If | x ℓ m 1 a x m 1 a b x 1 a b

If | x ℓ m 1 a x m 1 a...

If $\left|\begin{array}{cccc}x & \ell & m & 1 \\ a & x & m & 1 \\ a & b & x & 1 \\ a & b & c & 1\end{array}\right|=x^{3}-p x^{2}+q x-r$ then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is
(a) $\mathrm{pqr}$
(b) $\frac{\mathrm{p}}{\mathrm{r}}$
(c) $\frac{q}{r}$
(d) $\frac{p}{q}$

Key Concepts

Symmetric Sums of Roots

Symmetric sums of roots involve combinations of the roots of a polynomial (such as their sum, pairwise products, and triple product) that remain invariant under any permutation of the roots. In problems like this, expressing the sum of the reciprocals of roots in terms of these symmetric sums, and thus in the polynomial coefficients, is a critical technique.

Vieta's Formulas

Vieta's formulas provide relationships between the coefficients of a polynomial and sums and products of its roots. They are especially useful in relating elementary symmetric functions of the roots to the polynomial's coefficients, which is a key step in evaluating expressions like the sum of the reciprocals of the roots in terms of the coefficients.

Determinant

A determinant is a scalar value that can be computed from a square matrix and encapsulates important information about the matrix, such as its invertibility and the volume scaling factor of the linear transformation it represents. In this context, the determinant is expressed as a cubic polynomial in x, linking the matrix elements to the polynomial's coefficients.

Polynomial Identity

A polynomial identity expresses a determinant or any algebraic expression as a polynomial in a variable where the coefficients carry specific importance. Here, matching the determinant to a cubic polynomial allows us to relate the matrix entries to the coefficients p, q, and r of the polynomial.

Transcript

00:01 So in the given question we have a determinant that is given as if the determinant of xlm1 a x m1 a b xm1 ab x1 ab c1 is equal to if this determinant is equal to if this determinant is equal to x cube minus p x square plus q x minus r then what is then what is the value of 1 by a plus 1 by b plus 1 by c equal to right and in the option all the answers are in terms of p q and r right so this is what we are given in the question.

01:04 So what we can do over here? so first what we are going to do is to simplify the determinant.

01:11 So first we can do a set of row operations.

01:16 We can write we can change r1 to r1 minus r4.

01:22 We can change r2 to r2 minus r4.

01:29 Similarly we can change r3 to r3 minus r4.

01:34 So if we do these changes we will get a determinant that is we will get the determinant as x minus a l minus b l minus b m minus c and 0 in the first row second row we have 0 x minus b m minus c 0 third row we have 0 0 0 x minus c 0 3rd row we have 0 0 0 0 x minus c 0 and the fourth row is unchanged that is a, b, c 1.

02:14 And now what we are going to do is evaluate this determinant which would give us x minus a this is equal to x minus a times x minus b times x minus c and we can expand this determinant to get x cube minus a plus b plus c times x plus ab plus bc plus c a times x plus ab plus b c plus c times x minus a bc so this is what we get right so from this what we can do is we since we have a cubic equation we can and write if p q and r in the given question we have the cubic equation that is given as the answer as the value of the determinant so the cubic equation that is given in the question is x cube minus p x x square so over here it is x square so x cube minus px square plus q x minus r is equal to 0.

03:45 So we can compare both these, both this.

03:49 It is not equal to zero right...

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