Question 26. 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 $a=$

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Question 26. 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 $a=$

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∻Solution by Steps∻ ‖ step 1 ⋮ Given the matrix $A = \begin{pmatrix} 2 & 0 & -1 \ 4 & 2 & -2 \ 6 & 0 & a \end{pmatrix}$, we know that 1 is an eigenvalue of $A$. ‖ ‖ step 2 ⋮ To find the value of $a$, we need to solve the characteristic equation $\det(A - \lambda I) = 0$ for $\lambda = 1$. ‖ ‖ step 3 ⋮ Subtract $\lambda I$ from $A$: $$ A - \lambda I = \begin{pmatrix} 2-1 & 0 & -1 \ 4 & 2-1 & -2 \ 6 & 0 & a-1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & -1 \ 4 & 1 & -2 \ 6 & 0 & a-1 \end{pmatrix} $$ ‖ ‖ step 4 ⋮ Calculate the determinant of the matrix: $$ \det(A - I) = \begin{vmatrix} 1 & 0 & -1 \ 4 & 1 & -2 \ 6 & 0 & a-1 \end{vmatrix} $$ Expanding along the first row: $$ \det(A - I) = 1 \cdot \begin{vmatrix} 1 & -2 \ 0 & a-1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 4 & -2 \ 6 & a-1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 4 & 1 \ 6 & 0 \end{vmatrix} $$ ‖ ‖ step 5 ⋮ Simplify the determinants: $$ \det(A - I) = 1 \cdot (1 \cdot (a-1) - 0 \cdot (-2)) - 1 \cdot (4 \cdot 0 - 6 \cdot 1) $$ $$ = (a-1) - (-6) $$ $$ = a - 1 + 6 $$ $$ = a + 5 $$ ‖ ‖ step 6 ⋮ Since 1 is an eigenvalue, $\det(A - I) = 0$: $$ a + 5 = 0 $$ $$ a = -5 $$ ‖ ∻Answer∻ ⚹ $a = -5$ ⚹ ∻Key Concept∻ ⚹Eigenvalues and characteristic equation⚹ ∻Explanation∻ ⚹To find the value of $a$ for which 1 is an eigenvalue, we solve the characteristic equation $\det(A - \lambda I) = 0$ for $\lambda = 1$. This involves calculating the determinant of the matrix $A - I$ and setting it equal to zero.⚹◊What is the relationship between eigenvalues of a matrix and its determinant?⍭ Generate me a similar question◊

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