Understanding the Problem: Calculating det(I + AA')

The question asks us to find the determinant of a matrix formed by adding the identity matrix \(I\) to the product of a given column matrix \(A\) and its transpose \(A'\). We are given the matrix \(A\) and told that \(I\) is the 3 × 3 identity matrix.

Let's break down the steps required to solve this problem:

1. Identify the given matrix \(A\) and the identity matrix \(I\).
2. Find the transpose of matrix \(A\), denoted as \(A'\).
3. Calculate the matrix product \(AA'\).
4. Calculate the matrix sum \(I + AA'\).
5. Find the determinant of the resulting matrix \(I + AA'\).

Identifying Matrices A and I

The given matrix \(A\) is a 3 × 1 column matrix:

\(A = \left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]\)

The identity matrix \(I\) is a 3 × 3 matrix with ones on the main diagonal and zeros elsewhere:

\(I = \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\)

Calculating the Transpose of A (A')

The transpose of a matrix is obtained by flipping the matrix over its diagonal; rows become columns and columns become rows. Since \(A\) is a 3 × 1 matrix, its transpose \(A'\) will be a 1 × 3 row matrix.

\(A' = \left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right]\)

Calculating the Matrix Product AA'

Now we multiply the 3 × 1 matrix \(A\) by the 1 × 3 matrix \(A'\). The resulting matrix will have dimensions (3 × 1) × (1 × 3), which is a 3 × 3 matrix.

\(AA' = \left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right]\)

To find the elements of the resulting 3 × 3 matrix, we multiply each element of \(A\) (which is a column vector) by each element of \(A'\) (which is a row vector). The element at row \(i\), column \(j\) of \(AA'\) is the product of the \(i\)-th element of \(A\) and the \(j\)-th element of \(A'\).

\(AA' = \left[\begin{array}{lll} 1 \times 1 & 1 \times 2 & 1 \times 3 \\ 2 \times 1 & 2 \times 2 & 2 \times 3 \\ 3 \times 1 & 3 \times 2 & 3 \times 3 \end{array}\right] = \left[\begin{array}{rrr} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array}\right]\)

Calculating the Matrix Sum I + AA'

Next, we add the 3 × 3 identity matrix \(I\) to the matrix \(AA'\) we just calculated. Matrix addition is done element-wise.

\(I + AA' = \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] + \left[\begin{array}{rrr} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array}\right]\)

\(I + AA' = \left[\begin{array}{lll} 1+1 & 0+2 & 0+3 \\ 0+2 & 1+4 & 0+6 \\ 0+3 & 0+6 & 1+9 \end{array}\right] = \left[\begin{array}{rrr} 2 & 2 & 3 \\ 2 & 5 & 6 \\ 3 & 6 & 10 \end{array}\right]\)

Calculating the Determinant of I + AA'

Now we need to find the determinant of the matrix \( \left[\begin{array}{rrr} 2 & 2 & 3 \\ 2 & 5 & 6 \\ 3 & 6 & 10 \end{array}\right] \). We can use the cofactor expansion method along the first row.

The determinant of a 3 × 3 matrix \(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\) is given by \(a(ei - fh) - b(di - fg) + c(dh - eg)\).

For our matrix \( \left[\begin{array}{rrr} 2 & 2 & 3 \\ 2 & 5 & 6 \\ 3 & 6 & 10 \end{array}\right] \), the determinant is:

\( \det(I + AA') = 2 \times \det \left[\begin{array}{rr} 5 & 6 \\ 6 & 10 \end{array}\right] - 2 \times \det \left[\begin{array}{rr} 2 & 6 \\ 3 & 10 \end{array}\right] + 3 \times \det \left[\begin{array}{rr} 2 & 5 \\ 3 & 6 \end{array}\right] \)

Calculate the 2 × 2 determinants:

• \( \det \left[\begin{array}{rr} 5 & 6 \\ 6 & 10 \end{array}\right] = (5 \times 10) - (6 \times 6) = 50 - 36 = 14 \)
• \( \det \left[\begin{array}{rr} 2 & 6 \\ 3 & 10 \end{array}\right] = (2 \times 10) - (6 \times 3) = 20 - 18 = 2 \)
• \( \det \left[\begin{array}{rr} 2 & 5 \\ 3 & 6 \end{array}\right] = (2 \times 6) - (5 \times 3) = 12 - 15 = -3 \)

Substitute these values back into the determinant expansion:

\( \det(I + AA') = 2 \times (14) - 2 \times (2) + 3 \times (-3) \)

\( \det(I + AA') = 28 - 4 - 9 \)

\( \det(I + AA') = 24 - 9 \)

\( \det(I + AA') = 15 \)

Alternative Method using Matrix Determinant Lemma

A useful property involving the determinant of a sum involving a rank-one matrix is the Matrix Determinant Lemma. For an invertible matrix \(M\) and column vectors \(u, v\), the lemma states:

\( \det(M + uv') = \det(M)(1 + v'M^{-1}u) \)

In our case, we have \( \det(I + AA') \). Here, \(M=I\) (the identity matrix, which is invertible), \(u=A\), and \(v'=A'\). The inverse of the identity matrix \(I\) is \(I\) itself (\(I^{-1} = I\)).

Applying the lemma:

\( \det(I + AA') = \det(I)(1 + A'I^{-1}A) \)

Since \( \det(I) = 1 \) and \( I^{-1} = I \):

\( \det(I + AA') = 1 \times (1 + A'IA) \)

Matrix multiplication is associative, so \( A'IA = (A'I)A \). Since \( AI = A \) and \( IA = A \), we have \( A'IA = A'A \).

So, \( \det(I + AA') = 1 + A'A \).

Let's calculate \( A'A \):

\( A'A = \left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right] \left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \)

This is a (1 × 3) × (3 × 1) matrix multiplication, resulting in a 1 × 1 matrix (a scalar).

\( A'A = [(1 \times 1) + (2 \times 2) + (3 \times 3)] \)

\( A'A = [1 + 4 + 9] \)

\( A'A = [14] \)

Treating the 1 × 1 matrix as its scalar value, \( A'A = 14 \).

Now substitute this value back into the formula \( \det(I + AA') = 1 + A'A \):

\( \det(I + AA') = 1 + 14 \)

\( \det(I + AA') = 15 \)

Both methods confirm the determinant is 15.

Final Answer Calculation

Based on our calculations, the value of det(I + AA') is 15.

Revision Table: Key Matrix Concepts

Additional Information: Determinant Properties and Lemma

The determinant is a fundamental concept in linear algebra with many important properties. For example, a square matrix is invertible if and only if its determinant is non-zero.

The Matrix Determinant Lemma, which we used as an alternative approach, is a powerful tool for finding the determinant of a matrix that is a sum of an invertible matrix and a rank-one matrix (like \(uv'\)). It simplifies the calculation significantly in certain cases. In our problem, \(AA'\) is a rank-one matrix because it is the product of a column vector and a row vector. Any column (or row) of \(AA'\) is a scalar multiple of the first column (or row). For example, in \( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array}\right] \), the second column \( \left[\begin{array}{l} 2 \\ 4 \\ 6 \end{array}\right] \) is 2 times the first column \( \left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \), and the third column \( \left[\begin{array}{l} 3 \\ 6 \\ 9 \end{array}\right] \) is 3 times the first column.

Calculating determinants manually for larger matrices can be tedious, but understanding methods like cofactor expansion and properties like the Matrix Determinant Lemma provides valuable techniques for solving matrix problems.