Comprehensive Exercises on Determinants

Approach: Observe the structure of the determinant and use the properties to simplify the computation.

Steps:

1. Observation: the third row is 7 times the first row, and the second row is 4 times the first row minus 1

2. Add $-4$ times the first row to the second row: $\begin{vmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 7 & 8 & 9 \end{vmatrix}$

3. Add $-7$ times the first row to the third row: $\begin{vmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{vmatrix}$

4. The third row is twice the second row, so the determinant is zero.

Answer: the value of the determinant is 0.

Approach: Expand along the first row and use the recurrence relation.

Steps:

1. Expand along the first row: $|A| = 2 \begin{vmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix}$

2. Continue expanding the first third-order determinant: $= 2 \left( 2 \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix} \right) - 1 \left( 1 \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} \right)$

3. $= 2(2(4-1) - 1(2-0)) - 1(1(4-1)) = 2(6-2) - 1(3) = 8 - 3 = 5$

Answer: the value of the determinant is 5.

Approach: Use the formula for the Vandermonde determinant.

Steps:

1. $x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4$

2. $|V| = \prod_{1 \leq i < j \leq 4} (x_j - x_i)$

3. $= (2-1)(3-1)(4-1)(3-2)(4-2)(4-3)$

4. $= 1 \times 2 \times 3 \times 1 \times 2 \times 1 = 12$

Answer: the value of the determinant is 12.

Approach: Use the properties of determinants.

Steps:

1. $|2A| = 2^3|A| = 8 \times 3 = 24$ (multiplying an n-th order matrix by $k$ multiplies the determinant by $k^n$)

2. $|A^T| = |A| = 3$ (invariance under transposition)

Answer: $|2A| = 24$, $|A^T| = 3$.

Approach: Compute step by step with two properties of determinants: swapping two rows changes the sign, and multiplying every entry of a row by a constant multiplies the determinant by that same constant.

Steps:

1. First operation: interchange the first row and the second row; by the property the determinant changes sign, giving $|A| = -2$.
2. Second operation: multiply every entry of the third row by 2; by the property the determinant is also multiplied by 2, giving $-2 \times 2 = -4$.
3. Taken together, the two operations multiply the original determinant by $-2$.

Answer: the value of the new determinant is $-4$.

Approach: Use the formula of Cramer’s rule.

Steps:

1. $|A| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 2 \\ 6 & -3 & 1 \end{vmatrix} = 1(5+6) - 2(2-12) + 3(-6-30) = 11 + 20 - 108 = -77$

2. $|A_1| = \begin{vmatrix} 6 & 2 & 3 \\ 4 & 5 & 2 \\ 2 & -3 & 1 \end{vmatrix} = 6(5+6) - 2(4-4) + 3(-12-10) = 66 + 0 - 66 = 0$

3. $|A_2| = \begin{vmatrix} 1 & 6 & 3 \\ 2 & 4 & 2 \\ 6 & 2 & 1 \end{vmatrix} = 1(4-4) - 6(2-12) + 3(4-24) = 0 + 60 - 60 = 0$

4. $|A_3| = \begin{vmatrix} 1 & 2 & 6 \\ 2 & 5 & 4 \\ 6 & -3 & 2 \end{vmatrix} = 1(10+12) - 2(4-24) + 6(-6-30) = 22 + 40 - 216 = -154$

5. $x = \frac{0}{-77} = 0$, $y = \frac{0}{-77} = 0$, $z = \frac{-154}{-77} = 2$

Answer: $x = 0$, $y = 0$, $z = 2$.

Approach: Use the adjugate matrix method.

Steps:

1. $|A| = 1 \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2 \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3 \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}$ $= 1(-24) - 2(-20) + 3(-5) = -24 + 40 - 15 = 1$

2. Compute the adjugate matrix: $\text{adj}(A) = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$

3. $A^{-1} = \frac{1}{1} \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$

Answer: $A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$.

Approach: Use a third-order determinant to compute the volume.

Steps:

1. $V = \left| \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix} \right|$

2. Expand along the first row: $= \left| 1(45-48) - 2(36-42) + 3(32-35) \right|$

3. $= \left| 1(-3) - 2(-6) + 3(-3) \right| = |-3 + 12 - 9| = |0| = 0$

Answer: the volume is 0 (the three vectors are coplanar).

Approach: Use the definition of the adjugate matrix and the properties of determinants.

Steps:

1. From $A \cdot \text{adj}(A) = |A|I$ we get: $|A \cdot \text{adj}(A)| = |A|^n$

2. Also $|A \cdot \text{adj}(A)| = |A| \cdot |\text{adj}(A)|$

3. Therefore $|A| \cdot |\text{adj}(A)| = |A|^n$

4. When $|A| \neq 0$, $|\text{adj}(A)| = |A|^{n-1}$

5. When $|A| = 0$, $\text{adj}(A) = 0$, so the conclusion holds as well.

Answer: the proof is complete.

Approach: Discuss the cases separately and use the properties of determinants.

Steps:

1. When $|A| \neq 0$:

   • From $A \cdot \text{adj}(A) = |A|I$ we get $|A \cdot \text{adj}(A)| = |A|^n$
   • Also $|A \cdot \text{adj}(A)| = |A| \cdot |\text{adj}(A)|$
   • Therefore $|\text{adj}(A)| = |A|^{n-1}$

2. When $|A| = 0$:

   • If $A = 0$, then $\text{adj}(A) = 0$, so $|\text{adj}(A)| = 0 = |A|^{n-1}$
   • If $A \neq 0$, then $\text{adj}(A) \neq 0$, but $|\text{adj}(A)| = 0 = |A|^{n-1}$

Answer: the proof is complete.

Approach: Use the multiplicativity of the determinant and the properties of the inverse matrix.

Steps:

1. $|A^2| = |A|^2 = 2^2 = 4$

2. $|A^{-1}| = \frac{1}{|A|} = \frac{1}{2}$

Answer: $|A^2| = 4$, $|A^{-1}| = \frac{1}{2}$.

Approach: Use the multiplicativity of the determinant and the property under transposition.

Steps:

1. $|A^T A| = |A^T| \cdot |A|$

2. $|A^T| = |A| = 3$

3. Therefore $|A^T A| = 3 \times 3 = 9$

Answer: $|A^T A| = 9$.

Approach: Use the linearity of the determinant: multiplying every row of the matrix by 2 is the same as multiplying the whole matrix $A$ by 2.

Steps:

1. $A$ is a third-order square matrix; multiplying all three of its rows by 2 is equivalent to multiplying the matrix $A$ as a whole by 2.
2. A third-order determinant satisfies $|2A| = 2^3|A|$, that is, the value of the determinant is multiplied by $2^3 = 8$.
3. Substituting $|A| = 1$ gives $8 \times 1 = 8$.

Answer: the value of the new determinant is 8.

Approach: Use the properties of determinants and linear dependence.

Steps:

1. Let the sum of the entries of each row be $s$

2. Add all the rows except the first to the first row, obtaining: $\begin{vmatrix} s & s & \cdots & s \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix}$

3. Factor out the common factor $s$ from the first row: $s \begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix}$

4. If $s = 0$, then the determinant is zero

5. If $s \neq 0$, then the first row consists entirely of 1s, and by row transformations the other rows can be made identical, so the determinant is zero

Answer: the proof is complete.

Approach: Use the definition of the adjugate matrix and the properties of determinants.

Steps:

1. From $A \cdot \text{adj}(A) = |A|I$ we get: $|A \cdot \text{adj}(A)| = |A|^n$

2. Also $|A \cdot \text{adj}(A)| = |A| \cdot |\text{adj}(A)|$

3. Therefore $|A| \cdot |\text{adj}(A)| = |A|^n$

4. Since $|A| \neq 0$, $|\text{adj}(A)| = |A|^{n-1}$

Answer: the proof is complete.