No. Determinants are only defined for square matrices.
No. Determinants are only defined for square matrices.
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
That's a special calculation done on square matrices - for example, on a 2 x 2 matrix, or on a 3 x 3 matrix. For details, see the Wikipedia article on "Determinant".
Gauss Elimination
In theory, a 2x2 determinant requires the evaluation of 2 products, a 3x3 determinant requires 6 products, a 4x4 determinant requires 24 products (note: that is the factorial function). The Rule of Sarrus is just a convenient memory aid for this specific case.
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.
A determinant is defined only for square matrices, so a 2x3 matrix does not have a determinant.Determinants are defined only for square matrices, so a 2x3 matrix does not have a determinant.
If it a 2x2 matrix, the determinant is 3*a - (-2)*5 = 3a + 10 = 7 So 3a = -3 so a = -1
If the matrix is { a1 b1 c1} {a2 b2 c2} {a3 b3 c3} then the determinant is a1b2c3 + b1c2a3 + c1a2b3 - (c1b2a3 + a1c2b3 + b1a2c3)
That's a special calculation done on square matrices - for example, on a 2 x 2 matrix, or on a 3 x 3 matrix. For details, see the Wikipedia article on "Determinant".
Assuming that the terms, a and AA, are commutative, It is 1 + a^3 + (AA)^3 - 3aAA
Gauss Elimination
First we need to ask what you mean by a matrix equalling a number? A matrix is a rectangular array of numbers all of which might be zero and this is called the zero matrix. We can take the determinant of a square matrix such as a 3x3 and this may be zero even without the entries being zero.
In theory, a 2x2 determinant requires the evaluation of 2 products, a 3x3 determinant requires 6 products, a 4x4 determinant requires 24 products (note: that is the factorial function). The Rule of Sarrus is just a convenient memory aid for this specific case.
Call your matrix A, the eigenvalues are defined as the numbers e for which a nonzero vector v exists such that Av = ev. This is equivalent to requiring (A-eI)v=0 to have a non zero solution v, where I is the identity matrix of the same dimensions as A. A matrix A-eI with this property is called singular and has a zero determinant. The determinant of A-eI is a polynomial in e, which has the eigenvalues of A as roots. Often setting this polynomial to zero and solving for e is the easiest way to compute the eigenvalues of A.
Yes it is possible. The resulting matrix would be of the 2x3 order.
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.