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Byju's Answer

Standard XII

Chemistry

Heat Capacity

Let A+2B=1 ...

Question

Let $A + 2 B = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & 0 6 & - 3 & 3 - 5 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣$ and $2 A - B = ∣ ∣ ∣ ∣ \begin{matrix} 2 & - 1 & 5 2 & - 1 & 6 0 & 1 & 2 \end{matrix} ∣ ∣ ∣ ∣ .$ It $T r (A)$ denotes the sum of all diagonal elements of the matrix $A,$ then $T_{r} (A) - T_{r} (B)$ has value equal to:

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Solution

The correct option is D $2$
$t_{r} (A + 2 B) \equiv t_{r} (A) + 2 t_{r} (B) = - 1 \dots (1)$
and $t_{r} (2 A - B) \equiv 2 t_{r} (A) - t_{r} (B) = 3 \dots (2)$
on solving $(1)$ and $(2),$ we get
$t_{r} (A) = 1, t_{r} (B) = - 1$
$∴ t_{r} (A) - t_{r} (B) = 1 + 1 = 2$

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Similar questions

Q. 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

Q. Let

A + 2 B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 0 6 & - 3 & 3 - 5 & 3 & 1 \end{matrix} ⎤ ⎥ ⎦

and

2 A - B = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 5 2 & - 1 & 6 0 & 1 & 2 \end{matrix} ⎤ ⎥ ⎦

, then

t r (A) - t r (B)

has the value equal to

Q. For

α, β, γ \in R

, let

A = ⎡ ⎢ ⎣ \begin{matrix} α^{2} & 6 & 8 3 & β^{2} & 9 4 & 5 & γ^{2} \end{matrix} ⎤ ⎥ ⎦

and

B = ⎡ ⎢ ⎣ \begin{matrix} 2 α & 3 & 5 2 & 2 β & 6 1 & 4 & 2 γ - 3 \end{matrix} ⎤ ⎥ ⎦

. If

T_{r} (A) = T_{r} (B)

then the value of

(\frac{1}{α} + \frac{1}{β} + \frac{1}{γ})

is-

T_{r} (A)

is a Trace(A) of a matrix

Q. Let

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

and

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)

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Q. Let A be a

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A^{2} = I

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Let A+2B=∣∣ ∣∣1206−33−531∣∣ ∣∣ and 2A−B=∣∣ ∣∣2−152−16012∣∣ ∣∣. It Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A)–Tr(B) has value equal to:

The correct option is D 2tr(A+2B)≡tr(A)+2tr(B)=–1 ⋯(1) and tr(2A–B)≡2tr(A)–tr(B)=3 ⋯(2) on solving (1) and (2), we get tr(A)=1, tr(B)=–1 ∴tr(A)–tr(B)=1+1=2

The correct option is D $2$
$t_{r} (A + 2 B) \equiv t_{r} (A) + 2 t_{r} (B) = - 1 \dots (1)$
and $t_{r} (2 A - B) \equiv 2 t_{r} (A) - t_{r} (B) = 3 \dots (2)$
on solving $(1)$ and $(2),$ we get
$t_{r} (A) = 1, t_{r} (B) = - 1$
$∴ t_{r} (A) - t_{r} (B) = 1 + 1 = 2$