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Diagonal Matrix

If λ=3, the...

Question

If $λ = 3,$ then $\frac{1}{7} (t r (A B) + t r (B A))$ is equal to
Note: $t r (A)$ denotes sum of diagonal elements of matrix $A$ .

A

34

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B

42

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C

84

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D

63

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Solution

The correct option is A $34$
For $λ = 3$
$A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 3 & 5 2 & 4 & 8 3 & 5 & 10 \end{matrix} ⎤ ⎥ ⎦$ and $B = ⎡ ⎢ ⎣ \begin{matrix} 3 & 2 & 4 3 & 2 & 5 2 & 1 & 4 \end{matrix} ⎤ ⎥ ⎦$

We only need diagonal elements
$A B = ∣ ∣ ∣ ∣ \begin{matrix} 22 & . & . . & 20 & . . & . & 77 \end{matrix} ∣ ∣ ∣ ∣$ and $B A = ∣ ∣ ∣ ∣ \begin{matrix} 19 & . & . . & 42 & . . & . & 58 \end{matrix} ∣ ∣ ∣ ∣$

$t r (A B) + t r (B A) = 2 \times 119 = 238$
$\frac{1}{7} (t r (A B) + t r (B A)) = \frac{1}{7} (238) = 34$

Hence, option A.

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Similar questions

A = ⎛ ⎜ ⎝ \begin{matrix} x^{2} & 6 & 8 3 & y^{2} & 9 4 & 5 & z^{2} \end{matrix} ⎞ ⎟ ⎠

B = ⎛ ⎜ ⎝ \begin{matrix} 2 x & 3 & 5 2 & 2 y & 6 1 & 4 & 2 z - 3 \end{matrix} ⎞ ⎟ ⎠

be two matrices and if

T r (A) = T r (B)

, then the value of

(x + y + z)

is equal to
(Note:

T r (P)

denotes trace of matrix

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)

A

3 \times 3

skew symmetric matrix with real entries and trace of

A^{2}

equals zero, then

Note: Trace of matrix

A

denotes the sum of diagonal elements of matrix

A

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 7 0 & 1 & 0 1 & - 2 & 1 \end{matrix} ⎤ ⎥ ⎦

B = ⎡ ⎢ ⎣ \begin{matrix} - x & 14 x & 7 x 0 & 1 & 0 x & - 4 x & - 2 x \end{matrix} ⎤ ⎥ ⎦

be two matrices such that

A B = (A B)^{- 1}

A B \neq I,

I

is an identity matrix of order

3 \times 3.

Then the value of

t r (A B + (A B)^{2} + (A B)^{3} + \dots + (A B)^{100})

(

t r (A)

denotes the trace of matrix

A,

i.e., sum of diagonal elements of

A .)

2 \times 2

matrix with non-zero entries and let

A^{2} = I

2 \times 2

identity matrix. Define Tr(A)

=

sum of diagonal elements of A and

| A | =

determinant of matrix A.
Statement-1 Tr(A)

= 0

| A | = 1

A

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real matrix with entries from

{0, 1}

| A | \neq 0

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