The trace of a square matrix is defined as th | vedclass

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(C) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$. The characteristic equation of $A$ is given by $\lambda^2 - 	ext{tr}(A)\lambda + \det(A)

Vedclass · Accepted Answer

$5$

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(C) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$. The characteristic equation of $A$ is given by $\lambda^2 - 	ext{tr}(A)\lambda + \det(A) = 0$. Given $	ext{tr}(A) = a+d = 3$ and let $\Delta = \det(A) = ad-bc$. So,the characteristic equation is $\lambda^2 - 3\lambda + \Delta = 0$. By the Cayley-Hamilton theorem,$A^2 - 3A + \Delta I = 0$,which implies $A^2 = 3A - \Delta I$. Multiplying by $A$,we get $A^3 = 3A^2 - \Delta A$. Substituting $A^2 = 3A - \Delta I$ into the equation: $A^3 = 3(3A - \Delta I) - \Delta A = 9A - 3\Delta I - \Delta A = (9 - \Delta)A - 3\Delta I$. Taking the trace on both sides: $	ext{tr}(A^3) = (9 - \Delta)	ext{tr}(A) - 3\Delta 	ext{tr}(I)$. Since $	ext{tr}(A^3) = -18$,$	ext{tr}(A) = 3$,and $	ext{tr}(I) = 2$ (for a $2 	imes 2$ matrix): $-18 = (9 - \Delta)(3) - 3\Delta(2)$. $-18 = 27 - 3\Delta - 6\Delta$. $-18 = 27 - 9\Delta$. $9\Delta = 27 + 18 = 45$. $\Delta = 5$. Thus,the determinant of $A$ is $5$.

The trace of a square matrix is defined as the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is $3$ and the trace of $A^3$ is $-18$,then the value of the determinant of $A$ is:

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