Let A = a i j n × n where a i j = i 2 - j 2 . Show that A is skew-symmetric matrix.

We know, for skew-symmetric matrix, A = - A ' , where A ' is the transpose of matrix. Given, a i j = i 2 - j 2 So, a j i = j 2 - i 2 = - i 2 - j 2 = - a i j Hence, the matrix A = a i j n × n is a skew-symmetric matrix.

Let A = a i j n × n where a i j = i 2 - j 2 . Show that A is skew-symmetric matrix.

We know, for skew-symmetric matrix, A = - A ' , where A ' is the transpose of matrix. Given, a i j = i 2 - j 2 So, a j i = j 2 - i 2 = - i 2 - j 2 = - a i j Hence, the matrix A = a i j n × n is a skew-symmetric matrix.

If C = 1 4 6 7 2 5 9 8 3 0 2 3 - 2 0 4 - 3 - 4 0 1 7 9 4 2 8 6 5 3 , then trace of C + C 3 + C 5 + … … … + C 99 is

Given, C = 1 4 6 7 2 5 9 8 3 0 2 3 - 2 0 4 - 3 - 4 0 1 7 9 4 2 8 6 5 3 ⇒ C = - 26 - 22 19 - 19 - 6 29 - 25 6 59 1 7 9 4 2 8 6 5 3 = 0 - 131 - 353 131 0 - 132 353 132 0 So, C is a skew-symmetric matrix. For skew-symmetric matrix, C = - C T ⇒ C n = - C T n ⇒ C n = - 1 n C T n If n is odd, then C n also skew-symmetric matrix and the trace is zero. So, trace of C + C 3 + C 5 + … … … + C 99 = 0

If the minor of three-one element (i.e. M 31 ) in the determinant 0 1 sec α tan α - sec α tan α 1 0 1 is 1 then find the value of α . 0 ≤ α ≤ π .

We know if given a square matrix A , by minor of an element a i j , we mean the value of the determinant obtained by deleting the i th row and j th column of A matrix. M 31 = tan α + sec 2 α ⇒ tan α + 1 + tan 2 α = 1 ⇒ tan α + tan 2 α = 0 ⇒ tan α 1 + tan α = 0 ⇒ tan α = 0 or 1 + tan α = 0 ⇒ α = 0 , π or tan α = − 1 ⇒ α = 0 , π or α = 3 π 4

Using the properties of determinants, evaluate 23 6 11 36 5 26 63 13 37

Given, 23 6 11 36 5 26 63 13 37 = 1 2 23 2 × 6 11 36 2 × 5 26 63 2 × 13 37 = 1 2 23 12 11 36 10 26 63 26 37 = 1 2 23 12 23 36 10 36 63 26 63 (Apply C 3 ↔ C 2 + C 3 ) = 0

Using the properties of determinants evaluate 0 c b - c 0 a - b - a 0

For the given matrix, the elements of principle diagonal is zero and for every elements a i j = - a j i , where i is the column and j is the row. We can say that the given matrix is a skew-symmetric matrix. So, A = - A T ⇒ A = - A T ⇒ A = - A ⇒ 2 A = 0 ⇒ | A | = 0

Using the properties of determinants evaluate 103 115 114 111 108 106 104 113 116 + 113 116 104 108 106 111 115 114 103

103 115 114 111 108 106 104 113 116 + 113 116 104 108 106 111 115 114 103 = 103 115 114 111 108 106 104 113 116 + 104 113 116 111 108 106 103 115 114 (interchange the third column & first column and first column & second column of second matrix) = 103 115 114 111 108 106 104 113 116 - 103 115 114 111 108 106 104 113 116 (Interchange the first row and the third row of second matrix) = 0

Using the properties of determinants evaluate 13 + 3 2 5 5 15 + 26 5 10 3 + 65 15 5

Let Δ = 13 + 3 2 5 5 15 + 26 5 10 3 + 65 15 5 = 3 2 5 5 15 5 10 3 15 5 + 13 2 5 5 26 5 10 65 15 5 = 3 × 5 × 5 1 2 1 5 5 2 3 3 5 + 13 × 5 × 5 1 2 1 2 5 2 5 3 5 = 3 × 5 1 2 1 5 5 2 3 3 5 + 0 = 5 3 1 2 1 5 5 2 3 3 5 Expanding the determinant we get ∆ = 5 3 2 - 5 3

Prove that: (i) 1 1 1 a b c a 3 b 3 c 3 = a - b b - c c - a a + b + c (ii) a b + c a 2 b c + a b 2 c a + b c 2 = - a + b + c a - b b - c c - a (iii) b + c a a b c + a b c c a + b = 4 a b c (iv) 1 a 2 a 4 1 b 2 b 4 1 c 2 c 4 = a + b b + c c + a 1 1 1 a b c a 2 b 2 c 2

LHS = 1 1 1 a b c a 3 b 3 c 3 = 0 0 1 a - b b - c c a 3 - b 3 b 3 - c 3 c 3 (applying C 1 → C 1 - C 2 & C 2 → C 2 - C 3 ) = 0 0 1 a - b b - c c ( a - b ) ( a 2 + b 2 + a b ) ( b - c ) ( b 2 + c 2 + b c ) c 3 ( using x 3 - y 3 = ( x - y ) ( x 2 + y 2 + x y ) ) = a - b b - c 0 0 1 1 1 c ( a 2 + b 2 + a b ) ( b 2 + c 2 + b c ) c 3 ( taking ( a - b ) & ( b - c ) out from C 1 & C 2 respectively ) Now, expand along R 1 , = a - b b - c 0 - 0 + 1 × 1 1 a 2 + b 2 + a b b 2 + c 2 + b c = ( a - b ) ( b - c ) ( b 2 + c 2 + b c ) - ( a 2 + b 2 + a b ) = ( a - b ) ( b - c ) c 2 - a 2 + b c - a b = ( a - b ) ( b - c ) ( c - a ) ( c + a ) + b ( c - a ) = ( a - b ) ( b - c ) ( c - a ) ( c + a + b ) = ( b - c ) ( c - a ) ( a - b ) ( a + b + c ) = RHS Hence, proved.

Prove that a b + c a 2 b c + a b 2 c a + b c 2 = - a + b + c a - b b - c c - a

Let us consider L . H . S = a b + c a 2 b c + a b 2 c a + b c 2 and R . H . S = - a + b + c × a - b b - c c - a By applying C 2 → C 1 + C 2 , we get a a + b + c a 2 b a + b + c b 2 c a + b + c c 2 = a + b + c a 1 a 2 b 1 b 2 c 1 c 2 Now, by applying C 1 ↔ C 2 , we get - a + b + c 1 a a 2 1 b b 2 1 c c 2 Again, by applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , we get - a + b + c 1 a a 2 0 b - a b 2 - a 2 0 c - a c 2 - a 2 Taking b - a , c - a common from R 2 , R 3 we get - a + b + c b - a c - a 1 a a 2 0 1 b + a 0 1 c + a Expanding along first row, = - a + b + c b - a c - a 1 × c + a - b - a - a 0 + a 2 0 = - a + b + c × a - b b - c c - a = R . H . S Hence, proved.

Let M be a 2 × 2 symmetric matrix with integer entries. Then M is invertible if

the product of entries in the main diagonal of M is not the square of an integer

the first column of M is the transpose of the second row of M

the second row of M is the transpose of first column of M

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-1>Q 8

EASY

JEE Main/Advance

IMPORTANT

Earn 100

Let $A = {[a_{i j}]}_{n \times n}$ where $a_{i j} = i^{2} - j^{2} .$ Show that $A$ is skew-symmetric matrix.

Important Questions on Matrices and Determinants

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-1>Q 9

If

C = [\begin{matrix} 1 & 4 & 6 \\ 7 & 2 & 5 \\ 9 & 8 & 3 \end{matrix}] [\begin{matrix} 0 & 2 & 3 \\ - 2 & 0 & 4 \\ - 3 & - 4 & 0 \end{matrix}] [\begin{matrix} 1 & 7 & 9 \\ 4 & 2 & 8 \\ 6 & 5 & 3 \end{matrix}],

then trace of

C + C^{3} + C^{5} + \dots \dots \dots + C^{99}

is

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-1>Q 10

If the minor of three-one element (i.e.

M_{31}

) in the determinant

|\begin{matrix} 0 & 1 & \sec α \\ \tan α & - \sec α & \tan α \\ 1 & 0 & 1 \end{matrix}|

is

1

then find the value of

α . (0 \leq α \leq π) .

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-1>Q 11

Using the properties of determinants, evaluate $|\begin{matrix} 23 & 6 & 11 \\ 36 & 5 & 26 \\ 63 & 13 & 37 \end{matrix}|$

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-1>Q 12

Using the properties of determinants evaluate

$|\begin{matrix} 0 & c & b \\ - c & 0 & a \\ - b & - a & 0 \end{matrix}|$

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-1>Q 13

Using the properties of determinants evaluate

$|\begin{matrix} 103 & 115 & 114 \\ 111 & 108 & 106 \\ 104 & 113 & 116 \end{matrix}| + |\begin{matrix} 113 & 116 & 104 \\ 108 & 106 & 111 \\ 115 & 114 & 103 \end{matrix}|$

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-1>Q 14

Using the properties of determinants evaluate

$|\begin{matrix} \sqrt{13} + \sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{26} & 5 & \sqrt{10} \\ 3 + \sqrt{65} & \sqrt{15} & 5 \end{matrix}|$

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-1>Q 15

Prove that:

(i) $|\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c)$

(ii) $|\begin{array}{l} a & b + c & a^{2} \\ b & c + a & b^{2} \\ c & a + b & c^{2} \end{array}| = - (a + b + c) (a - b) (b - c) (c - a)$

(iii) $|\begin{matrix} b + c & a & a \\ b & c + a & b \\ c & c & a + b \end{matrix}| = 4 a b c$

(iv) $|\begin{array}{l} 1 & a^{2} & a^{4} \\ 1 & b^{2} & b^{4} \\ 1 & c^{2} & c^{4} \end{array}| = (a + b) (b + c) (c + a) |\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix}|$