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To find the value of [tex]\( a \)[/tex] such that the determinant of the given matrix is [tex]\(-19\)[/tex], we need to compute the determinant and solve for [tex]\( a \)[/tex]. The matrix is:
[tex]\[ \left[\begin{array}{ccc} -6 & 7 & 1 \\ a & -3 & 4 \\ -6 & 4 & -3 \end{array}\right] \][/tex]
The determinant of a [tex]\(3 \times 3\)[/tex] matrix [tex]\( \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \)[/tex] is given by:
[tex]\[ \text{det} = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \][/tex]
Let's substitute the given matrix elements into this formula:
[tex]\[ \begin{array}{l} a_{11} = -6, \quad a_{12} = 7, \quad a_{13} = 1, \\ a_{21} = a, \quad a_{22} = -3, \quad a_{23} = 4, \\ a_{31} = -6, \quad a_{32} = 4, \quad a_{33} = -3 \end{array} \][/tex]
Now, compute the determinant step by step:
[tex]\[ \begin{vmatrix} -6 & 7 & 1 \\ a & -3 & 4 \\ -6 & 4 & -3 \end{vmatrix} = -6 [(-3)(-3) - (4)(4)] - 7 [(a)(-3) - (4)(-6)] + 1 [(a)(4) - (-3)(-6)] \][/tex]
Compute the individual terms within the brackets:
[tex]\[ -6 [(9) - (16)] = -6 (-7) = 42 \][/tex]
For the second term:
[tex]\[ -7 [(a)(-3) - (4)(-6)] = -7 [-3a + 24] = -7(-3a + 24) = 21a - 168 \][/tex]
For the third term:
[tex]\[ 1 [(4a) - (9)] = 4a - 9 \][/tex]
Combine all the terms:
[tex]\[ \text{det} = 42 + 21a - 168 + 4a - 9 \][/tex]
Simplify this expression:
[tex]\[ \text{det} = 21a + 4a + 42 - 168 - 9 = 25a - 135 \][/tex]
We are given that the determinant is [tex]\(-19\)[/tex]. Set the determinant equation equal to [tex]\(-19\)[/tex] and solve for [tex]\( a \)[/tex]:
[tex]\[ 25a - 135 = -19 \][/tex]
Add 135 to both sides:
[tex]\[ 25a = 116 \][/tex]
Divide by 25:
[tex]\[ a = \frac{116}{25} = 4.64 \][/tex]
Since none of the options A, B, C, or D match exactly, there might be an error in the process of solving or in the provided problem options. However, according to our solution, the calculated value of [tex]\( a \)[/tex] is approximately [tex]\( 4.64 \)[/tex].
[tex]\[ \left[\begin{array}{ccc} -6 & 7 & 1 \\ a & -3 & 4 \\ -6 & 4 & -3 \end{array}\right] \][/tex]
The determinant of a [tex]\(3 \times 3\)[/tex] matrix [tex]\( \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \)[/tex] is given by:
[tex]\[ \text{det} = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \][/tex]
Let's substitute the given matrix elements into this formula:
[tex]\[ \begin{array}{l} a_{11} = -6, \quad a_{12} = 7, \quad a_{13} = 1, \\ a_{21} = a, \quad a_{22} = -3, \quad a_{23} = 4, \\ a_{31} = -6, \quad a_{32} = 4, \quad a_{33} = -3 \end{array} \][/tex]
Now, compute the determinant step by step:
[tex]\[ \begin{vmatrix} -6 & 7 & 1 \\ a & -3 & 4 \\ -6 & 4 & -3 \end{vmatrix} = -6 [(-3)(-3) - (4)(4)] - 7 [(a)(-3) - (4)(-6)] + 1 [(a)(4) - (-3)(-6)] \][/tex]
Compute the individual terms within the brackets:
[tex]\[ -6 [(9) - (16)] = -6 (-7) = 42 \][/tex]
For the second term:
[tex]\[ -7 [(a)(-3) - (4)(-6)] = -7 [-3a + 24] = -7(-3a + 24) = 21a - 168 \][/tex]
For the third term:
[tex]\[ 1 [(4a) - (9)] = 4a - 9 \][/tex]
Combine all the terms:
[tex]\[ \text{det} = 42 + 21a - 168 + 4a - 9 \][/tex]
Simplify this expression:
[tex]\[ \text{det} = 21a + 4a + 42 - 168 - 9 = 25a - 135 \][/tex]
We are given that the determinant is [tex]\(-19\)[/tex]. Set the determinant equation equal to [tex]\(-19\)[/tex] and solve for [tex]\( a \)[/tex]:
[tex]\[ 25a - 135 = -19 \][/tex]
Add 135 to both sides:
[tex]\[ 25a = 116 \][/tex]
Divide by 25:
[tex]\[ a = \frac{116}{25} = 4.64 \][/tex]
Since none of the options A, B, C, or D match exactly, there might be an error in the process of solving or in the provided problem options. However, according to our solution, the calculated value of [tex]\( a \)[/tex] is approximately [tex]\( 4.64 \)[/tex].
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