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Chapter 8: Problem 4

Find the products \(A B\) and \(B A\) to determine whether \(B\) is the multiplicative inverse of \(A\). $$A=\left[\begin{array}{rr} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{array}\right], \quad B=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]$$

Short Answer

Expert verified
No, 'B' is not the multiplicative inverse of 'A' as only 'AB' equals the Identity matrix, but 'BA' doesn't.

Step by step solution

01

Compute the product AB

Multiplying matrix 'A' and 'B' we have \(AB = \left[\begin{array}{rr}-2 & 1 \ \frac{3}{2} & -\frac{1}{2}\end{array}\right] \times \left[\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right] = \left[\begin{array}{rr} -2*1 + 1*3 & -2*2 + 1*4 \ \frac{3}{2}*1 + -\frac{1}{2}*3 & \frac{3}{2}*2 + -\frac{1}{2}*4\end{array}\right] = \left[\begin{array}{rr}1 & 0 \ 0 & 1\end{array}\right]\)
02

Compute the product BA

Multiplying matrix 'B' and 'A' we have \(BA = \left[\begin{array}{rr}1 & 2 \ 3 & 4\end{array}\right] \times \left[\begin{array}{ll}-2 & 1 \ \frac{3}{2} & -\frac{1}{2}\end{array}\right] = \left[\begin{array}{rr}1*-2 + 2*\frac{3}{2} & 1*1 + 2*-\frac{1}{2} \ 3*-2 + 4*\frac{3}{2} & 3*1 + 4*-\frac{1}{2}\end{array}\right] = \left[\begin{array}{rr}-1 & 0 \ 0 & -1\end{array}\right]\)
03

Check if \(AB\) or \(BA\) equals the Identity matrix

The product \(AB\) gives the Identity matrix but the product \(BA\) does not, hence matrix 'B' is not the multiplicative inverse of 'A'

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Most popular questions from this chapter

Use a graphing utility to find the multiplicative inverse of each matrix. Check that the displayed inverse is correct. $$\left[\begin{array}{rrr}-2 & 1 & -1 \\\\-5 & 2 & -1 \\\3 & -1 & 1\end{array}\right]$$

Solve: \(3^{2 x-8}=27 .\) (Section 3.4, Example 1)

Prove the following statement: If \(A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right], a \neq 0, b \neq 0, c \neq 0\) then \(A^{-1}=\left[\begin{array}{ccc}\frac{1}{a} & 0 & 0 \\ 0 & \frac{1}{b} & 0 \\ 0 & 0 & \frac{1}{c}\end{array}\right]\)

Find \(A^{-1}\) by forming \([A | I]\) and then using row operations to obtain \([I | B],\) where \(A^{-1}=[B] .\) Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $$A=\left[\begin{array}{lll} 2 & 4 & -4 \\ 1 & 3 & -4 \\ 2 & 4 & -3 \end{array}\right]$$

The figure shows the letter \(L\) in a rectangular coordinate system. GRAPH CAN'T COPY. The figure can be represented by the matrix $$B=\left[\begin{array}{llllll}0 & 3 & 3 & 1 & 1 & 0 \\\0 & 0 & 1 & 1 & 5 & 5\end{array}\right]$$ Each column in the matrix describes a point on the letter. The order of the columns shows the direction in which a pencil must move to draw the letter. The \(L\) is completed by connecting the last point in the matrix, \((0,5),\) to the starting point, \((0,0) .\) Use these ideas to solve Exercises \(53-60\) a. If \(A=\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right],\) find \(A B\) b. Graph the object represented by matrix \(A B\). What effect does the matrix multiplication have on the letter \(L\) represented by matrix \(B ?\)
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