Find the value of {bmatrix} 1 & 3 & 4 \\ 5 & 4 & 2 \\ 5 & 1 & 0 {bmatrix}

Correct Answer is : −32-32−32

Correct Answer is : −32-32−32

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Matrices

Question : 26 of 36

Marks: +1, -0

\begin{bmatrix} 1 & 3 & 4 \\ 5 & 4 & 2 \\ 5 & 1 & 0 \end{bmatrix}

Solution:

Concept:

The determinant of a 3x3 matrix is computed by expanding along any row or column using minors and cofactors.

Explanation:

Given matrix

A = \begin{bmatrix} 1 & 3 & 4 \\ 5 & 4 & 2 \\ 5 & 1 & 0 \end{bmatrix}

Expand along first row:

\det(A) = 1 \cdot \begin{vmatrix} 4 & 2 \\ 1 & 0 \end{vmatrix} - 3 \cdot \begin{vmatrix} 5 & 2 \\ 5 & 0 \end{vmatrix} + 4 \cdot \begin{vmatrix} 5 & 4 \\ 5 & 1 \end{vmatrix}

Compute each 2x2 determinant:

\begin{vmatrix} 4 & 2 \\ 1 & 0 \end{vmatrix} = (4 \cdot 0) - (2 \cdot 1) = 0 - 2 = -2

\begin{vmatrix} 5 & 2 \\ 5 & 0 \end{vmatrix} = (5 \cdot 0) - (2 \cdot 5) = 0 - 10 = -10

\begin{vmatrix} 5 & 4 \\ 5 & 1 \end{vmatrix} = (5 \cdot 1) - (4 \cdot 5) = 5 - 20 = -15

Now substitute:

\det(A) = 1 \cdot (-2) - 3 \cdot (-10) + 4 \cdot (-15) = -2 + 30 - 60 = -32

Answer:

The determinant is

-32

, which corresponds to option A.

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