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  • Find non-zero values of x satisfying the matrix equation: X bracket 2x 2 3 x

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

\mathrm{x}\left[\begin{array}{cc} 2 \mathrm{x} & 2 \\ 3 & \mathrm{x} \end{array}\right]+2\left[\begin{array}{cc} 8 & 5 \mathrm{x} \\ 4 & 4 \mathrm{x} \end{array}\right]=2\left[\begin{array}{cc} \left(\mathrm{x}^{2}+8\right) & 24 \\ (10) & 6 \mathrm{x} \end{array}\right]

Answers (1)

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

Also,

Addition or subtraction of any two matrices is possible only if they have the same order.

If a given matrix has m rows and n columns, then the order of the matrix is m x n.

We have matrix equation,

\\x\left[\begin{array}{cc}2 x & 2 \\ 3 & x\end{array}\right]+2\left[\begin{array}{cc}8 & 5 x \\ 4 & 4 x\end{array}\right]=2\left[\begin{array}{cc}\left(x^{2}+8\right) & 24 \\ 10 & 6 x\end{array}\right]$ \\Take matrix $\left[\begin{array}{cc}2 x & 2 \\ 3 & x\end{array}\right]$ \\And multiply it with $\mathrm{x}$\\ $x\left[\begin{array}{cc}2 x & 2 \\ 3 & x\end{array}\right]=\left[\begin{array}{cc}x \times 2 x & x \times 2 \\ x \times 3 & x \times x\end{array}\right]$

\Rightarrow \mathrm{x}\left[\begin{array}{cc}2 \mathrm{x} & 2 \\ 3 & \mathrm{x}\end{array}\right]=\left[\begin{array}{cc}2 \mathrm{x}^{2} & 2 \mathrm{x} \\ 3 \mathrm{x} & \mathrm{x}^{2}\end{array}\right]$
Take matrix  \left[\begin{array}{ll}8 & 5 x \\ 4 & 4 x\end{array}\right]$
Multiply it with 2 ,
 \\ 2\left[\begin{array}{ll}8 & 5 x \\ 4 & 4 x\end{array}\right]=\left[\begin{array}{lll}2 \times 8 & 2 \times 5 x \\ 2 \times 4 & 2 \times 4 x\end{array}\right]$ \\$\Rightarrow 2\left[\begin{array}{ll}8 & 5 x \\ 4 & 4 x\end{array}\right]=\left[\begin{array}{cc}16 & 10 x \\ 8 & 8 x\end{array}\right]_{\ldots .(i i)}$
Take matrix \left[\begin{array}{cc}\left(x^{2}+8\right) & 24 \\ 10 & 6 x\end{array}\right]$

Multiply it with 2,

 \\ 2\left[\begin{array}{cc}\left(x^{2}+8\right) & 24 \\ 10 & 6 x\end{array}\right]=\left[\begin{array}{cc}2 \times\left(x^{2}+8\right) & 2 \times 24 \\ 2 \times 10 & 2 \times 6 x\end{array}\right]$ \\$\Rightarrow 2\left[\begin{array}{cc}\left(x^{2}+8\right) & 24 \\ 10 & 6 x\end{array}\right]=\left[\begin{array}{cc}2\left(x^{2}+8\right) & 48 \\ 20 & 12 x\end{array}\right] \ldots$..(iii)
By adding equation (i) and (ii) and make it equal to equation (iii), we get
 \\ \mathrm{x}\left[\begin{array}{cc}2 \mathrm{x} & 2 \\ 3 & \mathrm{x}\end{array}\right]+2\left[\begin{array}{ll}8 & 5 \mathrm{x} \\ 4 & 4 \mathrm{x}\end{array}\right]=2\left[\begin{array}{cc}\left(\mathrm{x}^{2}+8\right) & 24 \\ 10 & 6 \mathrm{x}\end{array}\right]$ \\$\Rightarrow\left[\begin{array}{cc}2 x^{2} & 2 x \\ 3 x & x^{2}\end{array}\right]+\left[\begin{array}{cc}16 & 10 x \\ 8 & 8 x\end{array}\right]=\left[\begin{array}{cc}2\left(x^{2}+8\right) & 48 \\ 20 & 12 x\end{array}\right]$
By adding left side of the matrix equation as they have same order, we get
\Rightarrow\left[\begin{array}{cc}2 x^{2}+16 & 2 x+10 x \\ 3 x+8 & x^{2}+8 x\end{array}\right]=\left[\begin{array}{cc}2\left(x^{2}+8\right) & 48 \\ 20 & 12 x\end{array}\right]$

We need to find the value of x by comparing the elements in the two matrices.

If,

\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]=\left[\begin{array}{ll} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}\right]

Then,

\\a\textsubscript{11} = b\textsubscript{11} \\a\textsubscript{12} = b\textsubscript{12} \\a\textsubscript{21} = b\textsubscript{21} \\a\textsubscript{22} = b\textsubscript{22}

So,

 \\2x\textsuperscript{2} + 16 = 2(x\textsuperscript{2} + 8) $ \ldots $ (i) \\2x + 10x = 48 $ \ldots $ (ii) \\3x + 8 = 20 $ \ldots $ (iii) \\x\textsuperscript{2} + 8x = 12x $ \ldots $ (iv)

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

So, take equation (i).

\\2x\textsuperscript{2} + 16 = 2x\textsuperscript{2} + 16

We cannot find the value of x from this equation as they are similar. 

Now, take equation (ii).

2x + 10x = 48

$ \Rightarrow $ 12x = 48

$ \Rightarrow $ x = 4

From equation (iii),

3x + 8 = 20

\\ \Rightarrow $ 3x = 20 - 8 \\$ \Rightarrow $ 3x = 12

$ \Rightarrow $ x = 4

From equation (iv),

 \\x\textsuperscript{2} + 8x = 12x \\$ \Rightarrow $ x\textsuperscript{2} = 12x -8x \\$ \Rightarrow $ x\textsuperscript{2} = 4x \\$ \Rightarrow $ x\textsuperscript{2} - 4x = 0 \\$ \Rightarrow $ x(x - 4) = 0 \\$ \Rightarrow $ x = 0 or (x - 4) = 0 \\$ \Rightarrow $ x = 0 or x = 4 \\$ \Rightarrow $ x = 4 ($\because$ x = 0 $ does not satisfy equations (ii) and (iii))

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

x = 4

Hence, the value of x is 4.

Posted by

infoexpert22

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