Let M = A = a b c d : a , b , c , d ∈ ± 3 , ± 2 , ± 1 , 0 . Define f : M → Z , as f A = det A , for all A ∈ M where Z is set of all integers. Then the number of A ∈ M such that f A = 15 is equal to .

Question

Let M = A = a b c d : a , b , c , d &#8712; &#177; 3 , &#177; 2 , &#177; 1 , 0 . Define f : M &#8594; Z , as f A = det A , for all A &#8712; M where Z is set of all integers. Then the number of A &#8712; M such that f A = 15 is equal to .

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Correct answer: 0

. A = a d - b c = 15 where a , b , c , d &#8712; &#177; 3 , &#177; 2 , &#177; 1 , 0 Case I : a d = 9 &#160; &#38; &#160; b c = - 6 For a d possible pairs are 3 , 3 , - 3 , - 3 . For b c possible pairs are 3 , - 2 , - 3 , 2 , - 2 , 3 , 2 , - 3 So, total number of matrices in case I = 2 &#215; 4 = 8 Case II : a d = 6 &#160; &#38; &#160; b c = - 9 Similarly, total number of matrices in case II = 2 &#215; 4 = 8 Hence, total number of matrices are 16 .

Let M = A = a b c d : a , b , c , d ∈ ± 3 , ± 2 , ± 1 , 0 . Define f : M → Z , as f A = det A , for all A ∈ M where Z is set of all integers. Then the number of A ∈ M such that f A = 15 is equal to .

Correct answer

Step-by-step solution

More from Matrices

Let M = A = a b c d : a , b , c , d &#8712; &#177; 3 , &#177; 2 , &#177; 1 , 0 . Define f : M &#8594; Z , as f A = det A , for all A &#8712; M where Z is set of all integers. Then the number of A &#8712; M such that f A = 15 is equal to .

Correct answer

Step-by-step solution

More from Matrices

Let M = A = a b c d : a , b , c , d ∈ ± 3 , ± 2 , ± 1 , 0 . Define f : M → Z , as f A = det A , for all A ∈ M where Z is set of all integers. Then the number of A ∈ M such that f A = 15 is equal to .