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Chapter 6: Problem 48

If you are given two matrices, \(A\) and \(B\), explain how to determine if \(B\) is the multiplicative inverse of \(A\).

Short Answer

Expert verified
Matrix B is the multiplicative inverse of matrix A if AB = I and BA = I, where I is the Identity matrix of the same order as that of A or B.

Step by step solution

01

Identify provided matrices

Recognize the provided matrices A and B that are to be evaluated. Always verify that both the matrices are of the same order (i.e., have the same number of rows and columns), as the inverse of a matrix (if it exists) will always have the same order as the original matrix.
02

Multiply A and B

Multiply matrix A with matrix B (AB). Perform matrix multiplication by multiplying elements of the rows of the first matrix by the elements of the corresponding column of the other matrix and then add them. Repeat this process until all rows and columns are covered.
03

Multiply B and A

Multiply matrix B with matrix A (BA) following the same procedure used in multiplication of AB.
04

Compare with Identity Matrix

Compare both the multiplied matrices (AB and BA) with the Identity matrix of the equivalent order. If either of them or both comes out to be the Identity matrix (all diagonal elements are '1', and all non-diagonal elements are '0'), then the given matrix B is the multiplicative inverse of the matrix A.

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Most popular questions from this chapter

In Exercises \(1-4\) a. Give the order of each matrix. b. If \(A=\left[a_{i j}\right],\) identify \(a_{32}\) and \(a_{23}\) or explain why identification is not possible. $$ \left[\begin{array}{rrrr} 1 & -5 & \pi & e \\ 0 & 7 & -6 & -\pi \\ -2 & 1 & 11 & -1 \end{array}\right] $$

In Exercises \(27-36,\) find (if possible): \(\begin{array}{llll}\text { a. } A B & \text { and } & \text { b. } B A\end{array}\) $$ A=\left[\begin{array}{l} -1 \\ -2 \\ -3 \end{array}\right], \quad B=\left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right] $$

In Exercises \(9-16,\) find: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) .$$ A=\left[\begin{array}{lll} 6 & 2 & -3 \end{array}\right], \quad B=\left[\begin{array}{lll} 4 & -2 & 3 \end{array}\right] $$

Use Cramer's rule to solve each system. $$ \begin{aligned}x+y+z &=0 \\\2 x-y+z &=-1 \\\\-x+3 y-z &=-8\end{aligned} $$

In Exercises \(17-26,\) let $$ A=\left[\begin{array}{rr} -3 & -7 \\ 2 & -9 \\ 5 & 0 \end{array}\right] \text { and } B=\left[\begin{array}{rr} -5 & -1 \\ 0 & 0 \\ 3 & -4 \end{array}\right] $$ Solve each matrix equation for \(X\). $$ 3 X+2 A=B $$
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