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 .

If a diagonal matrix D = diag d 1 , d 2 , … … … , d n , then prove that f ( D ) = diag f d 1 , f d 2 , … … … f d n , where f ( x ) is a polynomial with scalar coefficient.

It is given that D = diag d 1 , d 2 , … … … , d n Now let f x = a 0 + a 1 x + a 2 x 2 + . . . . . . . a n x n f D = a 0 I + a 1 D + a 2 D 2 + . . . . . . . a n D n f D = a 0 × d i a g 1 , 1 , 1 . . . . . . . , 1 + a 1 × d i a g d 1 , d 2 , d 3 . . . . . . . , d n + a 2 × d i a g d 1 2 , d 2 2 , d 3 2 . . . . . . . , d n 2 + . . . . . . . a n × d i a g d 1 n , d 2 n , d 3 n . . . . . . . , d n n f D = d i a g a 0 + a 1 d 1 + a 2 d 1 2 + a 3 d 1 2 + . . . . . . . . . a n d 1 n , . . . . . . . . . , a 0 + a 1 d n + a 2 d 1 2 + a 3 d n 2 + . . . . . . . . . a n d n n f ( D ) = diag f d 1 , f d 2 , … … … f d n .

Given the matrix A = - 1 3 5 1 - 3 - 5 - 1 3 5 and X be the solution set of the equation A x = A , where x ∈ N - 1 . Evaluate ∏ x 3 + 1 x 3 - 1 , where the continued product extends ∀ x ∈ X .

It is given that A = - 1 3 5 1 - 3 - 5 - 1 3 5 . Now, A 2 = - 1 3 5 1 - 3 - 5 - 1 3 5 - 1 3 5 1 - 3 - 5 - 1 3 5 = - 1 3 5 1 - 3 - 5 - 1 3 5 = A . Similarly, A 3 = A 4 = . . . . . . = A x = A ∀ x ∈ N - 1 . Now, it is also given that A x = A for x = 2 , 3 , 4 , . . . . . . . . . . . , ∞ . ∴ ∏ x = 2 ∞ x 3 + 1 x 3 - 1 = ∏ x = 2 ∞ x + 1 x 2 - x + 1 x - 1 x 2 + x + 1 = ∏ x = 2 ∞ x + 1 x - 1 × x 2 - x + 1 x 2 + x + 1 = lim n → ∞ 3 1 × 4 2 × 5 3 × 6 4 × 7 5 × 8 6 × . . . . . . × n n - 2 × n + 1 n - 1 3 7 × 7 13 × 13 21 × . . . . . . × n 2 - n + 1 n 2 + n + 1 = lim n → ∞ n n + 1 2 3 n 2 + n + 1 = lim n → ∞ 3 n 2 + n 2 n 2 + n + 1 = 3 2 × lim n → ∞ n 2 + n n 2 n 2 + n + 1 n 2 = 3 2 × lim n → ∞ 1 + 1 n 1 + 1 n + 1 n 2 = 3 2 × 1 + 0 1 + 0 + 0 = 3 2

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 27

MEDIUM

JEE Main/Advance

IMPORTANT

Earn 100

Let $A, B, C$ be three $3 \times 3$ matrices with real entries. If $B A + B C + A C = I$ and $\det (A + B) = 0$ then find the value of $\det (A + B + C - B A C)$ .

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 28

If

|z_{1}| = |z_{2}| = 1,

then prove that

{[\begin{matrix} z_{1} & - z_{2} \\ z_{2} & z_{1} \end{matrix}]}^{- 1} {[\begin{matrix} z_{1} & z_{2} \\ - z_{2} & z_{1} \end{matrix}]}^{- 1} = [\begin{matrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{matrix}]

.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-4>Q 29

If

A

and

B

are two square matrices such that

B = - A^{- 1} B A,

then show that

{(A + B)}^{2} = A^{2} + B^{2}

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 9 - Matrices and Determinants>Exercise-4>Q 30

If $A = [\begin{array}{l} a & 1 & 0 \\ 1 & b & d \\ 1 & b & c \end{array}], B = [\begin{array}{l} a & 1 & 1 \\ 0 & d & c \\ f & g & h \end{array}], U = [\begin{array}{l} f \\ g \\ h \end{array}], V = [\begin{array}{l} a^{2} \\ 0 \\ 0 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}]$ and $A X = U$ has infinitely many solutions, prove that $B X = V$ has no unique solution. Also, prove that if $a f d$ $\neq 0,$ then $B X = V$ has no solution.