Value of | a^2+1 a b a c a d a b b^2+1 b c b d a c b

Value of | a^2+1 a b a c a d a b ...

Key Concepts

Outer Product

The outer product of two vectors creates a matrix where each element is the product of the corresponding entries of the vectors. This operation results in a rank-one matrix and is a foundational concept in understanding how low-rank changes affect the properties of matrices, including their determinants.

Rank-One Update

A rank-one update is the process of adding the outer product of two vectors (a rank-one matrix) to an existing matrix. This kind of modification is significant because it changes the matrix in a controlled and simple manner, often allowing the use of specialized formulas like the matrix determinant lemma to efficiently compute determinants.

Determinant

The determinant is a scalar value computed from a square matrix that provides key insights into the linear transformation represented by the matrix, such as whether the transformation is invertible, and how it scales volumes. It's a fundamental concept in linear algebra with many useful properties regarding matrix operations.

Matrix Determinant Lemma

The matrix determinant lemma gives a way to calculate the determinant of a matrix that has been altered by a rank-one update. Specifically, for an invertible matrix A and column vectors u and v, it states that det(A + uv^T) = det(A) (1 + v^T A^(-1) u). When A is the identity matrix, as in many cases, the formula simplifies considerably.

Transcript

00:01 So in the given question we are told to find the value of a determinant of a matrix that is given as it is the matrix is given as a square plus 1, ab ac ad, the second row is ab b square plus 1 b c bd, the third row is ac b c c square plus 1 cd and the third row is ad bd cd and d square plus 1.

00:46 So this is the matrix that we need to evaluate right so first the method of solving such problems is to take as many common factors outside the determinant and simplify the determinant and once the determinant is simplified we can easily find the value of the determinant.

01:10 So first what we are going to do is to take the common factors we have the common factor a in the first row, b in the second row c in the third row and d in the fourth row so it will be a b c d times s a plus a square plus 1 by a b c d b c d ab c d a ab c d a b b c d a b square plus 1 by b c d and in the fourth row we will have have a, b, c, d squared plus 1 by d.

02:10 So this is what we have as the simplified determinant in the after the first step, right? now what we are going to do is we can take the common factors, we can multiply the first, second, third and fourth columns by a, b, c and d separately now right so we are going to multiply each columns with a b c and d and what we would get from this is when we multiply the first column with a we have a square plus 1 and 1 and in the rest we have a square and the rest of the column is all a square right next we have b square b squared plus 1 b square b square b square next is c square c square plus 1 c square and then d square d square d square plus 1 d d d square b square plus 1 so now let's add let's add let's change this first column of this matrix by adding c1, c2, c3 and c4.

03:46 And what we would have is the determinant of a square plus b square plus c square plus 1, b square, c square, d square, and a square plus b square plus b square plus c square plus 1, c, b square plus 1, c square, d square.

04:12 Next we have a square plus b square plus c square plus 1, b square plus 1, b square plus 1 d square.

04:21 And in the fourth row we have a square, a square plus b square plus b square, d square plus 1.

04:36 And now we can take the common factor which is a square plus b square plus one outside the determinant.

04:46 So we will have a square plus b squared plus c squared plus one outside the determinant and the first column of the determinant are all ones and the rest is the same that is b square c square d square b square plus 1 b square b square c square c square plus 1 c square d square d square plus 1 so the rest is the same right and now what we do so now in the next step what we do is let's take let's do some row operations right so we have a lot of rows with the same elements...

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