Question

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Answer 1

Answer:

5

Step-by-step explanation:

Given ,

$A=\begin{bmatrix}2 & x\\0 & \dfrac{-1}{2} \end{bmatrix}$

$B=\begin{bmatrix}1 & 4\\0 & -1 \end{bmatrix}$

AB = BA

To Find :-

Value of 'x'

Solution :-

Finding value of 'AB' :-

$=\begin{bmatrix}2& x \\0 & \dfrac{-1}{2}\end{bmatrix}\times \begin{bmatrix}1 & 4\\0 & -1\end{bmatrix}$

Multiplying 1st row in 'A' with 1st column in 'B' :-

= (2 × 1) + (x × 0)

= 2 + 0

= 2

[ ∴ '2' will be in the 1st row 1st column in matrix 'AB' ]

Multiplying 1st row in 'A' with '2'nd column in 'B' :-

= (2 × 4) + (x × (-1) )

= 8 - x

[ ∴ '8 - x' will be in the 1st row 2nd column in the matrix 'AB' ]

Multiplying 2nd row in 'A' with '1'st column in 'B' :-

= (0 × 1) + (-1/2 × 0)

= 0 + 0

= 0

[ ∴ '0' will be in the 2nd row , 1st column in the matrix 'AB' ]

Multiplying 2nd row in 'A' with 2nd column in 'B' :-

= (0 × 4) + (-1/2 × - 1)

= 0 + 1/2

= 1/2

[ ∴ 1/2 will be in the 2nd row , 2nd column in 'AB' ]

$\implies AB=\begin{bmatrix}2 & 8-x \\0 & \dfrac{1}{2} \end{bmatrix}$

Finding value of 'BA' :-

$=\begin{bmatrix}1 & 4 \\0 & -1 \end{bmatrix}\times \begin{bmatrix}2 & x \\0& \dfrac{-1}{2}\end{bmatrix}$

Multiplying 1st row in 'B' with 1st column in 'A' :-

= (1 × 2) + (4 × 0)

= 2 + 0

= 2

[  ∴ '2' will be in the 1st row 1st column in matrix 'BA' ]

Multiplying 1st row in 'B' with '2'nd column in 'A' :-

= (1 × x) + (4 × -1/2 )

= x + (-2)

= x - 2

[ ∴ 'x - 2' will be in the 1st row 2nd column in the matrix 'BA' ]

Multiplying 2nd row in 'B' with '1'st column in 'A' :-

= (0 × 2) + (-1 × 0)

= 0 + 0

= 0

[ ∴ '0' will be in the 2nd row , 1st column in the matrix 'BA' ]

Multiplying 2nd row in 'B' with 2nd column in 'A' :-

= (0 × x) + (-1 × -1/2)

= 0 + 1/2

= 1/2

[ ∴ '0' will be in the 2nd row , 1st column in the matrix 'BA' ]

$\implies BA=\begin{bmatrix}2 & x-2 \\0 & \dfrac{1}{2} \end{bmatrix}$

AB = BA

$\implies \begin{bmatrix}2 & 8-x \\0 & \dfrac{1}{2} \end{bmatrix}=\begin{bmatrix}2 & x-2 \\0 & \dfrac{1}{2} \end{bmatrix}$

We can see that all terms are equal expect '1' , So we equating those both terms :-

8 - x = x - 2

8 + 2 = x + x

10 = 2x

x = 10/2

x = 5

∴ Value of 'x' = 5

Answer 2

Answer :-

Value of x is 5.

Step-by-step explanation :-

Given :

If A = $\left[\begin{array}{ccc}2&x&\\0&\frac{-1}{2} \end{array}\right]$, B = $\left[\begin{array}{ccc}1&4\\0&-1\end{array}\right]$, find x if AB = BA ?

Solution :

AB = $\left[\begin{array}{ccc}2&x&\\0&\frac{-1}{2} \end{array}\right]$ $\times$  $\left[\begin{array}{ccc}1&4\\0&-1\end{array}\right]$

AB = $\left[\begin{array}{ccc}(2\times1)+(x\times0)&(2\times4)+(x\times-1)\\(0\times1)+(\frac{-1}{2}\times0)&(0\times4)+(\frac{-1}{2}\times-1) \\\end{array}\right]$

AB = $\left[\begin{array}{ccc}2&8-x\\0&\frac{1}{2}\end{array}\right]$

∴ AB = $\left[\begin{array}{ccc}2&8-x\\0&\frac{1}{2}\end{array}\right]$

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BA = $\left[\begin{array}{ccc}1&4\\0&-1\end{array}\right]$ $\times$  $\left[\begin{array}{ccc}2&x&\\0&\frac{-1}{2} \end{array}\right]$

BA = $\left[\begin{array}{ccc}(1\times2)+(4\times0)&(1\times x)+(4\times\frac{-1}{2}) \\(0\times2)+(-1\times0)&(0\times x )+(-1\times\frac{-1}{2}) \\\end{array}\right]$

BA = $\left[\begin{array}{ccc}2&x-2\\0&\frac{1}{2}\end{array}\right]$

∴ BA = $\left[\begin{array}{ccc}2&x-2\\0&\frac{1}{2}\end{array}\right]$

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Now, AB = BA

$\left[\begin{array}{ccc}2&8-x\\0&\frac{1}{2}\end{array}\right]$ =  $\left[\begin{array}{ccc}2&x-2\\0&\frac{1}{2}\end{array}\right]$

So, [8 - x] = [x - 2]

⇒ 8 + 2 = x + x

⇒ 10 = 2x

⇒ x = 10/2

∴ x = 5