Solving Matrix Determinant and Adjoint Problem

The problem asks us to find the value of $(\alpha + \beta)$ given a 3 × 3 matrix $A$ with determinant $|A| = 4$, and the equation $|2\text{adj}(3A)| = 2^\alpha 3^\beta$. This requires us to use several key properties of determinants and adjoints of matrices.

Understanding Matrix Properties

Let $A$ be a square matrix of order $n$. We will use the following properties:

• Determinant of a scalar multiple: $|kA| = k^n |A|$, where $k$ is a scalar.
• Determinant of the adjoint: $|\text{adj}(A)| = |A|^{n-1}$.
• Adjoint of a scalar multiple: $\text{adj}(kA) = k^{n-1} \text{adj}(A)$ (This property can be helpful but we can solve it without explicitly using it by applying the definition step-by-step).

In this problem, the order of matrix $A$ is $n=3$, and we are given $|A|=4$.

Step-by-Step Calculation

We need to evaluate $|2\text{adj}(3A)|$ and express it in the form $2^\alpha 3^\beta$. Let's break it down:

1. Calculate the Determinant of 3A:

First, consider the matrix $3A$. Since $A$ is a 3 × 3 matrix, $3A$ is also a 3 × 3 matrix. Using the property $|kA| = k^n |A|$ with $k=3$ and $n=3$:

$$|3A| = 3^3 |A|$$

Given $|A|=4$, we have:

$$|3A| = 3^3 \times 4 = 27 \times 4 = 108$$

2. Calculate the Determinant of adj(3A):

Next, consider the adjoint of the matrix $3A$. The matrix $(3A)$ is a 3 × 3 matrix. Using the property $|\text{adj}(B)| = |B|^{n-1}$ for a 3 × 3 matrix $B$, where $B = 3A$ and $n=3$:

$$|\text{adj}(3A)| = |3A|^{3-1} = |3A|^2$$

We found $|3A| = 108$. So,

$$|\text{adj}(3A)| = 108^2$$

Let's express 108 in terms of its prime factors: $108 = 4 \times 27 = 2^2 \times 3^3$.

$$|\text{adj}(3A)| = (2^2 \times 3^3)^2 = (2^2)^2 \times (3^3)^2 = 2^{2 \times 2} \times 3^{3 \times 2} = 2^4 \times 3^6$$

3. Calculate the Determinant of 2adj(3A):

Now, consider the matrix $2\text{adj}(3A)$. Since $\text{adj}(3A)$ is the adjoint of a 3 × 3 matrix, it is also a 3 × 3 matrix. Let $C = \text{adj}(3A)$. We need to find $|2C|$. Using the property $|kC| = k^n |C|$ with $k=2$ and $n=3$:

$$|2\text{adj}(3A)| = 2^3 |\text{adj}(3A)|$$

We found $|\text{adj}(3A)| = 2^4 \times 3^6$. Substituting this value:

$$|2\text{adj}(3A)| = 2^3 \times (2^4 \times 3^6) = 2^3 \times 2^4 \times 3^6$$

Using the rule of exponents $a^m \times a^n = a^{m+n}$:

$$|2\text{adj}(3A)| = 2^{3+4} \times 3^6 = 2^7 \times 3^6$$

4. Compare with $2^\alpha 3^\beta$ to find $\alpha$ and $\beta$:

We are given that $|2\text{adj}(3A)| = 2^\alpha 3^\beta$. We calculated $|2\text{adj}(3A)| = 2^7 \times 3^6$.

Comparing the two expressions:

$$2^7 \times 3^6 = 2^\alpha 3^\beta$$

By comparing the powers of 2 and 3 on both sides, we find:

• $\alpha = 7$
• $\beta = 6$

5. Calculate $(\alpha + \beta)$:

Finally, we need to find the value of $(\alpha + \beta)$.

$$\alpha + \beta = 7 + 6 = 13$$

Conclusion

The value of $(\alpha + \beta)$ is 13.

Matrix Properties Revision Table

Additional Information on Matrix Adjoints

The adjoint of a square matrix $A$, denoted as $\text{adj}(A)$, is the transpose of the cofactor matrix of $A$. For a 3 × 3 matrix:

If $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$,

The cofactor $C_{ij}$ of the element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the determinant of the minor matrix obtained by removing the $i$-th row and $j$-th column of $A$.

The cofactor matrix is $\text{Cof}(A) = \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix}$.

The adjoint is the transpose of the cofactor matrix:

$$\text{adj}(A) = (\text{Cof}(A))^T = \begin{pmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{pmatrix}$$

The properties used in the solution are derived from this definition and the properties of determinants.