Let R=\(\begin{pmatrix}a&3&b\\c&2&d\\0&5&0\end{pmatrix}\): a,b,c,d ∈ {0,3,5,7,11,13,17,19}. Then the number of invertible matrices in R is
Approach Solution - 1
The absolute value of matrix R is negative 5, and the absolute value of matrix P is also negative 5. The determinant ∣R∣ can be equal to zero in the following cases:
(i) Two of the values a, b, c, d are zeros, which can be (a and b), (b and d), (d and c), or (c and a) \(→\) 4 × 72 ways = 196.
(ii) Any three of the values a, b, c, d are zeros \(→\) 4C3×7=28.
(iii) All four of the values a, b, c, d are zeros \(→\) 1.
(iv) All four of the values a, b, c, d are non-zero but the same number \(→\) 7.
(v) When two are alike and the other two are alike (non-zero) \(→\) 7C2×2×2=84.
The number of invertible matrices = 84 – 196 – 28 – 1 – 7 – 84 = 3780.
So, the answer is 3780.
Approach Solution -2
Given :
Total matrices = 84 = 4096
|R| = 5(bc – ad)
No. of non-invertible matrices : bc = ad
Case I :
If a, b, c, d ≠ 0, then cases = 7C2 . 2! 2! + 7C1(1) = 91
Case II :
if ad = bc = 0, then cases = 15C1 . 15C1 = 225
( ad & bc can take any combination from 0 × 0, 0 × 3, 0 × 5, ……, 0 × 19, 3 × 0, 5 × 0, ….. 19 × 0)
Number of invertible matrices :
= 4096 – (91 + 225)
= 3780
So, the correct answer is 3780.
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Concepts Used:
Linear Inequalities in One Variable
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The addition property of inequality says that adding the same number to each side of the inequality gives an equivalent inequality.
If x>y, then x+z>y+z If x>y, then x+z>y+z
If x<y, then x+z<y+z If x<y, then x+z<y+z
The subtraction property of inequality tells us that subtracting the same number from both sides of an inequality produces an equivalent inequality.
If x>y, then x−z>y−z If x>y, then x−z>y−z
If x<y, then x−z<y−z Ifx<y, then x−z<y−z
The multiplication property of inequality tells us that multiplication on both sides of an inequality with a positive number gives an equivalent inequality.
If x>y and z>0, then xz>yz If x>y and z>0, then xz>yz
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