If 1 an eigenvalue of the matrix $A$, where $A=\left[\begin{array}{ccc}2 & 0 & -1 \ 4 & 2 & -2 \ 6 & 0 & a\end{array} ight]$, then $a=$

Question

If 1 an eigenvalue of the matrix $A$, where $A=\left[\begin{array}{ccc}2 & 0 & -1 \ 4 & 2 & -2 \ 6 & 0 & a\end{array}ight]$, then $a=$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the matrix $A = \left[\begin{array}{ccc}2 & 0 & -1 \ 4 & -2 & 2 \ 6 & 0 & a\end{array}\right]$, we need to find the value of $a$ such that $1$ is an eigenvalue of $A$. ‖ ‖ step 2 ⋮ To find the eigenvalues, we solve the characteristic equation $\det(A - \lambda I) = 0$. For $\lambda = 1$, we have $A - I = \left[\begin{array}{ccc}1 & 0 & -1 \ 4 & -3 & 2 \ 6 & 0 & a-1\end{array}\right]$. ‖ ‖ step 3 ⋮ Calculate the determinant of $A - I$: $$ \det(A - I) = \left|\begin{array}{ccc}1 & 0 & -1 \ 4 & -3 & 2 \ 6 & 0 & a-1\end{array}\right| $$ Expanding along the first row: $$ \det(A - I) = 1 \cdot \left|\begin{array}{cc}-3 & 2 \ 0 & a-1\end{array}\right| - 0 \cdot \left|\begin{array}{cc}4 & 2 \ 6 & a-1\end{array}\right| + (-1) \cdot \left|\begin{array}{cc}4 & -3 \ 6 & 0\end{array}\right| $$ $$ = 1 \cdot (-3(a-1)) - 1 \cdot (4 \cdot 0 - (-3) \cdot 6) $$ $$ = -3(a-1) + 18 $$ $$ = -3a + 3 + 18 $$ $$ = -3a + 21 $$ ‖ ‖ step 4 ⋮ Set the determinant equal to zero and solve for $a$: $$ -3a + 21 = 0 $$ $$ -3a = -21 $$ $$ a = 7 $$ ‖ ∻Answer∻ ⚹ $a = 7$ ⚹ ∻Key Concept∻ ⚹Eigenvalues of a matrix⚹ ∻Explanation∻ ⚹To find the eigenvalues of a matrix, we solve the characteristic equation $\det(A - \lambda I) = 0$. In this problem, we set $\lambda = 1$ and solve for $a$ such that the determinant of $A - I$ is zero.⚹◊What is the characteristic equation for finding the eigenvalues of a matrix?⍭ Generate me a similar question◊

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