Determinants

To optimize speed, we expand along the second row since it contains a zero placeholder element: \[ |A| = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23} \] Calculate the non-zero positional terms systematically using sign allocations: \[ a_{21} = 0 \implies 0 \cdot C_{21} = 0 \] \[ a_{22} = -1 \implies C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 3 \\ 2 & 5 \end{vmatrix} = 1 \cdot (1(5) - 3(2)) = 5 - 6 = -1 \] \[ a_{23} = 4 \implies C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = -1 \cdot (1(1) - 2(2)) = -1 \cdot (1 - 4) = 3 \] Combine the evaluated components: \[ |A| = 0 + (-1)(-1) + 4(3) = 1 + 12 = 13 \] Now, let us calculate the requested standalone cofactor $C_{32}$: \[ C_{32} = (-1)^{3+2} M_{32} = -1 \cdot \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -1 \cdot (1(4) - 3(0)) = -4 \] Final Answer: $|A| = 13$ and $C_{32} = -4$.

According to the properties of Laplace expansions, if you multiply elements of any row by the corresponding cofactors of a different row, their sum total vanishes completely to zero. Here, the elements belong to Row 1 ($a_{11}, a_{12}, a_{13}$), but the cofactors belong to Row 2 ($C_{21}, C_{22}, C_{23}$). Therefore, by orthogonality properties of determinants, $E = 0$, regardless of the value of $\Delta$. Final Answer: $0$.

Expand the $2 \times 2$ determinants on both sides using the cross-multiplication formula: \[ \text{Left Side} = x(x) - 2(3) = x^2 - 6 \] \[ \text{Right Side} = 2x(2) - (-1)(5) = 4x + 5 \] Equate the two expressions to form a standard quadratic equation: \[ x^2 - 6 = 4x + 5 \implies x^2 - 4x - 11 = 0 \] Solve using the standard quadratic formula since it does not factor cleanly with integers: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-11)}}{2(1)} = \frac{4 \pm \sqrt{16 + 44}}{2} = \frac{4 \pm \sqrt{60}}{2} \] Simplify the radical term: $\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$. \[ x = \frac{4 \pm 2\sqrt{15}}{2} = 2 \pm \sqrt{15} \] Final Answer: $x = 2 \pm \sqrt{15}$.

Apply the scalar multiple property alongside the transposition invariant property: Since $A$ is of order $3 \times 3$, pulling the scalar 2 out of the determinant brackets requires raising it to the third power ($n=3$): \[ |2A^T| = 2^3 \cdot |A^T| \] Using the property $|A^T| = |A|$, substitute the given value into the expression: \[ |2A^T| = 8 \cdot |A| = 8 \times (-4) = -32 \] Final Answer: $-32$.

Let us use row operations to reduce the large values into manageable numbers. Perform two row operations simultaneously: Perform $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$: \[ R_2 \text{ row becomes: } \begin{vmatrix} 2025-2024 & 2026-2025 & 2027-2026 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \end{vmatrix} \] \[ R_3 \text{ row becomes: } \begin{vmatrix} 2026-2025 & 2027-2026 & 2028-2027 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \end{vmatrix} \] Update the rows inside the determinant: \[ \Delta = \begin{vmatrix} 2024 & 2025 & 2026 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \] Notice that the second row ($R_2$) and the third row ($R_3$) are completely identical. According to determinant properties, if two rows are identical, the value of the determinant is 0. Final Answer: $0$.

Apply the multiplicative property of determinants ($|XY| = |X||Y|$) to split the terms: \[ |A^2 B^{-1}| = |A^2| \cdot |B^{-1}| \] Using the power property $|A^k| = |A|^k$ and the inverse property $|B^{-1}| = \frac{1}{|B|}$: \[ |A^2 B^{-1}| = |A|^2 \cdot \frac{1}{|B|} \] Substitute the given numerical metrics into the formula: \[ |A^2 B^{-1}| = (3)^2 \cdot \frac{1}{-2} = 9 \times \left(-\frac{1}{2}\right) = -4.5 \] Final Answer: $-\frac{9}{2}$.

Apply the standard determinant identity for nested adjoint matrices: \[ |\text{adj}(\text{adj}(A))| = |A|^{(n-1)^2} \] Identify the given parameters: order $n = 3$ and base determinant $|A| = 4$. Substitute these parameters into the exponent formula: \[ \text{Exponent} = (3 - 1)^2 = 2^2 = 4 \] Now evaluate the final value: \[ |\text{adj}(\text{adj}(A))| = 4^4 = 256 \] Final Answer: $256$.

For three coordinate points to be collinear, they must form a triangle with an area of zero. Set up the coordinate matrix determinant and set it equal to 0: \[ \det \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 1 \\ k & 8 & 1 \end{bmatrix} = 0 \] Let us use row operations to introduce zeros before expanding. Perform $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - kR_1$: \[ \begin{vmatrix} 1 & 2 & 1 \\ 0 & 0 & -1 \\ 0 & 8-2k & 1-k \end{vmatrix} = 0 \] Expand along the first column: \[ 1 \cdot \begin{vmatrix} 0 & -1 \\ 8-2k & 1-k \end{vmatrix} = 0 \implies 0 - (-1)(8-2k) = 0 \implies 8 - 2k = 0 \implies 2k = 8 \implies k = 4 \] Final Answer: $k = 4$.

Set up the standard area evaluation matrix using the given vertices: \[ \Delta = \frac{1}{2} \left| \det \begin{bmatrix} 0 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & 4 & 1 \end{bmatrix} \right| \] Expand the determinant along the first row since it contains two zero placeholders: \[ \det = 1 \cdot \begin{vmatrix} 3 & 0 \\ 0 & 4 \end{vmatrix} = 1 \cdot (3(4) - 0(0)) = 12 \] Now substitute this value back into the area formula and take the absolute value: \[ \Delta = \frac{1}{2} |12| = 6 \text{ square units} \] This matches the standard geometric area formula for a right-angled triangle ($\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6$). Final Answer: $6$.

Step 1: Calculate the base system coefficient determinant $\Delta$: \[ \Delta = \begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} = 2(-1) - 3(3) = -2 - 9 = -11 \] Since $\Delta = -11 \neq 0$, the system is consistent and has a unique solution.
Step 2: Calculate the modified variable determinants $\Delta_x$ and $\Delta_y$ by substituting the constant vector $\begin{bmatrix} 8 \\ 1 \end{bmatrix}$ into the columns: \[ \Delta_x = \begin{vmatrix} 8 & 3 \\ 1 & -1 \end{vmatrix} = 8(-1) - 3(1) = -8 - 3 = -11 \] \[ \Delta_y = \begin{vmatrix} 2 & 8 \\ 3 & 1 \end{vmatrix} = 2(1) - 8(3) = 2 - 24 = -22 \] Step 3: Calculate the final values of $x$ and $y$ using the determinant ratios: \[ x = \frac{\Delta_x}{\Delta} = \frac{-11}{-11} = 1 \] \[ y = \frac{\Delta_y}{\Delta} = \frac{-22}{-11} = 2 \] The unique solution point is $(1, 2)$. Final Answer: Consistent with a unique solution: $x = 1, y = 2$.

Let us find the base coefficient determinant $\Delta$ first: \[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 0 & 0 & 0 \end{vmatrix} = 0 \] (The rows are proportional, so $\Delta = 0$). For the system to have no solution, Cramer's Rule requires the base determinant to be zero ($\Delta = 0$) and at least one modified variable determinant to be non-zero.
Let us evaluate $\Delta_x$ by substituting the constant vector column into the first column: \[ \Delta_x = \begin{vmatrix} 4 & 1 & 1 \\ \mu & 2 & 2 \\ 0 & 0 & 0 \end{vmatrix} = 0 \] Notice that the second and third columns are identical, which forces $\Delta_x = 0$ for any value of $\mu$. The same is true for $\Delta_y$ and $\Delta_z$.
Instead, let us look at the structural consistency of the rows directly. The equations are parallel planes: \[ R_2 \implies 2(x+y+z) = \mu \implies x+y+z = \frac{\mu}{2} \] The first equation states $x+y+z = 4$. For these two statements to contradict each other and produce no solution, the constant values must be different: \[ \frac{\mu}{2} \neq 4 \implies \mu \neq 8 \] If $\mu \neq 8$, the planes are parallel and never intersect, so the system has no solution. Final Answer: $\mu \neq 8$.

• If $\Delta \neq 0$: The system has a unique solution. Calculating the ratios yields $x = \frac{0}{\Delta} = 0, y = 0, z = 0$. This is the trivial solution (zero solution).
• If $\Delta = 0$: Since all modified determinants are also zero, the system satisfies the condition for infinitely many solutions. This allows for non-zero combinations of variables, yielding non-trivial solutions.

Set up the characteristic determinant equation $|A - \lambda I| = 0$: \[ \det \left( \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \right) = 0 \implies \begin{vmatrix} 1-\lambda & 2 \\ 2 & 4-\lambda \end{vmatrix} = 0 \] Expand the determinant to form the characteristic polynomial: \[ (1 - \lambda)(4 - \lambda) - 2(2) = 0 \] \[ 4 - \lambda - 4\lambda + \lambda^2 - 4 = 0 \implies \lambda^2 - 5\lambda = 0 \] This is the characteristic equation of the matrix. Let us solve it to find the eigenvalues: \[ \lambda(\lambda - 5) = 0 \implies \lambda = 0 \quad \text{or} \quad \lambda = 5 \] Notice that the sum of the eigenvalues ($0 + 5 = 5$) equals the trace of the matrix ($\text{tr}(A) = 1 + 4 = 5$), and their product ($0 \times 5 = 0$) equals the determinant ($|A| = 4 - 4 = 0$). This confirms standard eigenvalue properties. Final Answer: Equation: $\lambda^2 - 5\lambda = 0$, Eigenvalues: $0, 5$.

Step 1: Find the characteristic equation $|A - \lambda I| = 0$: \[ \begin{vmatrix} 3-\lambda & 1 \\ 2 & 4-\lambda \end{vmatrix} = 0 \implies (3-\lambda)(4-\lambda) - 2 = 0 \] \[ 12 - 3\lambda - 4\lambda + \lambda^2 - 2 = 0 \implies \lambda^2 - 7\lambda + 10 = 0 \] Step 2: According to the Cayley-Hamilton Theorem, substitute the matrix $A$ into this polynomial equation to form a valid matrix identity: \[ A^2 - 7A + 10I = O \] Step 3: To isolate $A^{-1}$, multiply the entire matrix equation by $A^{-1}$: \[ A^{-1}(A^2) - 7A^{-1}(A) + 10A^{-1}(I) = A^{-1}(O) \] \[ A - 7I + 10A^{-1} = O \] Rearrange the terms to solve for $10A^{-1}$: \[ 10A^{-1} = 7I - A \implies A^{-1} = \frac{1}{10}(7I - A) \] Step 4: Substitute the matrix entries into the expression: \[ 7I - A = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ -2 & 3 \end{bmatrix} \] \[ A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & -1 \\ -2 & 3 \end{bmatrix} \] This matches the result found using the standard adjoint method. Final Answer: $A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & -1 \\ -2 & 3 \end{bmatrix}$.

Isolate the highest power term $A^2$ from the given matrix identity: \[ A^2 = 5A - 6I \] To find an expression for $A^3$, multiply the entire equation by the matrix $A$: \[ A \cdot (A^2) = A \cdot (5A - 6I) \implies A^3 = 5A^2 - 6A \] Now substitute our expression for $A^2$ ($5A - 6I$) back into this equation to eliminate the quadratic term: \[ A^3 = 5(5A - 6I) - 6A \] Expand and combine like terms: \[ A^3 = 25A - 30I - 6A = 19A - 30I \] This linear expression allows us to compute $A^3$ quickly without performing full matrix multiplications. Final Answer: $19A - 30I$.