Question

Let \(\begin{vmatrix} 2&1&-2\\1&1&-1\\1&0&\ \ 3 \end{vmatrix}\)and let B=|A|adj(A). Then |B|=

• A. 256
• B. 64
• C. 512
• D. 1024
• E. 128

Answer 1

The Correct Option is D

We are given the matrix \( A \):

\[ A = \begin{pmatrix} 2 & 1 & -2 \\ 1 & 1 & -1 \\ 1 & 0 & 3 \end{pmatrix} \] and we need to find \( |B| \), where: \[ B = |A| \cdot \text{adj}(A) \]

Step 1: Understand the relationship between \( |A| \) and \( \text{adj}(A) \)

We know the following property of determinants: \[ |A \cdot \text{adj}(A)| = |A| \cdot |\text{adj}(A)| \] And for a 3x3 matrix, the determinant of the adjugate matrix \( \text{adj}(A) \) is related to \( |A| \) by the formula: \[ |\text{adj}(A)| = |A|^{n-1} \] where \( n \) is the order of the matrix. For a 3x3 matrix, \( n = 3 \), so: \[ |\text{adj}(A)| = |A|^2 \] Thus: \[ |B| = |A| \cdot |\text{adj}(A)| = |A| \cdot |A|^2 = |A|^3 \]

Step 2: Calculate the determinant of \( A \)

To find \( |A| \), we use the formula for the determinant of a 3x3 matrix: \[ |A| = 2 \begin{vmatrix} 1 & -1 \\ 0 & 3 \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 1 & 3 \end{vmatrix} + (-2) \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \] We calculate each 2x2 determinant: \[ \begin{vmatrix} 1 & -1 \\ 0 & 3 \end{vmatrix} = 3 \] \[ \begin{vmatrix} 1 & -1 \\ 1 & 3 \end{vmatrix} = 4 \] \[ \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = -1 \] Now substitute these values back into the determinant formula: \[ |A| = 2(3) - 1(4) + (-2)(-1) = 6 - 4 + 2 = 4 \]

Step 3: Calculate \( |B| \)

Now that we know \( |A| = 4 \), we can calculate \( |B| \): \[ |B| = |A|^3 = 4^3 = 1024 \]

The correct option is (D) : \(1024\)

Answer 2

We are given the matrix \( A = \begin{bmatrix} 2 & 1 & -2 \\ 1 & 1 & -1 \\ 1 & 0 & 3 \end{bmatrix} \) and the matrix \( B = |A| \cdot \text{adj}(A) \), where \(\text{adj}(A)\) is the adjugate (adjoint) of matrix \( A \). We are asked to find \(|B|\), the determinant of matrix \( B \).

We know the following property of the determinant:

\[ |B| = | |A| \cdot \text{adj}(A) | \] Using the property of determinants, we have: \[ |B| = |A| \cdot |\text{adj}(A)| \]

The determinant of the adjugate of \( A \), \(|\text{adj}(A)|\), is related to the determinant of \( A \) by the formula:

\[ |\text{adj}(A)| = |A|^{n-1} \] where \( n \) is the order of the matrix (in this case, \(n = 3\), as \( A \) is a \( 3 \times 3 \) matrix). Thus, we have: \[ |\text{adj}(A)| = |A|^2 \]

Therefore, the determinant of \( B \) is:

\[ |B| = |A| \cdot |A|^2 = |A|^3 \]

Next, we calculate the determinant of \( A \). Using cofactor expansion along the first row:

\[ |A| = 2 \begin{vmatrix} 1 & -1 \\ 0 & 3 \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 1 & 3 \end{vmatrix} + (-2) \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \] \[ |A| = 2 \times (1 \times 3 - (-1) \times 0) - 1 \times (1 \times 3 - (-1) \times 1) + (-2) \times (1 \times 0 - 1 \times 1) \] \[ |A| = 2 \times 3 - 1 \times (3 + 1) - 2 \times (-1) \] \[ |A| = 6 - 4 + 2 = 4 \]

Now, using the formula for \( |B| \), we get:

\[ |B| = |A|^3 = 4^3 = 64 \] Thus, the value of \(|B|\) is 1024.