Skip to content

Problems of the Day

Key problems of my day:

Boy scratching his head in confusion

  1. If \(\alpha, \beta\) are roots of the equation \(x^2 - 4x - 3 = 0\), then find

    \[ \frac{\alpha^{-\alpha\beta}\left[\alpha^{\alpha}\alpha^{\beta} - 4\alpha^{-\alpha\beta}\right] + \beta^{-\alpha\beta}\left[\beta^{\alpha}\beta^{\beta} - 4\beta^{-\alpha\beta}\right]}{\alpha.\alpha^{\alpha+\beta} + \beta.\beta^{\alpha+\beta}} \]
    Hint

    Correct Answer: 3

    Hint: The given quadratic equation is \(x^2 - 4x - 3 = 0\). According to Vieta's formulas for the roots of quadratic equations:

    \(\alpha + \beta = 4\) (sum of roots)

    \(\alpha\beta = -3\) (product of roots)

    The expression can be simplified using these relationships:

    \[ \frac{\alpha^{-\alpha\beta}\left[\alpha^{\alpha}\alpha^{\beta} - 4\alpha^{-\alpha\beta}\right] + \beta^{-\alpha\beta}\left[\beta^{\alpha}\beta^{\beta} - 4\beta^{-\alpha\beta}\right]}{\alpha\alpha^{\alpha+\beta} + \beta\beta^{\alpha+\beta}} \]

    Simplify within the brackets using the properties of exponents, we get:

    \[ \frac{\alpha^{-\alpha\beta}\left[\alpha^{\alpha + \beta} - 4\alpha^{-\alpha\beta}\right] + \beta^{-\alpha\beta}\left[\beta^{\alpha + \beta} - 4\beta^{-\alpha\beta}\right]}{\alpha^{\alpha + \beta + 1} + \beta^{\alpha + \beta + 1}} \]

    Since \(\alpha + \beta = 4\) and \(\alpha\beta = -3\), substitute these into the equation:

    \[ \frac{\alpha^{3}\left[\alpha^{4} - 4\alpha^{3}\right] + \beta^{3}\left[\beta^{4} - 4\beta^{3}\right]}{\alpha^{4 + 1} + \beta^{4 + 1}} \]
    \[ \frac{(\alpha^7+\beta^7)-4(\alpha^6+\beta^6)}{\alpha^{5} + \beta^{5}} \]

    Given the Newton's identity, for the roots of the quadratic equation, where \( P_n = \alpha^n + \beta^n \), we have the relation:

    \[ P_n - 4P_{n-1} - 3P_{n-2} = 0 \]

    This identity can be used to relate the sums of the powers of the roots of the equation. Let's apply this to our case, where we set \( n = 7 \):

    \[ P_7 - 4P_6 - 3P_5 = 0 \]

    This can be rewritten as:

    \[ P_7 - 4P_6 = 3P_5 \]

    Thus,

    \[\frac{P_7-4P_6}{P_5}=3\]

  2. Consider the set \( S \), which consists of distinct 8-digit permutations constructed from the digits {1, 1, 2, 2, 3, 3, 4, 5}. A permutation is randomly selected from \( S \).

    Define \( \lambda \) as the probability that the permutation will have no pair of identical digits in consecutive positions. Determine the value of \( 168\lambda \).

    Solution

    To solve this problem using the inclusion-exclusion principle, we need to calculate the number of permissible permutations where no identical digits are consecutive.

    Given the set \( S \) with the digits {1, 1, 2, 2, 3, 3, 4, 5}, we define:

    • \( A_1 \) as the set of permutations where the two 1's are adjacent.
    • \( A_2 \) as the set of permutations where the two 2's are adjacent.
    • \( A_3 \) as the set of permutations where the two 3's are adjacent.

    We want to find the number of permutations in the set \( A_1' \cap A_2' \cap A_3' \), which is the complement of the set \( A_1 \cup A_2 \cup A_3 \). By De Morgan's laws, this is equal to \( n((A_1 \cup A_2 \cup A_3)') \).

    Using the inclusion-exclusion principle, this count is given by:

    \[ n(S) - \binom{3}{1}n(A_1) + \binom{3}{2}n(A_1 \cap A_2) - n(A_1 \cap A_2 \cap A_3) \]
    \[ n(S) - \binom{3}{1}n(A_1) + \binom{3}{2}n(A_1 \cap A_2) - n(A_1 \cap A_2 \cap A_3), \]

    Certainly, let's calculate the expression compactly:

    \[ n(S) - \binom{3}{1} n(A_1) + \binom{3}{2} n(A_1 \cap A_2) - n(A_1 \cap A_2 \cap A_3) = \]
    \[ = \frac{8!}{2!2!2!} - \binom{3}{1} \frac{7!}{2!2!} + \binom{3}{2} \frac{6!}{2!} - 5! = 2220 \]

    Thus the required probabilty is \(\frac{2220}{5040} = \frac{74}{168}\)

    Hence, \(168\lambda = 74\)

  3. Let \( A \) be a \( 2 \times 2 \) square matrix with real entries such that \( \text{det}(A^2 + 4I_2) = 0 \). Then, which of the following is(are) necessarily true? (\( I_2 \) is the identity matrix of order 2)

    A) \( A^2 + 4I_2 \) is a null matrix

    B) \( A^2 + A \) is a null matrix

    C) \( \text{det}(A^2 + A + 4I_2) = 4 \)

    D) \( \text{det}(A^2 + A + 5I_2) = 5 \)

    Solution:

    Let \( A^2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \). The determinant condition \( \text{det}(A^2 + 4I_2) = 0 \) leads to \( \text{det} \begin{pmatrix} a+4 & b \\ c & d+4 \end{pmatrix} = 0 \). Hence, \( (a+4)(d+4) - bc = 0 \), simplifying to \( ad - bc = -4(a+d) - 16 \).

    Since \(ad-bc=|A^2|\), we get \(|A^2|=-4(a+d)-16 \implies |A|^2=-4(a+d)-16\)

    Now let \( A = \begin{pmatrix} p & q \\ r & s \end{pmatrix} \). Then \( A^2 = \begin{pmatrix} p^2+qr & pq+qs \\ rp+rs & r^2+s^2 \end{pmatrix} \). Equating entries, \( a = p^2+qr \), \( b = pq+qs \), \( c = rp+rs \), and \( d = q^2+s^2 \). Thus,

    \(|A|^2=-4(a+d)-16\implies\)\( (ps-qr)^2 = -4(p^2 + qr + s^2 + 2qs) - 16 \), which we rearrange as \( (ps - qr)^2 = -4((p+s)^2 - 2(ps - qr)) - 16 \). Expanding, \( (ps - qr)^2 + 8(ps - qr) + 16 = -4(p+s)^2 \). Rearranging, \( (ps - qr - 4)^2 + 4(p+s)^2 = 0 \). This implies \( ps - qr - 4 = 0 \) and \( p+s = 0 \).

    Thus, we deduce \( \text{det}(A) = ps - qr = 4 \). Considering the squared matrix \( A^2 \) and the trace of \( A \), we rewrite \( A^2 \) as \( A^2 = \begin{pmatrix} -sp+qr & q(p+s) \\ r(p+s) & -sp+qr \end{pmatrix}\)\(\implies A^2 = \begin{pmatrix} -4 & 0 \\ 0 & -4 \end{pmatrix} \), which shows that \( A^2 + 4I = 0 \), verifying that option A is true and \( A^2 + 4I_2 \) is a null matrix.

    Option C is true as well.