Q: If M= bmatrix 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 bmatrix , then what is the value of |M|\,| adj M| ?

Given M = bmatrix 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 bmatrix The determinant of the diagonal matrix M is |M| = 2 2 2 = 8. For a square matrix of order n, the determinant of its adjoint is given by | adj M| = |M|^ n-1 . Since the order of M is n = 3, we have | adj M| = |M|^ 3-1 = |M|^2. The required value is |M| \, | adj M| = |M| |M|^2 = |M|^3. Substituting |M| = 8, we get 8^3 = 512. Answer: 512

Q: If vmatrix a-b & p-q & x-y \\ b-c & q-r & y-z \\ c-a & r-p & z-x vmatrix =k vmatrix a & b & c \\ p & q & r \\ x & y & z vmatrix , then what is the value of k ?

Let = vmatrix a-b & p-q & x-y \\ b-c & q-r & y-z \\ c-a & r-p & z-x vmatrix Applying the row operation R_1 R_1 + R_2 + R_3, we get: = vmatrix (a-b)+(b-c)+(c-a) & (p-q)+(q-r)+(r-p) & (x-y)+(y-z)+(z-x) \\ b-c & q-r & y-z \\ c-a & r-p & z-x vmatrix = vmatrix 0 & 0 & 0 \\ b-c & q-r & y-z \\ c-a & r-p & z-x vmatrix Since all elements of the first row are zero, the value of the determinant is 0. Comparing this with k vmatrix a & b & c \\ p & q & r \\ x & y & z vmatrix , we get k = 0.

Q: The value of the determinant vmatrix a & b & c \\ l & m & n \\ p & q & r vmatrix is equal to

Let = vmatrix a & b & c \\ l & m & n \\ p & q & r vmatrix Interchanging the first and third rows (R_1 R_3), the sign of the determinant changes: = - vmatrix p & q & r \\ l & m & n \\ a & b & c vmatrix Now, interchanging the second and third rows (R_2 R_3), the sign changes again: = - ( - vmatrix p & q & r \\ a & b & c \\ l & m & n vmatrix ) = vmatrix p & q & r \\ a & b & c \\ l & m & n vmatrix Answer: vmatrix p & q & r \\ a & b & c \\ l & m & n vmatrix

Q: If a, b, c are the sides of a triangle ABC and p is the perimeter of the triangle, then what is vmatrix p+c & a & b \\ c & p+a & b \\ c & a & p+b vmatrix equal to?

Let = vmatrix p+c & a & b \\ c & p+a & b \\ c & a & p+b vmatrix Applying the column operation C_1 C_1 + C_2 + C_3: = vmatrix p+a+b+c & a & b \\ p+a+b+c & p+a & b \\ p+a+b+c & a & p+b vmatrix Since p = a+b+c is the perimeter of the triangle, we have p+a+b+c = 2p. Substituting this into the determinant: = vmatrix 2p & a & b \\ 2p & p+a & b \\ 2p & a & p+b vmatrix Taking 2p common from the first column C_1: = 2p vmatrix 1 & a & b \\ 1 & p+a & b \\ 1 & a & p+b vmatrix Applying the row operations R_2

Question 1

Let p=(x+y+z) and q=xyz. If vmatrix x & 1 & 1 \ 1 & y & 1 \ 1 & 1 & z vmatrix is positive, then which one of the following is correct ?

Accepted Answer

Expanding the given determinant along the first row and setting it greater than zero: x(yz - 1) - 1(z - 1) + 1(1 - y) > 0 xyz - x - z + 1 + 1 - y > 0 xyz - (x + y + z) + 2 > 0 Substituting p = x + y + z and q = xyz: q - p + 2 > 0 q + 2 > p Answer: q+2>p

Question 2

Let f(x)= vmatrix 3x^2 & x & - x \ 6 & -1 & 0 \ q & q^2 & q^3 vmatrix where q is any constant, then what is d^2 dx^2 (f(x) ) at x=0 equal to ?

Accepted Answer

Given the function f(x) as a determinant where only the first row contains the variable x: f(x)= vmatrix 3x^2 & x & - x \ 6 & -1 & 0 \ q & q^2 & q^3 vmatrix To differentiate the determinant with respect to x, we differentiate the elements of the first row while keeping the other rows unchanged. Differentiating once with respect to x: f'(x)= vmatrix 6x & - x & - x \ 6 & -1 & 0 \ q & q^2 & q^3 vmatrix Differentiating again with respect to x: f''(x)= vmatrix 6 & - x & x \ 6 & -1 & 0 \ q & q^2

Question 3

If M= bmatrix 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 bmatrix , then what is the value of |M|\,| adj M| ?

Accepted Answer

Given M = bmatrix 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 bmatrix The determinant of the diagonal matrix M is |M| = 2 2 2 = 8. For a square matrix of order n, the determinant of its adjoint is given by | adj M| = |M|^ n-1 . Since the order of M is n = 3, we have | adj M| = |M|^ 3-1 = |M|^2. The required value is |M| \, | adj M| = |M| |M|^2 = |M|^3. Substituting |M| = 8, we get 8^3 = 512. Answer: 512

Question 4

If vmatrix a-b & p-q & x-y \ b-c & q-r & y-z \ c-a & r-p & z-x vmatrix =k vmatrix a & b & c \ p & q & r \ x & y & z vmatrix , then what is the value of k ?

Accepted Answer

Let = vmatrix a-b & p-q & x-y \ b-c & q-r & y-z \ c-a & r-p & z-x vmatrix Applying the row operation R_1 R_1 + R_2 + R_3, we get: = vmatrix (a-b)+(b-c)+(c-a) & (p-q)+(q-r)+(r-p) & (x-y)+(y-z)+(z-x) \ b-c & q-r & y-z \ c-a & r-p & z-x vmatrix = vmatrix 0 & 0 & 0 \ b-c & q-r & y-z \ c-a & r-p & z-x vmatrix Since all elements of the first row are zero, the value of the determinant is 0. Comparing this with k vmatrix a & b & c \ p & q & r \ x & y & z vmatrix , we get k = 0.

Question 5

If A is a square matrix such that |A|=-2, then |AA^ T |, where A^ T is the transpose of A, is equal to

Accepted Answer

Using the property of determinants, |AB| = |A||B|. Therefore, |AA^ T | = |A||A^ T |. Since the determinant of a matrix is equal to the determinant of its transpose, |A^ T | = |A|. Substituting this into the equation gives |AA^ T | = |A||A| = |A|^2. Given |A| = -2, we get |AA^ T | = (-2)^2 = 4. Answer: 4

Question 6

The system of equations 2x - 3y - 5 = 0, 15y - 10x + 50 = 0

Accepted Answer

The given system of equations is: 2x - 3y - 5 = 0 -10x + 15y + 50 = 0 Comparing with a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0, we have: a_1 = 2, b_1 = -3, c_1 = -5 a_2 = -10, b_2 = 15, c_2 = 50 Calculating the ratios of the coefficients: a_1 a_2 = 2 -10 = - 1 5 b_1 b_2 = -3 15 = - 1 5 c_1 c_2 = -5 50 = - 1 10 Since a_1 a_2 = b_1 b_2 c_1 c_2 , the lines represented by the equations are parallel and distinct. Thus, the system of equations has no solution and is inconsistent.

Question 7

The value of the determinant vmatrix a & b & c \ l & m & n \ p & q & r vmatrix is equal to

Accepted Answer

Let = vmatrix a & b & c \ l & m & n \ p & q & r vmatrix Interchanging the first and third rows (R_1 R_3), the sign of the determinant changes: = - vmatrix p & q & r \ l & m & n \ a & b & c vmatrix Now, interchanging the second and third rows (R_2 R_3), the sign changes again: = - ( - vmatrix p & q & r \ a & b & c \ l & m & n vmatrix ) = vmatrix p & q & r \ a & b & c \ l & m & n vmatrix Answer: vmatrix p & q & r \ a & b & c \ l & m & n vmatrix

Question 8

If a, b, c are the sides of a triangle ABC and p is the perimeter of the triangle, then what is vmatrix p+c & a & b \ c & p+a & b \ c & a & p+b vmatrix equal to?

Accepted Answer

Let = vmatrix p+c & a & b \ c & p+a & b \ c & a & p+b vmatrix Applying the column operation C_1 C_1 + C_2 + C_3: = vmatrix p+a+b+c & a & b \ p+a+b+c & p+a & b \ p+a+b+c & a & p+b vmatrix Since p = a+b+c is the perimeter of the triangle, we have p+a+b+c = 2p. Substituting this into the determinant: = vmatrix 2p & a & b \ 2p & p+a & b \ 2p & a & p+b vmatrix Taking 2p common from the first column C_1: = 2p vmatrix 1 & a & b \ 1 & p+a & b \ 1 & a & p+b vmatrix Applying the row operations R_2

Question 9

Consider the following statements: Statement-I: If X is an n n matrix, then (mX) = m^n (X), where m is a scalar. Statement-II: If Y is a matrix obtained from X by multiplying any row or column by a sc

Accepted Answer

Statement-I states that for an n n matrix X, (mX) = m^n (X). This is a standard property of determinants and is correct. Statement-II states that if a single row or column of a matrix is multiplied by a scalar m, the determinant of the resulting matrix is m times the determinant of the original matrix. This is also a fundamental property of determinants and is correct. Multiplying a matrix X by a scalar m is equivalent to multiplying each of its n rows by m. By applying the property given in Sta

Question 10

If = vmatrix a & b & c \ d & e & f \ g & h & i vmatrix and A, B, C, D, G are the cofactors of the elements a, b, c, d, g respectively, then what is bB + cC - dD - gG equal to?

Accepted Answer

The sum of the products of elements of any row or column with their corresponding cofactors is equal to the determinant . Expanding the determinant along the first row, we get: aA + bB + cC = Expanding the determinant along the first column, we get: aA + dD + gG = Subtracting the second equation from the first equation, we get: (aA + bB + cC) - (aA + dD + gG) = - bB + cC - dD - gG = 0 Answer: 0

Question 11

If A^2 + B^2 + C^2 = 0, then what is the value of vmatrix 1 & C & B \ C & 1 & A \ B & A & 1 vmatrix ?

Accepted Answer

Given A^2 + B^2 + C^2 = 0. Since the square of any real number is non-negative, the sum of squares can be zero if and only if each number is zero. Thus, A = 0, B = 0, and C = 0. Substituting these values into the determinant, we get A = 0 = 1, B = 0 = 1, and C = 0 = 1. The given determinant becomes: vmatrix 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 vmatrix Since all rows are identical, the value of the determinant is 0. Answer: 0

Question 12

Consider the following statements in respect of the determinant = vmatrix k(k+2) & 2k+1 & 1 \ 2k+1 & k+2 & 1 \ 3 & 3 & 1 vmatrix : I. is positive if k > 0. II. is negative if k III. is zero if k = 0

Accepted Answer

Given determinant is = vmatrix k(k+2) & 2k+1 & 1 \ 2k+1 & k+2 & 1 \ 3 & 3 & 1 vmatrix Applying row operations R_1 R_1 - R_2 and R_2 R_2 - R_3: = vmatrix k^2-1 & k-1 & 0 \ 2k-2 & k-1 & 0 \ 3 & 3 & 1 vmatrix Expanding along the third column: = 1 vmatrix k^2-1 & k-1 \ 2k-2 & k-1 vmatrix = (k^2-1)(k-1) - (2k-2)(k-1) = (k-1) [ k^2 - 1 - 2k + 2 ] = (k-1) (k^2 - 2k + 1) = (k-1)^3 Evaluating the given statements: I. For k = 1 2 > 0, = ( 1 2 - 1 )^3 = - 1 8 II. For k III. For k = 0, = (0-1)^3 = -1 0

Question 13

If vmatrix 2 & 3+i & -1 \ 3-i & 0 & i-1 \ -1 & -1-i & 1 vmatrix = A + iB where i = -1 , then what is A + B equal to?

Accepted Answer

Given the determinant: vmatrix 2 & 3+i & -1 \ 3-i & 0 & i-1 \ -1 & -1-i & 1 vmatrix = A + iB Expanding the determinant along the first row, we get: D = 2 vmatrix 0 & i-1 \ -1-i & 1 vmatrix - (3+i) vmatrix 3-i & i-1 \ -1 & 1 vmatrix - 1 vmatrix 3-i & 0 \ -1 & -1-i vmatrix Evaluating each 2 2 minor: First minor: vmatrix 0 & i-1 \ -1-i & 1 vmatrix = (0)(1) - (i-1)(-1-i) = 0 - (-(i^2 - 1)) = 0 - (-(-1 - 1)) = -2 Second minor: vmatrix 3-i & i-1 \ -1 & 1 vmatrix = (3-i)(1) - (i-1)(-1) = 3 - i +

Question 14

Consider the following in respect of a non-singular matrix M: I. |M^2| = |M|^2 II. |M| = |M^ -1 | III. |M| = |M^T| How many of the above are correct?

Accepted Answer

For a non-singular matrix M, |M| 0. Using the property of determinants |AB| = |A||B|, we have |M^2| = |M M| = |M||M| = |M|^2. Thus, Statement I is correct. Using the property M M^ -1 = I, taking the determinant on both sides gives |M M^ -1 | = |I| |M||M^ -1 | = 1 |M^ -1 | = 1 |M| . Since |M| is not necessarily equal to 1 |M| , Statement II is incorrect. The determinant of a matrix is always equal to the determinant of its transpose, so |M| = |M^T|. Thus, Statement III is correct. Exactly two of

Question 15

In obtaining the solution of the system of equations x + y + z = 7, x + 2y + 3z = 16 and x + 3y + 4z = 22 by Cramer's rule, the value of y is obtained by dividing D by D_2, where D = vmatrix 1 & 1 & 1

Accepted Answer

The given system of linear equations is: x + y + z = 7 x + 2y + 3z = 16 x + 3y + 4z = 22 By Cramer's rule, the value of y is given by y = D_2 D , where D_2 is the determinant obtained by replacing the second column of the coefficient matrix D with the column of constant terms. D_2 = vmatrix 1 & 7 & 1 \ 1 & 16 & 3 \ 1 & 22 & 4 vmatrix Expanding the determinant along the first row: D_2 = 1(16 4 - 3 22) - 7(1 4 - 3 1) + 1(1 22 - 16 1) D_2 = 1(64 - 66) - 7(4 - 3) + 1(22 - 16) D_2 = 1(-2) - 7(1) +

Determinants — Previous Year Questions & Solutions