The purpose of this question is to fully understand how to answer the following question and show the full working on paper without the aid of a computer. (Note that I am refreshing my Linear Algebra knowledge).
Question - Evaluate the determinate |A| of
A = $\begin{bmatrix}t-2 & 4 & 3\\ 1 & t+1 & -2\\0 & 0 & t-4 \end{bmatrix}$
Determine those values of t for which |A| = 0.
My Approach:
step 1 - Calculate the minors for row 3
Note I am only using M${_3,_3}$ because A ${_3,_1}$ & A${_3,_2}$ both have value of zero.
M${_3,_3}\begin{bmatrix}t-2 & 4\\1 & t+1\\\end{bmatrix}$
$\therefore$ the cofactor value will be derived from:
(-1)$^{3 + 3}$((t - 2 ${\times}$ t + 1) - (4 ${\times}$ 1)) = -t - 3
$\therefore$ |A| = (t - 4) ${\times}$ (-t - 3) = -t$^2$ + t + 12
Questions
Is the above correct so far?
Having evaluated the determinant I now need to determine those values of t for which |A| is zero.
My approach for 2:
Use the formula: ax$^3$ + bx$^2$ + cx + d
I insert -t$^2$ + t + 12
such that : x$^3$ - x$^2$ + x + 12 = 0
Then solve.
Is this the correct approach to this problem and how would I solve on paper?
\begin{eqnarray} |A|&=& \left| \begin{matrix}t-2 & 4 & 3\\ 1 & t+1 & -2\\0 & 0 & t-4 \end{matrix}\right| \\ &=& (t-4)\times \left| \begin{matrix}t-2 & 4 \\ 1 & t+1\end{matrix}\right| \\ &=& (t-4)\left[(t-2)(t+1)-4\times 1 \right]\\ &=& (t-4) \left[t^2+t-2t-2-4 \right]\\ &=& (t-4)\left[t^2-t-6 \right]\\ &=& (t-4)(t+2)(t-3)\\ \end{eqnarray}