (-4)(9)) = 8(-36 + 36) = 8(0) = 0 $$

Correct Option: (a) 0

Q 79. If \( \begin{vmatrix} x & c+x & b+x \\ c+x & x & a+x \\ b+x & a+x & x \end{vmatrix} = 0 \), then the value of \( x \) is—

(a)

\( \frac{2abc}{(a-b-c)^2} \)
[Note: Formally \( \frac{2abc}{a^2+b^2+c^2-2(ab+bc+ca)} \)]

(b)

\( \frac{2abc}{(a-b+c)^2} \)

(c)

\( \frac{2abc}{(a+b-c)^2} \)

(d)

\( \frac{2abc}{(a+b+c)^2} \)

Ans 79.

Applying row operations \( R_1 \to R_1 - R_2 \) and \( R_2 \to R_2 - R_3 \) simplifies the matrix. Expanding the determinant directly results in a linear equation in \( x \) (the \( x^3 \) terms cancel out).

The expansion yields:

$$ x(2ab + 2bc + 2ca - a^2 - b^2 - c^2) + 2abc = 0 $$

Solving for \( x \):

$$ x = \frac{-2abc}{2ab + 2bc + 2ca - a^2 - b^2 - c^2} $$

$$ x = \frac{2abc}{a^2 + b^2 + c^2 - 2ab - 2bc - 2ca} $$

*(Note: Option (a) in typical prints represents this exact algebraic denominator).*
Correct Option: (a)