If | a^2+2 a-5 2 a+4 1 2 a-4 a+5 1 -2 6 1 |>0, then (a)

If | a^2+2 a-5 2 a+4 1 2 a-4 a+5 1...

Key Concepts

Cofactor Expansion

Cofactor expansion, also known as Laplace's expansion, is a method for calculating the determinant of a matrix by expanding along a row or column. This technique reduces the computation of a larger determinant into computations of smaller determinants, which is particularly useful when the matrix has entries that are functions of a parameter.

Polynomial Inequalities

When a determinant contains a variable parameter, its evaluation often yields a polynomial expression. Solving inequalities involving this polynomial requires determining its roots and analyzing the sign of the expression over various intervals. This process is key in identifying the conditions under which the original determinant inequality holds.

Determinant Evaluation

The determinant of a square matrix is a scalar quantity that provides important information about the matrix, such as whether it is invertible or singular. Calculating the determinant involves methods such as cofactor expansion or the rule of Sarrus (for 3x3 matrices) and is essential for understanding how the matrix scales volume in a corresponding space.

Parameter-Dependent Matrices

In many problems, matrix elements are functions of a parameter, and studying how the determinant changes with the parameter is important in contexts such as stability analysis or identifying transitions in system behavior. Analyzing parameter-dependent determinants leads to understanding the relationships and conditions that the parameter must satisfy for the matrix to exhibit certain properties.

Transcript

00:01 So in the given question we have we are given the determinant of a matrix that is given us a square plus 2a minus 5 2a plus 4 1 the second row is 2a minus 4 a plus 5 1 and the third row is minus 2 a minus 4 a plus 5 1 and the third row is minus 2 6 1 and we are told that the value of this determinant is if the value of this determinant is greater than 0 then we should find which of the given options is the correct option so the options are a greater than 1 a equal to 1 the third option is a less than 1 and the 4th option is a equal to 0 so these are the options that we have and we are asked to find which is the correct answer out of these options.

01:13 So let's do a column, a row operations, a row operation on the given matrix, right? so let's take, let's change r1 to r1 minus r3, r1 minus r3 and we can change r2 to r2 minus r3 as well.

01:39 So from this what we would have is we would have the determinant as a square plus 2a minus 3, 2a minus 2 0, 2a minus 2a minus 1 0 minus 2 6 1.

02:03 The third rule is unchanged.

02:05 So this is what we have and now we can easily evaluate this determinant right? so this determinant is then equal to a square minus a square plus 2a minus 3 times a minus 1 minus 2 a minus 2 2 which is 2 a minus 2 squared and this let's write we can simplify this a square plus 2a minus 3 by writing it as a square let's take this quadratic equation separately right so after taking this quadratic equation separately that is a square plus a square plus 2a minus 3 what we can write is you can write a square minus plus 3a minus a minus a minus 3 we can write 2a as 3a minus a minus a from which we can take a over here as a common factor and from here we can take minus 1 as a common factor and then we can write the simplified form of a square plus 2a minus 3 as a plus 3 times a minus 1 right so let's use this substitution this simplified result as the substitution for a square plus 2a minus 3 in the expansion over here so you would have a plus 3 times a minus 1 times a minus 1 minus 2 a minus 2 squared and this is equal to a plus 3 times a minus 1 squared minus 2a minus 2 squared so this is what we have after simplifying and now what we can do is we can write let's take write 2 a minus 2 as 2 times a minus 1 squared and this would be equal to we can write a plus 3 times a minus 1 squared minus 2 square times a minus 1 squared which is equal to 4 times a minus 1 squared right so now we can take a minus 1 as a common factor right so if we take a minus 1 as a to a minus 1 squared times a minus 1 a plus 3 minus 1 so we got a minus 1 again so we would have the simplified result as a minus 1 q right so this is what we got as the as the simplified result so the value of the determinant is a minus 1 q is a minus 1 cube and we are told that a minus 1 in the determinant is greater than 0 so we are told in the problem that the determinant is greater than 0 so what we can write from this is that a minus 1 whole cube would be greater than 0 only if we can see if let's take a positive number which is greater than 1.

06:22 So if you take 1 it will be 0 over here which means the determinant won't be greater than 0...

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