Given Matrix A:
$A = \begin{bmatrix} 2 & 4 & 5 \\ 0 & 7 & 4 \\ 3 & 11 & 8 \end{bmatrix}$
To find the determinant of a 3x3 matrix, we use:
$\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$
Where the matrix is structured as:
$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$
From matrix A, we identify:
Now apply the formula:
$\text{det}(A) = 2(7 \cdot 8 - 4 \cdot 11) - 4(0 \cdot 8 - 4 \cdot 3) + 5(0 \cdot 11 - 7 \cdot 3)$
Step-by-step:
So we get:
$\text{det}(A) = 2(56 - 44) - 4(0 - 12) + 5(0 - 21)$
$= 2(12) - 4(-12) + 5(-21)$
$= 24 + 48 - 105 = -33$
Check divisibility:
$-33$ is divisible by $11$, since $-33 = 11 \cdot (-3)$
Final Answer: Option (A): det A is divisible by 11
Step 1: Calculate the Determinant of A
The determinant of a 3x3 matrix is calculated as:
\(|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})\)
Plugging in the values from matrix A:
\(|A| = 2(7 \cdot 8 - 11 \cdot 4) - 0(4 \cdot 8 - 11 \cdot 5) + 3(4 \cdot 4 - 7 \cdot 5)\)
\(|A| = 2(56 - 44) - 0 + 3(16 - 35)\)
\(|A| = 2(12) + 3(-19) = 24 - 57 = -33\)
Step 2: Check Divisibility by 11
Since \(|A| = -33\), it is divisible by 11 because \(-33 = -3 \cdot 11\).
Step 3: Check if det A = 0
\(|A| = -33 \neq 0\), so det A is not 0.
Step 4: Check if A is an Orthogonal Matrix
For a matrix to be orthogonal, its transpose multiplied by itself must equal the identity matrix, i.e., \(AA^T = I\). In the above case this doesn't hold therefore it isn't orthogonal
\(A^T = \begin{pmatrix} 2 & 4 & 5 \\ 0 & 7 & 4 \\ 3 & 11 & 8 \end{pmatrix}\)
\({A^T} \) This option implies it.
Conclusion:
Since the determinant of A is -33 and is divisible by 11, the correct statement is:
det A is divisible by 11.
Read More: Properties of Determinants