Differentiation (Part-38) | S N De

Group-A: General MCQ (Expansion of Determinant Related)

Q 1. If \( A = \begin{bmatrix} 1 & 2 & 5 \\ 0 & 0 & 0 \\ -1 & 3 & 2 \end{bmatrix} \), then the value of \( |A| \) will be—

(a) 10

(b) 0

(c) -6

(d) 1

Ans 1.

The second row of the matrix \( A \) consists entirely of zeros.
Expanding along the second row:

$$ |A| = -0 + 0 - 0 = 0 $$

Since an entire row is zero, the value of the determinant is \( 0 \).
Correct Option: (b) 0

Q 2. If \( A = \begin{bmatrix} 3 & 0 & 1 \\ 1 & 0 & 5 \\ 2 & 0 & 7 \end{bmatrix} \), then the value of \( |A| \) will be—

(a) 0

(b) 6

(c) 35

(d) -1

Ans 2.

The second column of matrix \( A \) contains only zeros.
Expanding along the second column:

$$ |A| = 0 $$

If all elements of any row or column in a matrix are zero, its determinant value is \( 0 \).
Correct Option: (a) 0

Q 3. If \( I_3 \) is an identity matrix of order \( 3 \times 3 \), then the value of \( |I_3| \) will be—

(a) 1

(b) 0

(c) -1

(d) None of these

Ans 3.

The identity matrix of order 3 is:

$$ I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Its determinant is:

$$ |I_3| = 1(1 - 0) - 0 + 0 = 1 $$

The determinant of any identity matrix is always \( 1 \).
Correct Option: (a) 1

Q 4. If \( D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{bmatrix} \), then \( |D| = ? \)

(a) 0

(b) 1

(c) 15

(d) 9

Ans 4.

Matrix \( D \) is a diagonal matrix. The determinant of a diagonal matrix is the product of its main diagonal elements:

$$ |D| = 3 \times 1 \times 5 = 15 $$

Correct Option: (c) 15

Q 5. If \( A = \begin{bmatrix} 2 & 1 & -2 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{bmatrix} \), then the value of \( |A| \) will be—

(a) 0

(b) 1

(c) 12

(d) -1

Ans 5.

Matrix \( A \) is an upper triangular matrix. The determinant of an upper triangular matrix equals the product of its diagonal elements:

$$ |A| = 2 \times 2 \times 3 = 12 $$

Correct Option: (c) 12

Q 6. If \( C = \begin{bmatrix} -1 & 0 & 0 \\ 2 & 3 & 0 \\ 0 & 1 & 5 \end{bmatrix} \), then the value of \( |C| \) will be—

(a) 15

(b) 0

(c) 7

(d) -15

Ans 6.

Matrix \( C \) is a lower triangular matrix. The determinant of a triangular matrix is the product of its diagonal entries:

$$ |C| = (-1) \times 3 \times 5 = -15 $$

Correct Option: (d) -15

Q 7. If \( X = \begin{bmatrix} x-y & 0 & 0 \\ 0 & x^2 & 0 \\ 2(x-y) & 0 & x^2+xy+y^2 \end{bmatrix} \), then the value of \( |X| \) will be—

(a) \( x^3 - y^3 \)

(b) \( 3(x - y) \)

(c) \( x^2(x^3 - y^3) \)

(d) \( x^2(x - y) \)

Ans 7.

Notice that matrix \( X \) is a lower triangular matrix (all entries above the main diagonal are zero).
Therefore, its determinant is the product of its main diagonal elements:

$$ |X| = (x - y) \cdot x^2 \cdot (x^2 + xy + y^2) $$

Rearranging the factors:

$$ |X| = x^2 \cdot [(x - y)(x^2 + xy + y^2)] $$

Using the algebraic formula \( (x - y)(x^2 + xy + y^2) = x^3 - y^3 \):

$$ |X| = x^2(x^3 - y^3) $$

Correct Option: (c) \( x^2(x^3 - y^3) \)

Q 8. If \( A \) is a square matrix of order \( 3 \times 3 \), then the value of \( \det(kA) \) will be—

(a) \( k^2 \det(A) \)

(b) \( k^3 \det(A) \)

(c) \( k \det(A) \)

(d) \( (k^2 - k) \det(A) \)

Ans 8.

For any square matrix \( A \) of order \( n \times n \), scaling the matrix by a scalar \( k \) scales every row by \( k \).
Hence, the determinant property gives:

$$ \det(kA) = k^n \det(A) $$

Here, order \( n = 3 \), so:

$$ \det(kA) = k^3 \det(A) $$

Correct Option: (b) \( k^3 \det(A) \)

Q 9. If \( A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \), then the value of \( \det \begin{bmatrix} 2a_1 & 2b_1 & 2c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \) will be—

(a) \( 2 \det(A) \)

(b) \( 4 \det(A) \)

(c) \( 8 \det(A) \)

(d) \( \det(A) \)

Ans 9.

Let \( B = \begin{bmatrix} 2a_1 & 2b_1 & 2c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \).
Taking the common factor \( 2 \) out from the first row (\( R_1 \)):

$$ \det(B) = 2 \times \det \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} = 2 \det(A) $$

Correct Option: (a) \( 2 \det(A) \)

Q 10. The value of the determinant \( \begin{vmatrix} 1 & 2 & 5 \\ -1 & -2 & -5 \\ 3 & 5 & 7 \end{vmatrix} \) is—

(a) 10

(b) 1

(c) 0

(d) -1

Ans 10.

Let \( \Delta = \begin{vmatrix} 1 & 2 & 5 \\ -1 & -2 & -5 \\ 3 & 5 & 7 \end{vmatrix} \).
Taking out \( -1 \) as a factor from the second row (\( R_2 \)):

$$ \Delta = (-1) \begin{vmatrix} 1 & 2 & 5 \\ 1 & 2 & 5 \\ 3 & 5 & 7 \end{vmatrix} $$

Since row 1 (\( R_1 \)) and row 2 (\( R_2 \)) are identical, the value of the determinant is \( 0 \):

$$ \Delta = (-1) \times 0 = 0 $$

Correct Option: (c) 0

Q 11. If \( A \) is a skew-symmetric matrix of order 3, then \( \det A \) is—

(a) 0

(b) 1

(c) A perfect square

(d) None of these

Ans 11.

By definition of a skew-symmetric matrix, \( A^T = -A \).
Taking the determinant on both sides:

$$ \det(A^T) = \det(-A) $$

Since \( \det(A^T) = \det(A) \) and \( A \) is of order \( n = 3 \), we have \( \det(-A) = (-1)^3 \det(A) \):

$$ \det(A) = -\det(A) $$

$$ 2 \det(A) = 0 \implies \det(A) = 0 $$

The determinant of an odd-order skew-symmetric matrix is always zero.
Correct Option: (a) 0

Q 12. \( \begin{vmatrix} 0 & b & -c \\ -b & 0 & a \\ c & -a & 0 \end{vmatrix} = \)

(a) \( a + b + c \)

(b) \( abc \)

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

(d) 0

Ans 12.

Let \( \Delta = \begin{vmatrix} 0 & b & -c \\ -b & 0 & a \\ c & -a & 0 \end{vmatrix} \).
Notice that the corresponding matrix is a skew-symmetric matrix of order 3 (since \( a_{ij} = -a_{ji} \) and diagonal elements are zero).
Expanding the determinant directly:

$$ \Delta = 0(0 + a^2) - b(0 - ac) + (-c)(ab - 0) $$

$$ \Delta = 0 + abc - abc = 0 $$

Correct Option: (d) 0

Q 13. If \( \omega \) is an imaginary cube root of unity, then \( \begin{vmatrix} 1 & \omega^2 & \omega \\ \omega & 1 & \omega^2 \\ \omega^2 & \omega & 1 \end{vmatrix} = \)

(a) -1

(b) \( \omega^3 \)

(c) 0

(d) \( -\omega^2 \)

Ans 13.

Apply column operation \( C_1 \to C_1 + C_2 + C_3 \):

$$ \begin{vmatrix} 1+\omega^2+\omega & \omega^2 & \omega \\ \omega+1+\omega^2 & 1 & \omega^2 \\ \omega^2+\omega+1 & \omega & 1 \end{vmatrix} $$

We know that for the imaginary cube root of unity, \( 1 + \omega + \omega^2 = 0 \).
Substituting this value into the first column:

$$ \begin{vmatrix} 0 & \omega^2 & \omega \\ 0 & 1 & \omega^2 \\ 0 & \omega & 1 \end{vmatrix} = 0 $$

Since all entries in the first column are zero, the determinant equals \( 0 \).
Correct Option: (c) 0

Q 14. \( \begin{vmatrix} 1 & \omega & 1+\omega \\ 1+\omega & 1 & \omega \\ \omega & 1+\omega & 1 \end{vmatrix} = \)

(a) 4

(b) 2

(c) 0

(d) 3

Ans 14.

Apply column operation \( C_1 \to C_1 + C_2 + C_3 \):

\( \text{First element of } C_1 = 1 + \omega + (1+\omega) = 2 + 2\omega \)

\( \text{Second element of } C_1 = (1+\omega) + 1 + \omega = 2 + 2\omega \)

\( \text{Third element of } C_1 = \omega + (1+\omega) + 1 = 2 + 2\omega \)

The determinant becomes:

$$ \begin{vmatrix} 2(1+\omega) & \omega & 1+\omega \\ 2(1+\omega) & 1 & \omega \\ 2(1+\omega) & 1+\omega & 1 \end{vmatrix} $$

Taking \( 2(1+\omega) \) common from column 1 (\( C_1 \)):

$$ = 2(1+\omega) \begin{vmatrix} 1 & \omega & 1+\omega \\ 1 & 1 & \omega \\ 1 & 1+\omega & 1 \end{vmatrix} $$

Now apply row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):

$$ = 2(1+\omega) \begin{vmatrix} 1 & \omega & 1+\omega \\ 0 & 1-\omega & -1 \\ 0 & 1 & -\omega \end{vmatrix} $$

Expanding along column 1 (\( C_1 \)):

$$ = 2(1+\omega) \left[ 1 \cdot ((1-\omega)(-\omega) - (-1)(1)) \right] $$

$$ = 2(1+\omega) \left[ -\omega + \omega^2 + 1 \right] $$

Since \( 1 + \omega^2 = -\omega \):

$$ = 2(1+\omega) \left[ -\omega - \omega \right] = 2(1+\omega)(-2\omega) = -4\omega(1+\omega) $$

$$ = -4(\omega + \omega^2) $$

Since \( 1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1 \):

$$ = -4(-1) = 4 $$

Correct Option: (a) 4