Consider an odd order square symmetric matrix A = a i j n × n . It's elements in any row are 1 , 2 , … … , n in some order, then prove that a 11 , a 22 , … … … . , a n n are numbers 1 , 2 , 3 , … … , n in some order.

We have, A = a i j n × n It is a symmetric matrix. Now, 3 × 3 symmetric matrix is A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 A T = a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 For symmetric matrix: A = A T ⇒ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 = a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 Hence, a 12 = a 21 , a 31 = a 13 , a 32 = a 33 . And, for diagonal elements, a 11 , a 22 , a 33 we can put any real number 1 , 2 , 3 , but we cannot put zero at the diagonal position, otherwise it will become skew symmetric matrix. Similarly, we have for n × n matrix.

Consider an odd order square symmetric matrix A = a i j n × n . It's elements in any row are 1 , 2 , … … , n in some order, then prove that a 11 , a 22 , … … … . , a n n are numbers 1 , 2 , 3 , … … , n in some order.

We have, A = a i j n × n It is a symmetric matrix. Now, 3 × 3 symmetric matrix is A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 A T = a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 For symmetric matrix: A = A T ⇒ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 = a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 Hence, a 12 = a 21 , a 31 = a 13 , a 32 = a 33 . And, for diagonal elements, a 11 , a 22 , a 33 we can put any real number 1 , 2 , 3 , but we cannot put zero at the diagonal position, otherwise it will become skew symmetric matrix. Similarly, we have for n × n matrix.

Let A = x + λ x x x x + λ x x x x + λ , then prove that A - 1 exists if 3 x + λ ≠ 0 , λ ≠ 0 .

Given the matrix A = x + λ x x x x + λ x x x x + λ . Inverse of matrix A exist if A ≠ 0 . So, x + λ x x x x + λ x x x x + λ ≠ 0 . Applying R 1 → R 1 + R 2 + R 3 , we get 3 x + λ 3 x + λ 3 x + λ x x + λ x x x x + λ ≠ 0 ⇒ 3 x + λ 1 1 1 x x + λ x x x x + λ ≠ 0 Applying C 2 → C 2 - C 1 & C 3 → C 3 - C 1 , we get 3 x + λ 1 0 0 x λ 0 x 0 λ ≠ 0 ⇒ 3 x + λ 1 λ 2 ≠ 0 ⇒ 3 x + λ ≠ 0 , λ ≠ 0

Prove that if A and B are n × n matrices, then d e t I n - A B = d e t I n - B A .

Given that A & B are n × n order square matrices and I n be an identity matrix of order n × n . Then A B & B A are also n × n order square matrices. Since I n and I n - B A are square matrices, then d e t I n - B A = d e t I n 0 B I n - B A = d e t I n A B I n I n - A 0 I n = d e t I n A B I n d e t I n - A 0 I n Since d e t I n - A 0 I n = d e t I n d e t I n - 0 = d e t I n d e t I n = 1 Then d e t I n - B A = d e t I n A B I n Now d e t I n A B I n = d e t I n - A B 0 B I n = d e t I n - A B d e t I n - 0 = d e t I n - A B Hence, d e t I n - A B = d e t I n - B A

If z 1 = z 2 = 1 , then prove that z 1 - z 2 z 2 z 1 - 1 z 1 z 2 - z 2 z 1 - 1 = 1 2 0 0 1 2 .

Given z 1 = z 2 = 1 . By L . H . S , z 1 - z 2 z 2 z 1 - 1 z 1 z 2 - z 2 z 1 - 1 = z 1 z 2 - z 2 z 1 z 1 - z 2 z 2 z 1 - 1 = z 1 z 1 + z 2 z 2 - z 1 z 2 + z 2 z 1 - z 2 z 1 + z 1 z 2 z 2 z 2 + z 1 z 1 - 1 ∵ B - 1 A - 1 = A B - 1 We know that for any complex number z , z 2 = z z . ⇒ L . H . S . = z 1 2 + z 2 2 0 0 z 2 2 + z 1 2 - 1 = 2 0 0 2 - 1 We also know that for a matrix A , A - 1 = A d j A A . ⇒ L . H . S . = 1 4 2 0 0 2 = 1 2 0 0 1 2 = R . H . S .

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-4>Q 20

MEDIUM

JEE Main/Advance

IMPORTANT

Earn 100

Consider an odd order square symmetric matrix $A = {[a_{i j}]}_{n \times n}$ . It's elements in any row are $1, 2, \dots \dots, n$ in some order, then prove that $a_{11}, a_{22}, \dots \dots \dots ., a_{n n}$ are numbers $1, 2, 3, \dots \dots, n$ in some order.

Important Questions on Matrices and Determinants

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-4>Q 21

Let $A = [\begin{array}{l} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}]; B = [\begin{matrix} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \end{matrix}]$ and $C = 3 A + 7 B$