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If the determinant |(x+a,p+u,l+f),(y+b,q...

If the determinant `|(x+a,p+u,l+f),(y+b,q+v,m+g),(z+c,r+w,n+h)|` splits into exactly `k` determinants of order 3, each element of which contains onlyone term, then the value of `k` is 8.

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To solve the problem, we need to analyze the determinant given and determine how many distinct determinants of order 3 can be formed from it. The determinant is given as: \[ \Delta = \begin{vmatrix} x + a & p + u & l + f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{vmatrix} \] ### Step 1: Identify the structure of the determinant The determinant consists of sums of variables. Each element in the determinant can be expressed as a sum of two terms. ### Step 2: Expand the determinant We can expand the determinant using the properties of determinants. Each term in the determinant can be treated as a separate entity. ### Step 3: Determine the number of combinations Since each element in the determinant can be expressed as a sum of two terms, we can think of each element as contributing to two different determinants. For each row, we can choose either the first term or the second term from each element. Since there are 3 rows and each row has 2 choices (the first term or the second term), the total number of combinations is: \[ 2^3 = 8 \] ### Step 4: Conclusion Thus, the determinant can be split into exactly \( k = 8 \) determinants of order 3, each containing only one term. ### Final Answer The value of \( k \) is \( 8 \). ---
To solve the problem, we need to analyze the determinant given and determine how many distinct determinants of order 3 can be formed from it. The determinant is given as: \[ \Delta = \begin{vmatrix} x + a & p + u & l + f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{vmatrix} ...
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