Chapter 11: Q. 13 (page 740)

Find the value of the determinant.
$|\begin{array}{ccc} 4 & - 1 & 2 \\ 6 & - 1 & 0 \\ 1 & - 3 & 4 \end{array}|$

Short Answer

Expert verified

$|\begin{array}{ccc} 4 & - 1 & 2 \\ 6 & - 1 & 0 \\ 1 & - 3 & 4 \end{array}| = - 26$

Step by step solution

Step 1. Given information

We have been given:

$|\begin{array}{ccc} 4 & - 1 & 2 \\ 6 & - 1 & 0 \\ 1 & - 3 & 4 \end{array}|$

We have to find the value of this determinant.

Step 2. Use formula a11a12a13a21a22a23a31a32a33=a11a22a23a32a33-a12a21a23a31a33+a13a21a22a31a32.

$|\begin{array}{ccc} 4 & - 1 & 2 \\ 6 & - 1 & 0 \\ 1 & - 3 & 4 \end{array}| = 4 \cdot |\begin{array}{ll} - 1 0 \\ - 3 4 \end{array}| - (- 1) \cdot |\begin{array}{cc} 6 & 0 \\ 1 & 4 \end{array}| + 2 \cdot |\begin{array}{cc} 6 & - 1 \\ 1 & - 3 \end{array}|$

Step 3. Simplify

$4 \cdot |\begin{array}{ll} - 1 0 \\ - 3 4 \end{array}| - (- 1) \cdot |\begin{array}{cc} 6 & 0 \\ 1 & 4 \end{array}| + 2 \cdot |\begin{array}{cc} 6 & - 1 \\ 1 & - 3 \end{array}|$

$= 4 \cdot [(- 1) \cdot 4 - (- 3) \cdot 0] + 1 \cdot [6 \cdot 4 - 1 \cdot 0] + 2 \cdot [6 \cdot (- 3) - 1 \cdot (- 1)] = 4 \cdot (- 4) + 1 \cdot 24 + 2 \cdot (- 18 + 1) = - 16 + 24 + 2 \cdot (- 17) = 8 - 34 = - 26$

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Find the value of the determinant.
$|\begin{array}{ccc} 4 & - 1 & 2 \\ 6 & - 1 & 0 \\ 1 & - 3 & 4 \end{array}|$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Use formula a11a12a13a21a22a23a31a32a33=a11a22a23a32a33-a12a21a23a31a33+a13a21a22a31a32.

Step 3. Simplify

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Find the value of the determinant.4−126−101−34

Short Answer

Step by step solution

Step 1. Given information

Step 2. Use formula a11a12a13a21a22a23a31a32a33=a11a22a23a32a33-a12a21a23a31a33+a13a21a22a31a32.

Step 3. Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Geometry

Mechanics Maths

Decision Maths

Pure Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Find the value of the determinant.
$|\begin{array}{ccc} 4 & - 1 & 2 \\ 6 & - 1 & 0 \\ 1 & - 3 & 4 \end{array}|$