Given: [tex]CD=I[/tex], and
[tex]C=\left[\begin{array}{ccc}-2 & 4 & 1 \\ 1 & 0 & 3 \\ -2 & 3 & -1\end{array}\right] \quad D=\left[\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right][/tex]
Find the elements of [tex]D[/tex].
[tex]a=[/tex] $\square$
[tex]b=[/tex] $\square$
[tex]c=[/tex] $\square$
[tex]d=[/tex] $\square$
e = $\square$
[tex]f=[/tex] $\square$
[tex]g=[/tex] $\square$
[tex]h=[/tex] $\square$
Answer (1)
Determine that D is the inverse of matrix C since C D = I .
Calculate the determinant of C to ensure it's invertible: det ( C ) = 1 .
Find the adjugate of C by finding the matrix of minors, cofactors, and then transposing the cofactor matrix.
Since det ( C ) = 1 , D = C − 1 is equal to the adjugate of C , thus a = − 9 , b = 7 , c = 12 , d = − 5 , e = 4 , f = 7 , g = 3 , h = − 2 .
Explanation
Problem Analysis We are given two matrices, C and D , such that their product C D is the identity matrix I . This means that D is the inverse of C , i.e., D = C − 1 . Our goal is to find the elements of matrix D .
Given Matrices The given matrix C is: C = [ − 2 4 1 1 0 3 − 2 3 − 1 ] and the matrix D is: D = [ a b c d e f g h i ]
Determinant of C Since D = C − 1 , we need to find the inverse of matrix C . First, we calculate the determinant of C . det ( C ) = − 2 ( 0 ⋅ ( − 1 ) − 3 ⋅ 3 ) − 4 ( 1 ⋅ ( − 1 ) − 3 ⋅ ( − 2 )) + 1 ( 1 ⋅ 3 − 0 ⋅ ( − 2 )) det ( C ) = − 2 ( − 9 ) − 4 ( − 1 + 6 ) + 1 ( 3 ) det ( C ) = 18 − 4 ( 5 ) + 3 = 18 − 20 + 3 = 1
Matrix of Minors Since the determinant of C is 1 (which is non-zero), the inverse exists. Now we find the matrix of minors of C . M = [ 0 ⋅ ( − 1 ) − 3 ⋅ 3 1 ⋅ ( − 1 ) − 3 ⋅ ( − 2 ) 1 ⋅ 3 − 0 ⋅ ( − 2 ) 4 ⋅ ( − 1 ) − 1 ⋅ 3 − 2 ⋅ ( − 1 ) − 1 ⋅ ( − 2 ) − 2 ⋅ 3 − 4 ⋅ ( − 2 ) 4 ⋅ 3 − 1 ⋅ 0 − 2 ⋅ 3 − 1 ⋅ 1 − 2 ⋅ 0 − 4 ⋅ 1 ] = [ − 9 5 3 − 7 4 2 12 − 7 − 4 ]
Matrix of Cofactors Next, we find the matrix of cofactors by applying the sign changes to the matrix of minors. The sign changes follow the pattern: [ + − + − + − + − + ] So, the matrix of cofactors is: C o f = [ − 9 − 5 3 7 4 − 2 12 7 − 4 ]
Adjugate of C Now, we find the adjugate (transpose) of the cofactor matrix: a d j ( C ) = [ − 9 7 12 − 5 4 7 3 − 2 − 4 ]
Inverse Matrix Since det ( C ) = 1 , the inverse matrix C − 1 is simply the adjugate matrix: C − 1 = det ( C ) 1 a d j ( C ) = 1 1 [ − 9 7 12 − 5 4 7 3 − 2 − 4 ] = [ − 9 7 12 − 5 4 7 3 − 2 − 4 ] Therefore, D = C − 1 = [ − 9 7 12 − 5 4 7 3 − 2 − 4 ] .
Elements of D Finally, we can identify the elements of matrix D :
a = − 9 b = 7 c = 12 d = − 5 e = 4 f = 7 g = 3 $h = -2
Examples
In computer graphics, matrices are heavily used for transformations such as rotations, scaling, and translations of objects in 3D space. The inverse of a transformation matrix is used to undo a transformation, allowing you to revert an object back to its original position or orientation. This is crucial for interactive applications where users might want to undo actions or reset the scene. For example, if you rotate an object by a certain angle, multiplying by the inverse rotation matrix will bring the object back to its initial orientation. This principle extends to various fields like robotics, where controlling the movement and orientation of robotic arms requires precise matrix operations and their inverses.
