Question

if the matrix
x + 4 x x
x x+4 x
x x x+4
is singular. find x​

Answer 1

Answer:

x = -4/3.

​

Step-by-step explanation:

We are given that the following matrix . . .

$\left[\begin{array}{ccc}\sf x + 4& \sf x&\sf x\\ \sf x & \sf x + 4& \sf x\\ \sf x& \sf x& \sf x + 4\end{array}\right]$

. . . is a singular matrix.

​

Any matrix whose determinant comes out equal to 0 is known as a singular matrix. Therefore, we can say that -

​

$\Longrightarrow \left|\begin{array}{ccc}\sf x + 4& \sf x&\sf x\\ \sf x & \sf x + 4& \sf x\\ \sf x& \sf x& \sf x + 4\end{array}\right| = 0$

​

On expanding the determinant we get,

​

$\Longrightarrow \mathsf{(x + 4) \Big\{ (x+4)(x+4) - (x)(x) \Big\} - x \Big \{ x(x+4) - x(x) \Big \} + x \Big \{ x(x) - x(x+4) \Big\} = 0}$

​

$\Longrightarrow \mathsf{(x + 4) \Big\{x^2 + 4x + 4x + 16 - x^2\Big\} - x \Big \{x^2 + 4x - x^2 \Big \} + x \Big \{x^2 - x^2 -4x \Big\} = 0}$

​

$\Longrightarrow \mathsf{(x + 4) \Big\{8x + 16 \Big\} - x \Big \{4x \Big \} + x \Big \{-4x \Big\} = 0}$

​

$\Longrightarrow \mathsf{ 8x^2 + 16x + 32x + 64 - 4x^2 - 4x^2 = 0}$

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$\Longrightarrow \mathsf{ 8x^2 + 16x + 32x + 64 - 8x^2 = 0}$

​

$\Longrightarrow \mathsf{16x + 32x + 64 = 0}$

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$\Longrightarrow \mathsf{48x + 64 = 0}$

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$\Longrightarrow \mathsf{48x = -64}$

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$\Longrightarrow \mathsf{x = \dfrac{-64}{48}}$

​

$\Longrightarrow \mathsf{x = \dfrac{-16}{12}}$

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$\Longrightarrow \boxed{\mathbf{x = \dfrac{-4}{3}}}$

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Verification:

We can substitute x = -4/3 in our matrix to see if this value of x actually results in a singular matrix.

​

$\Longrightarrow \left|\begin{array}{ccc}\sf \dfrac{-4}{3} + 4& \sf \dfrac{-4}{3} &\sf \dfrac{-4}{3} \\ \\ \sf \dfrac{-4}{3} & \sf \dfrac{-4}{3} + 4 & \sf \dfrac{-4}{3} \\ \\ \sf \dfrac{-4}{3} & \sf \dfrac{-4}{3} & \sf \dfrac{-4}{3} + 4\end{array}\right| = 0$

​

$\Longrightarrow \left|\begin{array}{ccc}\sf \dfrac{8}{3} & \sf \dfrac{-4}{3} &\sf \dfrac{-4}{3} \\ \\ \sf \dfrac{-4}{3} & \sf \dfrac{8}{3}& \sf \dfrac{-4}{3} \\ \\ \sf \dfrac{-4}{3} & \sf \dfrac{-4}{3} & \sf \dfrac{8}{3}\end{array}\right| = 0$

​

Take -1/3 out since it's a common factor in each row.

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$\Longrightarrow \Big \{-\dfrac{1}{3} \times -\dfrac{1}{3} \times -\dfrac{1}{3} \Big \} \left|\begin{array}{ccc}\sf -8 & \sf 4 &\sf 4 \\ \\ \sf 4 & \sf -8 & \sf 4 \\ \\ \sf 4 & \sf 4 & \sf -8 \end{array}\right| = 0$

​

$\Longrightarrow \Big \{-\dfrac{1}{27} \Big \} \left|\begin{array}{ccc}\sf -8 & \sf 4 &\sf 4 \\ \\ \sf 4 & \sf -8 & \sf 4 \\ \\ \sf 4 & \sf 4 & \sf -8 \end{array}\right| = 0$

​

On expanding the determinant we get,

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$\Longrightarrow \mathsf{\Big \{-\dfrac{1}{27} \Big \} \Big \{ \Big -8\big\{-8(-8) - 4(4) \big\} - 4 \big\{4(-8) - 4(4) \big\} + 4 \big\{4(4) + 8(4) \big\}\Big \}}$

​

$\Longrightarrow \mathsf{\Big \{-\dfrac{1}{27} \Big \} \Big \{ \Big -8\big\{64 - 16 \big\} - 4 \big\{-32 - 16 \big\} + 4 \big\{16 + 32 \big\}\Big \}}$

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$\Longrightarrow \mathsf{\Big \{-\dfrac{1}{27} \Big \} \Big \{ \Big -8\big\{48 \big\} - 4 \big\{-48 \big\} + 4 \big\{48\big\}\Big \}}$

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$\Longrightarrow \mathsf{\Big \{-\dfrac{1}{27} \Big \} \Big \{ \Big -384 + 192 + 192 \Big \}}$

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$\Longrightarrow \mathsf{\Big \{-\dfrac{1}{27} \Big \} \Big \{ \Big -384 + 384 \Big \}}$

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$\Longrightarrow \mathsf{\Big \{-\dfrac{1}{27} \Big \} \Big \{ \Big 0 \Big \}}$

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$\Longrightarrow \mathsf{0}$

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Hence verified.

Answer 2

Appropriate Question:-

$\sf \: If\:the\:matrix \: \left[\begin{array}{ccc}x + 4&x&x\\x&x + 4&x\\x&x &x + 4\end{array}\right] \:is \: singular, \: find \: x$

$\large\underline{\sf{Solution-}}$

Given that,

$\sf \: \: \left[\begin{array}{ccc}x + 4&x&x\\x&x + 4&x\\x&x &x + 4\end{array}\right] \:is \: singular$

So,

$\implies\sf \: \left |\begin{array}{ccc}x + 4&x&x\\x&x + 4&x\\x&x &x + 4\end{array}\right | = 0$

$\boxed{\sf \: OP \: R_1 \: \to \: R_1 + R_2 + R_3 \: }$

$\sf \: \left |\begin{array}{ccc}3x + 4&3x + 4&3x + 4\\x&x + 4&x\\x&x &x + 4\end{array}\right | = 0$

Take out 3x + 4 common from first row.

$\sf \: (3x + 4)\left |\begin{array}{ccc}1&1&1\\x&x + 4&x\\x&x &x + 4\end{array}\right | = 0$

$\boxed{\sf \: OP \: C_2 \: \to \: C_2 - C_1 \: }$

$\sf \: (3x + 4)\left |\begin{array}{ccc}1&0&1\\x&4&x\\x&0 &x + 4\end{array}\right | = 0$

$\boxed{\sf \: OP \: C_3 \: \to \: C_3 - C_1 \: }$

$\sf \: (3x + 4)\left |\begin{array}{ccc}1&0&0\\x&4&0\\x&0 &4\end{array}\right | = 0$

On expanding along first row, we get

$\sf \: (3x + 4)(16 - 0) = 0$

$\sf \: 16 (3x + 4) = 0$

$\sf \:3x + 4 = 0$

$\sf \:3x = - 4$

$\implies\sf \: x \: = \: - \: \dfrac{4}{3}$