Related Formula
|AB| = |A||B|$|AB| = |A||B|$
Trace(B$B$) = Σ bᵢᵢ$\sum b_{ii}$
Core Logic
Since B = [B₁, B₂, B₃]$B = [B_1, B_2, B_3]$, the matrix multiplication AB$AB$ effectively applies A$A$ to each column of B$B$:
AB = [AB₁, AB₂, AB₃] = bmatrix 1 & 2 & 3 0 & 3 & 2 0 & 0 & 1 bmatrix$$AB = [AB_1, AB_2, AB_3] = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$$
Let this target matrix be C$C$. We know AB = C$AB = C$.
Thus, taking determinants on both sides:
|A| |B| = |C|$|A| |B| = |C|$
Step 1: Finding Determinants
Calculate determinant of A$A$:
|A| = 2(1 - 0) - 0 + 1(0 - 1) = 2 - 1 = 1$$|A| = 2(1 - 0) - 0 + 1(0 - 1) = 2 - 1 = 1$$
Calculate determinant of C$C$. Since C$C$ is an upper triangular matrix, its determinant is simply the product of its main diagonal:
|C| = 1 × 3 × 1 = 3$$|C| = 1 \times 3 \times 1 = 3$$
Therefore, 1 × |B| = 3 ⇒ |B| = 3$1 \times |B| = 3 \Rightarrow |B| = 3$.
So α = 3$\alpha = 3$.
Step 2: Finding Matrix B
To find β$\beta$ (the trace of B$B$), we must explicitly find B = A⁻¹C$B = A^{-1}C$.
Rather than finding the full inverse, manually solve A Bᵢ = Cᵢ$A B_i = C_i$:
For B₁ = [x, y, z]^T$B_1 = [x, y, z]^T$:
2x+z=1$2x+z=1$, x+y=0$x+y=0$, x+z=0 ⇒ x=1, z=-1, y=-1$x+z=0 \Rightarrow x=1, z=-1, y=-1$. Thus B₁ = [1, -1, -1]^T$B_1 = [1, -1, -1]^T$.
For B₂$B_2$:
2x+z=2$2x+z=2$, x+y=3$x+y=3$, x+z=0 ⇒ x=2, z=-2, y=1$x+z=0 \Rightarrow x=2, z=-2, y=1$. Thus B₂ = [2, 1, -2]^T$B_2 = [2, 1, -2]^T$.
For B₃$B_3$:
2x+z=3$2x+z=3$, x+y=2$x+y=2$, x+z=1 ⇒ x=2, z=-1, y=0$x+z=1 \Rightarrow x=2, z=-1, y=0$. Thus B₃ = [3, 0, -1]^T$B_3 = [3, 0, -1]^T$.
Step 3: Calculating Trace and Final Output
Constructing Matrix B:
B = bmatrix 1 & 2 & 3 -1 & 1 & 0 -1 & -2 & -1 bmatrix$$B = \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{bmatrix}$$
The diagonal elements are 1, 1, -1$1, 1, -1$.
Trace β = 1 + 1 - 1 = 1$\beta = 1 + 1 - 1 = 1$.
Finally compute α³ + β³$\alpha^3 + \beta^3$:
3³ + 1³ = 27 + 1 = 28$$3^3 + 1^3 = 27 + 1 = 28$$
Pattern Recognition
When given AXᵢ = Yᵢ$AX_i = Y_i$ for multiple columns, they collectively form A X = Y$A X = Y$. Using |A||X| = |Y|$|A||X| = |Y|$ bypasses full matrix inversion if you strictly need determinants. To grab the trace, solving equations systematically column-by-column is generally less error-prone than forming the full adjoint inverse matrix.
Chapter Mix
Class 12 Maths: Matrices