Matrix A has order 3 × n, and B has order m × 5; find the order of matrix C = 5A + 3B Step 1: Rules governing addition of matrices Addition is defined only when both matrices are the same order, and the same with our scenario of A versus B. Matrix A has order 3 × n and matrix B has order m × 5. ForRead more
Matrix A has order 3 × n, and B has order m × 5; find the order of matrix C = 5A + 3B
Step 1: Rules governing addition of matrices
Addition is defined only when both matrices are the same order, and the same with our scenario of A versus B.
Matrix A has order 3 × n and matrix B has order m × 5.
For the addition 5A + 3B to be possible, we must have n = m, meaning both matrices must have the same number of columns.
Step 2: Order of matrix C
Once the condition n = m is met, then the matrix C that results from it will be of the same order as that of A and B, which is 3 × 5.
- Matrix A has order 2 × 3 and matrix B has order 3 × 2. - The transpose of matrix B, B', has order 2 × 3. - For the sum A + B', both matrices have order 2 × 3, so the addition is possible. - However, matrix C has order 3 × 3, and for matrix multiplication to be defined, the number of columns in matRead more
– Matrix A has order 2 × 3 and matrix B has order 3 × 2.
– The transpose of matrix B, B’, has order 2 × 3.
– For the sum A + B’, both matrices have order 2 × 3, so the addition is possible.
– However, matrix C has order 3 × 3, and for matrix multiplication to be defined, the number of columns in matrix C (3) must match the number of rows in A + B’ (2).
– Since 3 ≠ 2, multiplying matrix C by A + B’ is not defined.
We are given that P is a 3 × 3 matrix such that P' = 2P + I, where P' is the transpose of P. Step 1: Take the transpose of both sides We take the transpose of both sides of the equation P' = 2P + I: (P')' = (2P + I)' Since the transpose of the transpose of a matrix is the matrix itself, we get: P =Read more
We are given that P is a 3 × 3 matrix such that P’ = 2P + I, where P’ is the transpose of P.
Step 1: Take the transpose of both sides
We take the transpose of both sides of the equation P’ = 2P + I:
(P’)’ = (2P + I)’
Since the transpose of the transpose of a matrix is the matrix itself, we get:
P = 2P’ + I
Step 2: Plug P’ = 2P + I into this equation
Next, plug the expression P’ = 2P + I into the equation:
P = 2(2P + I) + I
P = 4P + 2I + I
P = 4P + 3I
Step 3: Move terms around
Now we move the terms around in the equation.
P – 4P = 3I
-3P = 3I
P = -I
Mud houses do not get warmer in summer because mud is a bad conductor of heat. They resist the movement of heat that would make its interior warm with hot weather and cold in winter by either preventing the movement of heat within or outside in. Click here for more: https://www.tiwariacademy.com/nceRead more
Mud houses do not get warmer in summer because mud is a bad conductor of heat. They resist the movement of heat that would make its interior warm with hot weather and cold in winter by either preventing the movement of heat within or outside in.
We can solve this using Newton's Law of Cooling, which states, (dT/dt) = -k(T - Tₛ) (dT/dt) is the rate of change of temperature k is a constant T is the temperature of the body Tₛ is the temperature of the surroundings. We are given that the body cools from 80°C to 64°C in 5 minutes and again fromRead more
We can solve this using Newton’s Law of Cooling, which states,
(dT/dt) = -k(T – Tₛ)
(dT/dt) is the rate of change of temperature
k is a constant
T is the temperature of the body
Tₛ is the temperature of the surroundings.
We are given that the body cools from 80°C to 64°C in 5 minutes and again from 80°C to 52°C in 10 minutes. Applying Newton’s Law, we can derive the temperature of the surroundings.
We can use this formula to find the temperatures:
ln[(T₁ – Tₛ) / (T₂ – Tₛ)] = k(t₂ – t₁)
For the first cooling process, where it cools from 80 °C to 64 °C in 5 minutes:
ln[(80 – Tₛ) / (64 – Tₛ)] = k × 5
For the second cooling (from 80 °C to 52 °C in 10 minutes):
ln[(80 – Tₛ) / (52 – Tₛ)] = k × 10
This system of equations solves for the value of Tₛ.
After solving, we find that the temperature of the surroundings is 25 °C.
Given that matrices A and B are of order 3 x n and m x 5 respectively, then the order of matrix C = 5A + 3B is:
Matrix A has order 3 × n, and B has order m × 5; find the order of matrix C = 5A + 3B Step 1: Rules governing addition of matrices Addition is defined only when both matrices are the same order, and the same with our scenario of A versus B. Matrix A has order 3 × n and matrix B has order m × 5. ForRead more
Matrix A has order 3 × n, and B has order m × 5; find the order of matrix C = 5A + 3B
Step 1: Rules governing addition of matrices
Addition is defined only when both matrices are the same order, and the same with our scenario of A versus B.
Matrix A has order 3 × n and matrix B has order m × 5.
For the addition 5A + 3B to be possible, we must have n = m, meaning both matrices must have the same number of columns.
Step 2: Order of matrix C
Once the condition n = m is met, then the matrix C that results from it will be of the same order as that of A and B, which is 3 × 5.
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If order of matrix A is 2 x 3 of matrix B is 3 x 2, and of matrix C is 3 x 3, then which one of the following is not defined?
- Matrix A has order 2 × 3 and matrix B has order 3 × 2. - The transpose of matrix B, B', has order 2 × 3. - For the sum A + B', both matrices have order 2 × 3, so the addition is possible. - However, matrix C has order 3 × 3, and for matrix multiplication to be defined, the number of columns in matRead more
– Matrix A has order 2 × 3 and matrix B has order 3 × 2.
– The transpose of matrix B, B’, has order 2 × 3.
– For the sum A + B’, both matrices have order 2 × 3, so the addition is possible.
– However, matrix C has order 3 × 3, and for matrix multiplication to be defined, the number of columns in matrix C (3) must match the number of rows in A + B’ (2).
– Since 3 ≠ 2, multiplying matrix C by A + B’ is not defined.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
If P is a 3 x 3 matrix such that P’ = 2P + I, where P’ is the transpose of P, then
We are given that P is a 3 × 3 matrix such that P' = 2P + I, where P' is the transpose of P. Step 1: Take the transpose of both sides We take the transpose of both sides of the equation P' = 2P + I: (P')' = (2P + I)' Since the transpose of the transpose of a matrix is the matrix itself, we get: P =Read more
We are given that P is a 3 × 3 matrix such that P’ = 2P + I, where P’ is the transpose of P.
Step 1: Take the transpose of both sides
We take the transpose of both sides of the equation P’ = 2P + I:
(P’)’ = (2P + I)’
Since the transpose of the transpose of a matrix is the matrix itself, we get:
P = 2P’ + I
Step 2: Plug P’ = 2P + I into this equation
Next, plug the expression P’ = 2P + I into the equation:
P = 2(2P + I) + I
P = 4P + 2I + I
P = 4P + 3I
Step 3: Move terms around
Now we move the terms around in the equation.
P – 4P = 3I
-3P = 3I
P = -I
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
Mud houses are cooler in summer and warmer in winter because
Mud houses do not get warmer in summer because mud is a bad conductor of heat. They resist the movement of heat that would make its interior warm with hot weather and cold in winter by either preventing the movement of heat within or outside in. Click here for more: https://www.tiwariacademy.com/nceRead more
Mud houses do not get warmer in summer because mud is a bad conductor of heat. They resist the movement of heat that would make its interior warm with hot weather and cold in winter by either preventing the movement of heat within or outside in.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
A body cools from 80 °C to 64 °C in 5 minutes and same body cools from 80 °C to 52 °C in 10 minutes. What is the temperature of surrounding?
We can solve this using Newton's Law of Cooling, which states, (dT/dt) = -k(T - Tₛ) (dT/dt) is the rate of change of temperature k is a constant T is the temperature of the body Tₛ is the temperature of the surroundings. We are given that the body cools from 80°C to 64°C in 5 minutes and again fromRead more
We can solve this using Newton’s Law of Cooling, which states,
(dT/dt) = -k(T – Tₛ)
(dT/dt) is the rate of change of temperature
k is a constant
T is the temperature of the body
Tₛ is the temperature of the surroundings.
We are given that the body cools from 80°C to 64°C in 5 minutes and again from 80°C to 52°C in 10 minutes. Applying Newton’s Law, we can derive the temperature of the surroundings.
We can use this formula to find the temperatures:
ln[(T₁ – Tₛ) / (T₂ – Tₛ)] = k(t₂ – t₁)
For the first cooling process, where it cools from 80 °C to 64 °C in 5 minutes:
ln[(80 – Tₛ) / (64 – Tₛ)] = k × 5
For the second cooling (from 80 °C to 52 °C in 10 minutes):
ln[(80 – Tₛ) / (52 – Tₛ)] = k × 10
This system of equations solves for the value of Tₛ.
After solving, we find that the temperature of the surroundings is 25 °C.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/