Chapter 9: Q. 13 (page 627)

In Problems $7 - 14$ , find the value of each determinant.
$|\begin{matrix} A & B & C \\ - 1 & 3 & 5 \\ 5 & 0 & - 2 \end{matrix}|$

Short Answer

Expert verified

The value of the determinant $|\begin{matrix} A & B & C \\ - 1 & 3 & 5 \\ 5 & 0 & - 2 \end{matrix}|$ is, $- 6 A + 23 B - 15 C$ .

Step by step solution

Step 1 Given determinant is,

$|\begin{matrix} A & B & C \\ - 1 & 3 & 5 \\ 5 & 0 & - 2 \end{matrix}|$

Evaluate the determinant.

role="math" localid="1646807619254" $|\begin{matrix} A & B & C \\ - 1 & 3 & 5 \\ 5 & 0 & - 2 \end{matrix}| = A [3 (- 2) - 5 (0)] - B [- 2 (- 1) - 5 (5)] + C [- 1 (0) - 3 (5)]$

Step 2 Now simplify it.

$|\begin{matrix} A & B & C \\ - 1 & 3 & 5 \\ 5 & 0 & - 2 \end{matrix}| = = A [- 6 - 0] - B [2 - 25] + C [- 15] = A [- 6] - B [- 23] + C [- 15] = - 6 A + 23 B - 15 C$

Thus, the value of the determinant is, $- 6 A + 23 B - 15 C$ .

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In Problems $7 - 14$ , find the value of each determinant.
$|\begin{matrix} A & B & C \\ - 1 & 3 & 5 \\ 5 & 0 & - 2 \end{matrix}|$

Short Answer

Step by step solution

Step 1 Given determinant is,

Step 2 Now simplify it.

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In Problems 7-14, find the value of each determinant.ABC-13550-2

Short Answer

Step by step solution

Step 1 Given determinant is,

Step 2 Now simplify it.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Discrete Mathematics

Calculus

Pure Maths

Logic and Functions

Decision Maths

Study anywhere. Anytime. Across all devices.

In Problems $7 - 14$ , find the value of each determinant.
$|\begin{matrix} A & B & C \\ - 1 & 3 & 5 \\ 5 & 0 & - 2 \end{matrix}|$