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Chapter 2: Q2Q (page 93)

2. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{2}}\\{\bf{7}}&{\bf{4}}\end{aligned}} \right)\).

Short Answer

Expert verified

The inverse of \(\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)\) is \(\left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\{3.5}&{ - 1.5}\end{aligned}} \right)\).

Step by step solution

01

Check if the matrix is invertible

\(\begin{aligned}{c}\det \left( {\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)} \right) = 3\left( 4 \right) - 2\left( 7 \right)\\ = 12 - 14\\\det \left( {\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)} \right) = - 2 \ne 0\end{aligned}\)

This implies that\(\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)\)is invertible.

02

Use the formula

\({\left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{aligned}{*{20}{c}}d&{ - b}\\{ - c}&a\end{aligned}} \right)\) when \(ad - bc \ne 0\).

03

Write the inverse matrix

\(\begin{aligned}{c}{\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)^{ - 1}} = \frac{1}{{ - 2}}\left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 7}&3\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\{\frac{7}{2}}&{ - \frac{3}{2}}\end{aligned}} \right)\\{\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\{3.5}&{ - 1.5}\end{aligned}} \right)\end{aligned}\)

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Most popular questions from this chapter

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

38. Use at least three pairs of random \(4 \times 4\) matrices Aand Bto test the equalities \({\left( {A + B} \right)^T} = {A^T} + {B^T}\) and \({\left( {AB} \right)^T} = {A^T}{B^T}\). (See Exercise 37.) Report your conclusions. (Note:Most matrix programs use \(A'\) for \({A^{\bf{T}}}\).

In exercise 11 and 12, mark each statement True or False.Justify each answer.

a. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}&{{A_{\bf{2}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), with \({A_{\bf{1}}}\) and \({A_{\bf{2}}}\) the same sizes as \({B_{\bf{1}}}\) and \({B_{\bf{2}}}\), respectively then \(A + B = \left[ {\begin{array}{*{20}{c}}{{A_1} + {B_1}}&{{A_{\bf{2}}} + {B_{\bf{2}}}}\end{array}} \right]\).

b. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_{\bf{2}}}}\end{array}} \right]\), then the partitions of \(A\) and \(B\) are comfortable for block multiplication.

Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).
See all solutions

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