Question

Let I be the identity matrix of order 3 × 3 and for the matrix $ A = \begin{pmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{pmatrix} $, $ |A| = -1 $. Let B be the inverse of the matrix $ \text{adj}(A \cdot \text{adj}(A^2)) $. Then $ |(\lambda B + I)| $ is equal to _______

Hint:

Use the properties of adjoint and inverse of matrices, such as \( A \cdot \text{adj}(A) = |A| I \) and \( \text{adj}(A^{-1}) = (\text{adj}(A))^{-1} \), to simplify the expression for \( B \). Then, use the properties of determinants, such as \( |AB| = |A| |B| \), to find the value of \( |\lambda B + I| \). Be careful with matrix operations and determinant calculations.

Answer 1

Correct Answer: 38

The objective is to compute the determinant \(|(\lambda B + I)|\). Given is a 3x3 matrix \(A\) with determinant \(|A| = -1\), the 3x3 identity matrix \(I\), and \(B\), which is the inverse of the adjugate of a matrix product involving \(A\).

Concept Used:

The solution employs the following matrix and determinant properties for an \(n \times n\) matrix M:

• Determinant of a matrix.
• \( \text{adj}(M) = |M| \cdot M^{-1} \)
• \( |\text{adj}(M)| = |M|^{n-1} \)
• \( (M^{-1})^{-1} = M \)
• \( (\text{adj}(M))^{-1} = \frac{M}{|M|} \)
• \( |MN| = |M| |N| \)
• \( |M^{-1}| = \frac{1}{|M|} \)
• \( |kM| = k^n |M| \), where \(k\) is a scalar.

Step-by-Step Solution:

Step 1: Determine the value of \(\lambda\) from the given determinant of A.

The matrix A is provided as \( A = \begin{pmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{pmatrix} \). We are given \(|A| = -1\).

Calculating the determinant of A:

\[|A| = \lambda(5 \cdot 2 - 6 \cdot (-1)) - 2(4 \cdot 2 - 6 \cdot 7) + 3(4 \cdot (-1) - 5 \cdot 7)\]\[|A| = \lambda(10 + 6) - 2(8 - 42) + 3(-4 - 35)\]\[|A| = 16\lambda - 2(-34) + 3(-39)\]\[|A| = 16\lambda + 68 - 117\]\[|A| = 16\lambda - 49\]

Given \(|A| = -1\):

\[16\lambda - 49 = -1\]\[16\lambda = 48\]\[\lambda = 3\]

Step 2: Simplify the expression for matrix B.

B is defined as the inverse of \(\text{adj}(A \cdot \text{adj}(A^2))\). Let \(C = A \cdot \text{adj}(A^2)\). Then \(B = (\text{adj}(C))^{-1}\).

Using the property \((\text{adj}(M))^{-1} = \frac{M}{|M|}\):

\[B = \frac{C}{|C|} = \frac{A \cdot \text{adj}(A^2)}{|A \cdot \text{adj}(A^2)|}\]

Simplifying the numerator, \(A \cdot \text{adj}(A^2)\):

\[A \cdot \text{adj}(A^2) = A \cdot (|A^2| \cdot (A^2)^{-1}) = A \cdot (|A|^2 \cdot (A^{-1})^2) = |A|^2 \cdot A \cdot A^{-1} \cdot A^{-1} = |A|^2 \cdot I \cdot A^{-1} = |A|^2 A^{-1}\]

Simplifying the denominator, \(|A \cdot \text{adj}(A^2)|\):

\[|A \cdot \text{adj}(A^2)| = |A| \cdot |\text{adj}(A^2)| = |A| \cdot |A^2|^{3-1} = |A| \cdot |A^2|^2 = |A| \cdot (|A|^2)^2 = |A| \cdot |A|^4 = |A|^5\]

Substituting these back into the equation for B:

\[B = \frac{|A|^2 A^{-1}}{|A|^5} = \frac{1}{|A|^3} A^{-1}\]

Using the given value \(|A| = -1\):

\[B = \frac{1}{(-1)^3} A^{-1} = \frac{1}{-1} A^{-1} = -A^{-1}\]

Step 3: Evaluate the final expression \(|(\lambda B + I)|\).

With \(\lambda = 3\) and \(B = -A^{-1}\), the expression becomes:

\[|(\lambda B + I)| = |(3(-A^{-1}) + I)| = |-3A^{-1} + I|\]

Rewriting \(I\) as \(A \cdot A^{-1}\) for simplification:

\[|-3A^{-1} + A \cdot A^{-1}| = |(-3I + A) \cdot A^{-1}|\]

Applying the property \(|MN| = |M||N|\):

\[|A - 3I| \cdot |A^{-1}| = |A - 3I| \cdot \frac{1}{|A|}\]

Since \(\lambda = 3\), this expression is equivalent to \(|A - \lambda I| \cdot \frac{1}{|A|}\).

Step 4: Calculate the determinant \(|A - \lambda I|\).

For \(\lambda = 3\), the matrix A is \( A = \begin{pmatrix} 3 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{pmatrix} \).

\[A - \lambda I = A - 3I = \begin{pmatrix} 3 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{pmatrix} - \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 2 & 3 \\ 4 & 2 & 6 \\ 7 & -1 & -1 \end{pmatrix}\]

Computing its determinant:

\[|A - 3I| = 0(2 \cdot (-1) - 6 \cdot (-1)) - 2(4 \cdot (-1) - 6 \cdot 7) + 3(4 \cdot (-1) - 2 \cdot 7)\]\[|A - 3I| = 0 - 2(-4 - 42) + 3(-4 - 14)\]\[|A - 3I| = -2(-46) + 3(-18)\]\[|A - 3I| = 92 - 54 = 38\]

Step 5: Compute the final value.

\[|(\lambda B + I)| = |A - 3I| \cdot \frac{1}{|A|} = 38 \cdot \frac{1}{-1} = -38\]

The determinant \(|(\lambda B + I)|\) evaluates to -38.