Chapter 11: Q. 55 (page 741)

In Problems 51–56, solve for x,
$|\begin{array}{r} x & 2 & 3 \\ 1 & x & 0 \\ 6 & 1 & - 2 \end{array}| = 7$

Short Answer

Expert verified

The value of xis $0 or 9$ .

Step by step solution

Step 1. Given Information

We are given,

$|\begin{array}{r} x & 2 & 3 \\ 1 & x & 0 \\ 6 & 1 & - 2 \end{array}| = 7$

We have to find the value of x.

Step 2. Solving the determinant

Solving the determinant, we get

$|\begin{array}{r} x & 2 & 3 \\ 1 & x & 0 \\ 6 & 1 & - 2 \end{array}| = 7 x |\begin{matrix} x & 0 \\ 1 & - 2 \end{matrix}| - 2 |\begin{matrix} 1 & 0 \\ 6 & - 2 \end{matrix}| + 3 |\begin{array}{l} 1 & x \\ 6 & 1 \end{array}| = 7 x [(x) (- 2) - (0) (1)] - 2 [(1) (- 2) - (0) (6)] + 3 [(1) (1) - (x) (6)] = 7 x (- 2 x) - 2 (- 2) + 3 (1 - 6 x) = 7 - 2 x^{2} - 18 x + 7 = 7 - 2 x^{2} - 18 x = 0 2 x (x - 9) = 0 2 x = 0 x = 0$

$x - 9 = 0 x = 9$

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In Problems 51–56, solve for x,
$|\begin{array}{r} x & 2 & 3 \\ 1 & x & 0 \\ 6 & 1 & - 2 \end{array}| = 7$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solving the determinant

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In Problems 51–56, solve for x,x231x061-2=7

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solving the determinant

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

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Study anywhere. Anytime. Across all devices.

In Problems 51–56, solve for x,
$|\begin{array}{r} x & 2 & 3 \\ 1 & x & 0 \\ 6 & 1 & - 2 \end{array}| = 7$