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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

The inverse of \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\C&I&{\bf{0}}\\A&B&I\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\Z&I&{\bf{0}}\\X&Y&I\end{array}} \right]\). Find X, Y, and Z.

Suppose Ais an \(m \times n\) matrix and there exist \(n \times m\) matrices C and D such that \(CA = {I_n}\) and \(AD = {I_m}\). Prove that \(m = n\) and \(C = D\). (Hint: Think about the product CAD.)

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

Suppose \(AD = {I_m}\) (the \(m \times m\) identity matrix). Show that for any b in \({\mathbb{R}^m}\), the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) has a solution. (Hint: Think about the equation \(AD{\mathop{\rm b}\nolimits} = {\mathop{\rm b}\nolimits} \).) Explain why Acannot have more rows than columns.

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{9}}&{ - {\bf{1}}}&{\bf{3}}\\{ - {\bf{8}}}&{\bf{7}}&{ - {\bf{6}}}\\{ - {\bf{4}}}&{\bf{1}}&{\bf{8}}\end{aligned}} \right)\)
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