b) If [tex]A=\left[\begin{array}{ccc}2 & 4 & 3 \\ 1 & -2 & -2 \\ -3 & 3 & 2\end{array}\right][/tex] and [tex]B=\left[\begin{array}{ccc}2 & 1 & -2 \\ 4 & 13 & 7 \\ -3 & -18 & -8\end{array}\right][/tex] show that [tex]A B=K i[/tex] where [tex]k[/tex] is a constant and I is the identity matrix. Hence solve the equations
[tex]\begin{array}{l}
2 x+4 y+3 z=4 \\
X-2 y-2 z=-1 \\
-3 x+3 y+2 z=-2
\end{array}[/tex]
Answer (2)
Show that A B = k I by computing the matrix product A B and verifying that it is a scalar multiple of the identity matrix.
Express the system of equations in matrix form as A X = C .
Use the result A B = 11 I to find A − 1 = 11 1 B .
Solve for X using X = A − 1 C = 11 1 BC , which gives the solution x = 1 , y = − 1 , z = 2 .
Explanation
Problem Analysis First, we need to show that A B = k I , where A = 2 1 − 3 4 − 2 3 3 − 2 2 and B = 2 4 − 3 1 13 − 18 − 2 7 − 8 , k is a constant, and I is the identity matrix. We will then solve the system of equations using this result.
Calculating AB Let's compute the matrix product A B :
A B = 2 1 − 3 4 − 2 3 3 − 2 2 2 4 − 3 1 13 − 18 − 2 7 − 8 = 2 ( 2 ) + 4 ( 4 ) + 3 ( − 3 ) 1 ( 2 ) + ( − 2 ) ( 4 ) + ( − 2 ) ( − 3 ) − 3 ( 2 ) + 3 ( 4 ) + 2 ( − 3 ) 2 ( 1 ) + 4 ( 13 ) + 3 ( − 18 ) 1 ( 1 ) + ( − 2 ) ( 13 ) + ( − 2 ) ( − 18 ) − 3 ( 1 ) + 3 ( 13 ) + 2 ( − 18 ) 2 ( − 2 ) + 4 ( 7 ) + 3 ( − 8 ) 1 ( − 2 ) + ( − 2 ) ( 7 ) + ( − 2 ) ( − 8 ) − 3 ( − 2 ) + 3 ( 7 ) + 2 ( − 8 ) A B = 4 + 16 − 9 2 − 8 + 6 − 6 + 12 − 6 2 + 52 − 54 1 − 26 + 36 − 3 + 39 − 36 − 4 + 28 − 24 − 2 − 14 + 16 6 + 21 − 16 = 11 0 0 0 11 0 0 0 11 Thus, A B = 11 I , where k = 11 and I is the identity matrix.
Expressing the system in matrix form Now, we need to solve the system of equations: ⎩ ⎨ ⎧ 2 x + 4 y + 3 z = 4 x − 2 y − 2 z = − 1 − 3 x + 3 y + 2 z = − 2 We can express this system in matrix form as A X = C , where A = 2 1 − 3 4 − 2 3 3 − 2 2 , X = x y z , and C = 4 − 1 − 2 .
Finding the inverse of A Since A B = 11 I , we have A − 1 = 11 1 B . Multiplying both sides of A X = C by A − 1 , we get X = A − 1 C . Substituting A − 1 = 11 1 B into the equation X = A − 1 C , we get X = 11 1 BC .
Calculating BC Now, let's compute the matrix product BC :
BC = 2 4 − 3 1 13 − 18 − 2 7 − 8 4 − 1 − 2 = 2 ( 4 ) + 1 ( − 1 ) + ( − 2 ) ( − 2 ) 4 ( 4 ) + 13 ( − 1 ) + 7 ( − 2 ) − 3 ( 4 ) + ( − 18 ) ( − 1 ) + ( − 8 ) ( − 2 ) = 8 − 1 + 4 16 − 13 − 14 − 12 + 18 + 16 = 11 − 11 22 Therefore, X = 11 1 BC = 11 1 11 − 11 22 = 1 − 1 2 .
Final Solution Thus, the solution to the system of equations is x = 1 , y = − 1 , and z = 2 .
Examples
Matrix algebra is used extensively in computer graphics for transformations such as rotations, scaling, and translations of objects in 3D space. The matrix multiplication performed in this problem is fundamental to these transformations. For example, if you want to rotate a 3D object, you would multiply the object's vertex coordinates by a rotation matrix. Similarly, solving systems of equations is crucial in various engineering applications, such as circuit analysis, where you need to determine the currents and voltages in different parts of a circuit.
To show that the product of matrices A and B yields a scalar multiple of the identity matrix, we calculated the product and found that it equals 11I. Subsequently, we expressed the given system of equations in matrix form and used the inverse of A to solve for the variables, resulting in x = 1, y = -1, and z = 2. This demonstrates the relationship between matrix operations and solving linear equations.
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