1 - x - x -21 -a -a a -21 - b - b-2lendvmatrix Then- is x - a a factor to this determinant A- TrueB- False

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Standard XII

Mathematics

Special Determinants

1 & x & x ∧21...

Question

$Given f (x)= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} 1 & a & a^{2} 1 & b & b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ Then, is (x - a) a factor to this determinant$

True

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False

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Solution

The correct option is A
True

If we put x = a, f(a) = 0 . It’s important to see that the determinant f(x) is polynomial in $x^{2}$ . So when we put a value x = k for which the function goes to 0 then it’s a root of f(x) or (x-k) is factor for f(x). Therefore (x-a) is a factor for the determinant.

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Similar questions

$Given f (x)= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} 1 & a & a^{2} 1 & b & b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ Then, is (x - a) a factor to this determinant$

Q. The determinants

∣ ∣ ∣ ∣ \begin{matrix} 1 & a & b c 1 & b & c a 1 & c & a b \end{matrix} ∣ ∣ ∣ ∣

and

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} 1 & b & b^{2} 1 & c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

are not identically equal.

The highest common factor of a and b is $x$ . y is also a common factor of a and b if y is a number greater than $x$ .

If $x^{a}$ = $x^{b}$ , then a = b. State whether true or false.

Q. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} (a - 1)^{2} & a^{2} & (a + 1)^{2} (b - 1)^{2} & b^{2} & (b + 1)^{2} (c - 1)^{2} & c^{2} & (c + 1)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = k ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} 1 & b & b^{2} 1 & c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣,

then the value of

k

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Given f (x)=∣∣ ∣ ∣∣1xx21aa21bb2∣∣ ∣ ∣∣Then, is (x - a) a factor to this determinant

The correct option is A True If we put x = a, f(a) = 0 . It’s important to see that the determinant f(x) is polynomial in x2. So when we put a value x = k for which the function goes to 0 then it’s a root of f(x) or (x-k) is factor for f(x). Therefore (x-a) is a factor for the determinant.

$Given f (x)= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} 1 & a & a^{2} 1 & b & b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ Then, is (x - a) a factor to this determinant$