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Chapter 6: Problem 51

Explain how to find the multiplicative inverse for a \(3 \times 3\) invertible matrix.

Short Answer

Expert verified
The multiplicative inverse of a matrix can be found by augmenting the square matrix with the identity matrix, then row reducing until the original matrix becomes the identity matrix and the identity matrix transforms into the inverse matrix. The right half of the resultant matrix is the inverse matrix.

Step by step solution

01

Setting up the augmented matrix

Let the original 3x3 matrix be \(A\). Write \(A\) as an augmented matrix with the 3x3 identity matrix. This results in a \(3 \times 6\) matrix with \(A\) on the left and the identity matrix on the right.
02

Row reducing to row echelon form

Use the Gauss-Jordan elimination to row reduce the matrix to row echelon form. This is done by performing row operations.
03

Normalizing rows

Next, normalize each row by dividing every term by the leading term if the leading term is not 1.
04

Row reducing to reduced row echelon form

Perform further row operations until the matrix on the left side is turned into an identity matrix. At this stage, the original matrix has been transformed into its inverse on the right side of the augmented matrix.
05

Identifying the Inverse

Once the left side is the identity matrix, the right side will be the inverse of the original matrix. The right side of the augmented matrix is the multiplicative inverse of the matrix \(A\).

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Most popular questions from this chapter

Use Cramer's rule to solve each system. $$ \begin{aligned}&3 x+2 z=4\\\&5 x-y=-4\\\&4 y+3 z=22\end{aligned} $$

Evaluate each determinant. $$ \left|\begin{array}{rrr}4 & 0 & 0 \\\3 & -1 & 4 \\\2 & -3 & 5\end{array}\right| $$

Evaluate each determinant. $$ \left|\begin{array}{rrr}3 & 0 & 0 \\\2 & 1 & -5 \\\2 & 5 & -1\end{array}\right| $$

Evaluate each determinant. $$ \left|\begin{array}{rrr}1 & 2 & 3 \\\2 & 2 & -3 \\\3 & 2 & 1\end{array}\right| $$

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 \(X\). $$ X-B=A $$
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