Problems from Properties of determinants

Prove that a determinant with a repeated column is zero (prove it for order 3 or higher).

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Development:

The first step is to construct the matrix, in this case 3×3, in the most general way possible but with a repeated column. That one is A=(adabebcfc) let's now calculate the determinant (we can do it by means of the general method or using the rule of Sarrus) det(A)=|adabebcfc|=a⋅e⋅c+b⋅f⋅a+c⋅d⋅b −a⋅e⋅c−b⋅f⋅a−c⋅d⋅b=0

Solution:

det(A)=0

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Create any 3×3 matrix, calculate its transpose and then its determinant. Calculate also the determinant of the matrix without being transposed.

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Development:

First of all, we create the 3×3 matrix A=(101221−21−4) The construction of the transpose matrix is done by exchanging lines and columns, that is to say At=(12−202111−4) The determinant can be now computed: det(At)=|12−202111−4|=−8+0+2−(−4)−1−0=−3 Calculate also the det (A). The result should be the same, so calculating the det (A) can be a way to verify that we have not made any mistakes in the calculations. det(A)=|101221−21−4|=−8+2+0−(−4)−1−0=−3 The results are the same.

Solution:

det(A)=det(At)=−3

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Create a 4×4 matrix whose 4th column is a linear combination of the two first and calculate its determinant.

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Development:

We create a 4×4 matrix while leaving the 4th column empty (112 −103 −21−1 001 )

We demand that the 4th column is a linear combination of the columns C1 and C2.

C4=C1+C2 (There are infinite possibilities)

Then the 4×4 matrix is (1122−103−1−21−1−10010)

The determinant can be calculated now. Is it necessary to do it? By construction the property 2.c) is satisfied, so the determinant is empty.

Solution:

det(A)=0

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