Matrices & Determinants
Matrix Product — Finding B and Computing |B|
nta_pyq_2024_jan
Grade 12
Question:
Let $A=\begin{bmatrix}2&0&1\\1&1&0\\1&0&1\end{bmatrix}$, $B=[B_1,B_2,B_3]$, where $B_1,B_2,B_3$ are column matrices, and $AB_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$, $AB_2=\begin{bmatrix}2\\3\\0\end{bmatrix}$, $AB_3=\begin{bmatrix}3\\2\\1\end{bmatrix}$. If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^3+\beta^3$ is equal to
Step-by-Step Solution
Key Concept: Solve $AB_i=c_i$ for each column $B_i$ by row-reducing the augmented matrices. Assemble $B$, compute $|B|=\alpha$ and trace$(B)=\beta$, then evaluate $\alpha^3+\beta^3$.
Solve $AB_i=c_i$ for each column: $B_1=(1,-1,-1)^T$, $B_2=(2,1,-2)^T$, $B_3=(2,0,-1)^T$. $B=\begin{bmatrix}1&2&2\\-1&1&0\\-1&-2&-1\end{bmatrix}$. $\alpha=|B|=3$, $\beta=1+1-1=1$. $\alpha^3+\beta^3=27+1=28$.
Correct Answer: 28