Calculating the Determinant of a 4x4 Matrix

andrey21
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Find the determinant of the following 4x4 matrix

-1 0 0 0
-15 7 -15 5
-5 1 -6 1
6 -7 6 -5



Here is my attempt

Ive seen that I should do the following:

-1.(determinant of 3x3 matrix)-0.(determinant of 3x3 matrix) + 0.(determinant of 3x3 matrix)-0.(determinant of 3x3 matrix)

which would simplify to

-1.(determinat of 3x3 matrix)

IS THIS CORRECT??
 
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andrey21 said:
Find the determinant of the following 4x4 matrix

-1 0 0 0
-15 7 -15 5
-5 1 -6 1
6 -7 6 -5



Here is my attempt

Ive seen that I should do the following:

-1.(determinant of 3x3 matrix)-0.(determinant of 3x3 matrix) + 0.(determinant of 3x3 matrix)-0.(determinant of 3x3 matrix)

which would simplify to

-1.(determinat of 3x3 matrix)

IS THIS CORRECT??
Yes. To be more specific, it is -1 times the determinant of this submatrix:

7 -15 5
1 -6 1
-7 6 -5

I.e., the cofactor of the -1 entry in row 1, column 1.
 
Great thanks mark 44 I ended up with the correct answer of -18:)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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