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 $$3\times3$$, in the most general way possible but with a repeated column. That one is $$$A=\left(\begin{matrix} a & d & a \\ b & e & b \\ c & f & c \end{matrix} \right)$$$ let's now calculate the determinant (we can do it by means of the general method or using the rule of Sarrus) $$$det(A)=\left|\begin{matrix} a & d & a \\ b & e & b \\ c & f & c \end{matrix} \right|=\cancel{a\cdot e\cdot c}+\bcancel{b\cdot f \cdot a} + \xcancel{c \cdot d \cdot b}$$$ $$$ - \cancel{a \cdot e \cdot c} - \bcancel{b\cdot f \cdot a} - \xcancel{c \cdot d \cdot b} = 0$$$

Solution:

$$det(A)=0$$

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Create any $$3\times3$$ 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 $$3\times3$$ matrix $$$A=\left(\begin{matrix} 1 & 0 & 1 \\ 2 & 2 & 1 \\ -2 & 1 & -4 \end{matrix} \right)$$$ The construction of the transpose matrix is done by exchanging lines and columns, that is to say $$$A^t=\left(\begin{matrix} 1 & 2 & -2 \\ 0 & 2 & 1 \\ 1 & 1 & -4 \end{matrix} \right)$$$ The determinant can be now computed: $$$det(A^t)=\left|\begin{matrix} 1 & 2 & -2 \\ 0 & 2 & 1 \\ 1 & 1 & -4 \end{matrix} \right|=-8+0+2-(-4)-1-0=-3$$$ 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. $$$det(A)=\left|\begin{matrix} 1 & 0 & 1 \\ 2 & 2 & 1 \\ -2 & 1 & -4 \end{matrix} \right|=-8+2+0-(-4)-1-0=-3$$$ The results are the same.

Solution:

$$det(A)=det(A^t)=-3$$

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Create a $$4\times4$$ matrix whose 4th column is a linear combination of the two first and calculate its determinant.

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Development:

We create a $$4\times4$$ matrix while leaving the 4th column empty $$$\left(\begin{matrix} 1 & 1 & 2 & \fbox{ } \\ -1 & 0 & 3 & \fbox{ } \\ -2 & 1 & -1 & \fbox{ } \\ 0 & 0 & 1 & \fbox{ } \end{matrix} \right)$$$

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 $$4\times4$$ matrix is $$$\left(\begin{matrix} 1 & 1 & 2 & 2 \\ -1 & 0 & 3 & -1 \\ -2 & 1 & -1 & -1 \\ 0 & 0 & 1 & 0 \end{matrix} \right)$$$

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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