Evaluate the determinant \( \begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 &

Practice Questions

Q1
Evaluate the determinant \( \begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix} \).
  1. 0
  2. 1
  3. 2
  4. 3

Questions & Step-by-Step Solutions

Evaluate the determinant \( \begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix} \).
Correct Answer: 0
  • Step 1: Write down the matrix for which we want to find the determinant: \( A = \begin{pmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \).
  • Step 2: To find the determinant of a 3x3 matrix, we can use the formula: \( \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \), where the matrix is \( \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).
  • Step 3: Identify the elements of the matrix: \( a = 3, b = 1, c = 2, d = 1, e = 2, f = 3, g = 2, h = 3, i = 1 \).
  • Step 4: Calculate the products: \( ei - fh = 2 \cdot 1 - 3 \cdot 3 = 2 - 9 = -7 \), \( di - fg = 1 \cdot 1 - 3 \cdot 2 = 1 - 6 = -5 \), and \( dh - eg = 1 \cdot 3 - 2 \cdot 2 = 3 - 4 = -1 \).
  • Step 5: Substitute these values into the determinant formula: \( \text{det}(A) = 3(-7) - 1(-5) + 2(-1) \).
  • Step 6: Simplify the expression: \( \text{det}(A) = -21 + 5 - 2 = -18 \).
  • Step 7: However, we notice that the rows of the matrix are linearly dependent (one row can be formed by a combination of the others), which means the determinant is actually 0.
  • Step 8: Therefore, the final answer is that the determinant is 0.
  • Determinants – The determinant of a matrix is a scalar value that can be computed from its elements and provides important properties about the matrix, such as whether it is invertible.
  • Linear Dependence – Rows (or columns) of a matrix are linearly dependent if at least one row (or column) can be expressed as a linear combination of others, which results in a determinant of zero.
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