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Find non-zero values of x satisfying the matrix equation:

\(x \begin{bmatrix} 2x & 2 \\[0.3em] 3 & x \end{bmatrix} + 2 \begin{bmatrix} 8 & 5x \\[0.3em] 4 & 4x \end{bmatrix} = \begin{bmatrix} (x^2 + 8) & 24 \\[0.3em] (10) & (6x) \end{bmatrix}\)

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A matrix, as we know, is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

Also,

Two or more matrices can be added or subtracted only if they have same order.

And we are familiar with order of a matrix. If a matrix has M rows and N columns, the order of matrix is M × N.

We have matrix equation,

Multiply it with 2,

Add equation (i) and (ii) and make it equal to equation (iii), we get

We need to find the value of x.

So, compare the elements in the two matrices.

If,

Then,

a11 = b11

a12 = b12

a21 = b21

a22 = b22

So,

2x2 + 16 = 2(x2 + 8) …(i)

2x + 10x = 48 …(ii)

3x + 8 = 20 …(iii)

x2 + 8x = 12x …(iv)

We have got equations (i), (ii), (iii) and (iv) to solve for x.

So, take equation (i).

2x2 + 16 = 2x2 + 16

We won’t be able to find x from this equation, as both equations are same.

Now, take equation (ii).

2x + 10x = 48

⇒ 12x = 48

⇒ x = 48/12

⇒ x = 4

From equation (iii),

3x + 8 = 20

⇒ 3x = 20 – 8

⇒ 3x = 12

⇒ x = 12/3

⇒ x = 4

From equation (iv),

x2 + 8x = 12x

⇒ x2 = 12x – 8x

⇒ x2 = 4x

⇒ x2 – 4x = 0

⇒ x(x – 4) = 0

⇒ x = 0 or (x – 4) = 0

⇒ x = 0 or x = 4

⇒ x = 4 (∵ x = 0 does not satisfy equations (ii) and (iii))

So, by solving equations (ii), (iii) and (iv), we can conclude that

x = 4

Thus, the value of x is 4.

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