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
Correct Answer: 3780
Solution and Explanation
The determinant \( |R| \) can be equal to zero in the following cases:
Case (i): Two values are zero
If two of the values \( a, b, c, d \) are zero, there are 4 possible pairs: (a and b), (b and d), (d and c), or (c and a). Each case can occur in \( 7^2 \) ways, so the total number of such cases is:
\(4 \times 7^2 = 196\)
Case (ii): Any three values are zero
If any three of the values \( a, b, c, d \) are zero, the number of such cases is given by the combination \( \binom{4}{3} \times 7 \), which is:
\(\binom{4}{3} \times 7 = 28\)
Case (iii): All four values are zero
If all four values \( a, b, c, d \) are zero, there is only 1 such case:
\(1\)
Case (iv): All four values are non-zero but the same
If all four values \( a, b, c, d \) are non-zero and the same, there are 7 possible values for them:
\(7\)
Case (v): Two values are alike and the other two are alike (non-zero)
If two values are alike and the other two are alike, there are \( \binom{7}{2} \times 2 \times 2 \) such cases. This gives:
\(\binom{7}{2} \times 2 \times 2 = 84\)
Calculating the number of invertible matrices:
The total number of invertible matrices is given by the total number of cases minus the non-invertible cases:
\(84 - 196 - 28 - 1 - 7 - 84 = 3780\)
Thus, the number of invertible matrices is 3780.
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Concepts Used:
Linear Inequalities in One Variable
The graph of a linear inequality in one variable is a number line. We can use an open circle for < and > and a closed circle for ≤ and ≥.
Inequalities that have the same solution are commonly known as equivalents. There are several properties of inequalities as well as the properties of equality. All the properties below are also true for inequalities including ≥ and ≤.
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
If x<y and z>0, then xz<yz If x<y and z>0,then xz<yz