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\( x \) \( A \)

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Q1

∣ ∣ ∣ ∣ \begin{matrix} a + x & a - x & a - x a - x & a + x & a - x a - x & a - x & a + x \end{matrix} ∣ ∣ ∣ ∣ = 0

then

x

is

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Q2

using the properties of determinants, solve for x: |a+x a-x a-x a-x a+x a-x a-x a-x a+x| = 0

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Q3

If

∣ ∣ ∣ ∣ \begin{matrix} a + x & a - x & a - x a - x & a + x & a - x a - x & a - x & a + x \end{matrix} ∣ ∣ ∣ ∣ = 0

the value of

x

are

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Q4

Using properties of determinants, solve the following equation for x :
| a+x a-x a-x |
| a-x a+x a-x | = 0
| a-x a-x a+x |

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Q5

If

∣ ∣ ∣ ∣ \begin{matrix} a + x & a - x & a - x a - x & a + x & a - x a - x & a - x & a + x \end{matrix} ∣ ∣ ∣ ∣ = 0

then

x

is

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