To find that, you multiply the first element of the first row by the second element of the second row. You also multiply the first element of the second row with the second element of the first row. Then you subtract the products not add them.
Oh, what a lovely question! To solve by Cramer's rule, you'll first calculate the determinant of the coefficient matrix, then create matrices by replacing the x-column with the constants and solving for the determinant of that matrix. Finally, you'll divide these determinants to find the values of x and y. Just like painting a happy little tree, take your time and follow each step with care.
To answer this question, let me establish an example 3 x 3 matrix named "A": A= [a b c] [d e f] [g h i] The formula I will give you, called co-factor expansion, works for any size square matrix, so you could use it to find the determinant of a 2 x 2, 3 x 3, all the way up to an n x n matrix. To find the determinant, pick any row or column in the matrix. It will make your work much easier if you choose a row or column that has many zeroes in it. A general notation that is often used to find the determinant of a matrix is to use straight bars in place of the brackets surrounding the matrix contents. So, if I was to say mathematically that I was finding the determinant of the above example matrix, I could write it as: det(A)= |a b c| |d e f| |g h i| This notation will be used in the formula, so it is important to know this. For the sake of an arbitrary example, let us suppose I chose Row 1 of the matrix as my chosen row. To find the determinant of this matrix, I will perform the following calculation: (-1)2(a)|e f| + (-1)3(b)|d f| + (-1)4(c)|d e| |h i| |g i| |g h| This is the specific application of this general formula to the example matrix: (-1)i+j(aij)det(A1) In this formula, i and j are the row and column addresses, respectively, of a given matrix element. So, like in our specific application, when Row 1 was chosen as our subject row, the first term was (-1)1+1(A11)det(A1). The element "a" is in the first row, first column of the matrix, mean i=1 and j=1, therefore the superscript of (-1) is 1+1=2. A11 is simply the value held in the address i=1, j=1 of the matrix A. For this application, A11 was "a". det(A1) is the determinant of the submatrix A1. This submatrix has no formal nomenclature, I simply call it this for ease of explanation. A1 is the matrix created by "crossing out" the row and column that belong to the matrix element A11. In this application, that means it is the submatrix that is left after crossing out a, b, c, d, and g, which is simply the 2 x 2 matrix e,f;h,i. Performing this same process for the remainder of the matrix elements in Row 1 will yield the determinant of the matrix. So, the "generalized" form of the specific application above is: (-1)1+1(A11)det(A1) + (-1)1+2(A12)det(A2) + (-1)1+3(A13)det(A3) where A1 is the submatrix created by crossing out Row 1 and Column 1, A2 is the submatrix created by crossing out Row 1 and Column 2, and A3 is the submatrix created by crossing out Row 1 and Column 3. A final note is how to calculate the determinants of the submatrices. For a 3 x 3 matrix, its submatrices are all 2 x 2. For 2 x 2 matrices, a simple formula exists that makes this easy: |a b| = (ad) - (bc) |c d| For higher-dimension matrices, the submatrices also become larger, making the computation much more intensive.
A quadratic of the form ax² + bx + c = 0 has two equal (real) roots when the determinant of b²-4ac = 0. Thus, for: 2x² - (k - 1)x + 80 = 0 b² - 4ac = (-(k - 1))² - 4×2×80 = 0 → (k - 1)² = 640 → k - 1 = ±√640 → k = 1 ± √640 → k ≈ 26.298 or -24.298
Oh, dude, an eigenvector is like a fancy term in math for a vector that doesn't change direction when a linear transformation is applied to it. It's basically a vector that just chills out and stays the same way, no matter what you do to it. So, yeah, eigenvectors are like the cool, laid-back dudes of the math world.
For a matrix A, A is read as determinant of A and not, as modulus of A. ... sum of two or more elements, then the given determinant can be expressed as the sum
A single math equation does not have a determinant. A system of equations (3x3 , 4x4, etc.) will have a determinant. You can find a determinant of a system by converting the system into a corresponding matrix and finding its determinant.
Assuming that the terms, a and AA, are commutative, It is 1 + a^3 + (AA)^3 - 3aAA
To find the determinant of a matrix on a Casio fx-991MS calculator, you first need to enter the matrix into the calculator using the matrix mode. Then, navigate to the matrix menu and select the matrix you want to find the determinant of. Finally, choose the option to calculate the determinant, and the calculator will display the result. Remember that the determinant of a matrix is a scalar value that represents certain properties of the matrix.
233
There is no easy way to find the determinant; it's long and tedious. There are computer programs available (like MATLAB) that will find the determinant. You'll find there probably won't be a large matrix in an exam if you're required to find the determinant.
It isn't clear what you want to solve for. If you want to find the matrix, there is not a unique solution - there are infinitely many matrices with the same determinant.
If it a 2x2 matrix, the determinant is 3*a - (-2)*5 = 3a + 10 = 7 So 3a = -3 so a = -1
The determinant.The determinant is the part under the square root of the quadratic equation and is:b2-4ac where your quadratic is of the form: ax2+bx+cIf the determinant is less than zero then you have 'no real solutions' (as the square root of a negative number is imaginary.)If the determinant is = 0, then you have one real solution (because you can discount the square root of the quadratic equation)If the determinant is greater than zero you have two real solutions as you have (-b PLUS OR MINUS the square root of the determinant) all over 2aTo find the solutions where they exist you'll need to solve the quadratic formula or use another method.
Determine whether the determinant of M is zero, and if so, find the values of α1 and α2 for which this occurs.
What is the solution of this question By wronskian determinant method please find whether the solution exist or not where.... y1 = (squre of) cos x, y2 = 1 + cos 2 x.
This cannot be answered. This does not make any sense.