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Math  /  Algebra

QuestionSelect the correct answer.
Which is the inverse of this matrix? [12535911−2]\left[\begin{array}{ccc} 1 & 2 & 5 \\ 3 & 5 & 9 \\ 1 & 1 & -2 \end{array}\right] A. [−199−715−76−21−1]\left[\begin{array}{ccc}-19 & 9 & -7 \\ 15 & -7 & 6 \\ -2 & 1 & -1\end{array}\right] B. [−199−7−21−115−76]\left[\begin{array}{ccc}-19 & 9 & -7 \\ -2 & 1 & -1 \\ 15 & -7 & 6\end{array}\right] c. [5−76−21−1]\left[\begin{array}{ccc}5 & -7 & 6 \\ -2 & 1 & -1\end{array}\right] D. [−19915−7−21]\left[\begin{array}{cc}-19 & 9 \\ 15 & -7 \\ -2 & 1\end{array}\right] E. The matrix is noninvertible.

Studdy Solution

STEP 1

1. We are given a 3x3 matrix.
2. We need to find the inverse of the given matrix, if it exists.
3. The matrix is invertible if its determinant is non-zero.
4. The inverse of a matrix A A is denoted as A−1 A^{-1} .

STEP 2

1. Calculate the determinant of the matrix.
2. Determine if the matrix is invertible.
3. If invertible, calculate the inverse.
4. Compare the calculated inverse with the given options.

STEP 3

Calculate the determinant of the matrix.
Given matrix: A=[12535911−2] A = \left[\begin{array}{ccc} 1 & 2 & 5 \\ 3 & 5 & 9 \\ 1 & 1 & -2 \end{array}\right]
The determinant of a 3x3 matrix A A is calculated as: det(A)=a(ei−fh)−b(di−fg)+c(dh−eg) \text{det}(A) = a(ei-fh) - b(di-fg) + c(dh-eg)
For matrix A A : a=1,b=2,c=5,d=3,e=5,f=9,g=1,h=1,i=−2 a = 1, b = 2, c = 5, d = 3, e = 5, f = 9, g = 1, h = 1, i = -2
det(A)=1(5⋅(−2)−9⋅1)−2(3⋅(−2)−9⋅1)+5(3⋅1−5⋅1) \text{det}(A) = 1(5 \cdot (-2) - 9 \cdot 1) - 2(3 \cdot (-2) - 9 \cdot 1) + 5(3 \cdot 1 - 5 \cdot 1)
det(A)=1(−10−9)−2(−6−9)+5(3−5) \text{det}(A) = 1(-10 - 9) - 2(-6 - 9) + 5(3 - 5)
det(A)=1(−19)−2(−15)+5(−2) \text{det}(A) = 1(-19) - 2(-15) + 5(-2)
det(A)=−19+30−10 \text{det}(A) = -19 + 30 - 10
det(A)=1 \text{det}(A) = 1

STEP 4

Determine if the matrix is invertible.
Since the determinant is 1 1 , which is non-zero, the matrix is invertible.

STEP 5

Calculate the inverse of the matrix.
The inverse of a 3x3 matrix A A is given by: A−1=1det(A)⋅adj(A) A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)
Since det(A)=1\text{det}(A) = 1, we need to find the adjugate of A A .
The adjugate of a matrix is the transpose of its cofactor matrix.
Calculate the cofactor matrix of A A :
Cofactor of element a11 a_{11} : C11=det(591−2)=(5⋅(−2))−(9⋅1)=−10−9=−19 C_{11} = \text{det}\left(\begin{array}{cc} 5 & 9 \\ 1 & -2 \end{array}\right) = (5 \cdot (-2)) - (9 \cdot 1) = -10 - 9 = -19
Cofactor of element a12 a_{12} : C12=−det(391−2)=−(3⋅(−2)−9⋅1)=−(−6−9)=15 C_{12} = -\text{det}\left(\begin{array}{cc} 3 & 9 \\ 1 & -2 \end{array}\right) = -(3 \cdot (-2) - 9 \cdot 1) = -(-6 - 9) = 15
Cofactor of element a13 a_{13} : C13=det(3511)=(3⋅1)−(5⋅1)=3−5=−2 C_{13} = \text{det}\left(\begin{array}{cc} 3 & 5 \\ 1 & 1 \end{array}\right) = (3 \cdot 1) - (5 \cdot 1) = 3 - 5 = -2
Cofactor of element a21 a_{21} : C21=−det(251−2)=−(2⋅(−2)−5⋅1)=−(−4−5)=9 C_{21} = -\text{det}\left(\begin{array}{cc} 2 & 5 \\ 1 & -2 \end{array}\right) = -(2 \cdot (-2) - 5 \cdot 1) = -(-4 - 5) = 9
Cofactor of element a22 a_{22} : C22=det(151−2)=(1⋅(−2))−(5⋅1)=−2−5=−7 C_{22} = \text{det}\left(\begin{array}{cc} 1 & 5 \\ 1 & -2 \end{array}\right) = (1 \cdot (-2)) - (5 \cdot 1) = -2 - 5 = -7
Cofactor of element a23 a_{23} : C23=−det(1211)=−(1⋅1−2⋅1)=−(1−2)=1 C_{23} = -\text{det}\left(\begin{array}{cc} 1 & 2 \\ 1 & 1 \end{array}\right) = -(1 \cdot 1 - 2 \cdot 1) = -(1 - 2) = 1
Cofactor of element a31 a_{31} : C31=det(2559)=(2⋅9)−(5⋅5)=18−25=−7 C_{31} = \text{det}\left(\begin{array}{cc} 2 & 5 \\ 5 & 9 \end{array}\right) = (2 \cdot 9) - (5 \cdot 5) = 18 - 25 = -7
Cofactor of element a32 a_{32} : C32=−det(1539)=−(1⋅9−5⋅3)=−(9−15)=6 C_{32} = -\text{det}\left(\begin{array}{cc} 1 & 5 \\ 3 & 9 \end{array}\right) = -(1 \cdot 9 - 5 \cdot 3) = -(9 - 15) = 6
Cofactor of element a33 a_{33} : C33=det(1235)=(1⋅5)−(2⋅3)=5−6=−1 C_{33} = \text{det}\left(\begin{array}{cc} 1 & 2 \\ 3 & 5 \end{array}\right) = (1 \cdot 5) - (2 \cdot 3) = 5 - 6 = -1
The cofactor matrix is: [−1915−29−71−76−1] \left[\begin{array}{ccc} -19 & 15 & -2 \\ 9 & -7 & 1 \\ -7 & 6 & -1 \end{array}\right]
The adjugate (transpose of the cofactor matrix) is: adj(A)=[−199−715−76−21−1] \text{adj}(A) = \left[\begin{array}{ccc} -19 & 9 & -7 \\ 15 & -7 & 6 \\ -2 & 1 & -1 \end{array}\right]
Since det(A)=1\text{det}(A) = 1, the inverse is: A−1=adj(A)=[−199−715−76−21−1] A^{-1} = \text{adj}(A) = \left[\begin{array}{ccc} -19 & 9 & -7 \\ 15 & -7 & 6 \\ -2 & 1 & -1 \end{array}\right]

STEP 6

Compare the calculated inverse with the given options.
The correct answer is option A: [−199−715−76−21−1] \left[\begin{array}{ccc} -19 & 9 & -7 \\ 15 & -7 & 6 \\ -2 & 1 & -1 \end{array}\right]

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