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Chapter 2: Q33Q (page 93)

Use the algorithm from this section to find the inverse of

\(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}\end{aligned}} \right)\)and \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}\end{aligned}} \right)\)

Let \(A\) be the corresponding \(n \times n\) matrix, let \(B\) be its inverse. Guess the form of \(B\) and then prove that \(AB = I\) and \(BA = I\).

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}1&0&0\\{ - 1}&1&0\\0&{ - 1}&1\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\0&{ - 1}&1&0\\0&0&{ - 1}&1\end{aligned}} \right)\), and \(AB = BA = I\)

Step by step solution

01

Find the expression \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\) for the matrix of \({\bf{3}} \times {\bf{3}}\)

Form the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\1&1&0\\1&1&1\end{aligned}\,\,\,\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right)\)

At row three, subtract row one from row three, i.e., \({R_3} \to {R_3} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\1&1&0\\0&1&1\end{aligned}\,\,\,\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\{ - 1}&0&1\end{aligned}} \right)\)

At row two, subtract row one from row two, i.e., \({R_2} \to {R_2} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&1&1\end{aligned}\,\,\,\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\{ - 1}&1&0\\{ - 1}&0&1\end{aligned}} \right)\)

02

Apply row operation to \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

At row three, subtract row two from row three, i.e., \({R_3} \to {R_3} - {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}\,\,\,\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\{ - 1}&1&0\\0&{ - 1}&1\end{aligned}} \right)\)

So, the inverse matrix is \(\left( {\begin{aligned}{*{20}{c}}1&0&0\\{ - 1}&1&0\\0&{ - 1}&1\end{aligned}} \right)\).

03

Find the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\) for the matrix of \({\bf{4}} \times {\bf{4}}\)

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\1&1&0&0\\1&1&1&0\\1&1&1&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{aligned}} \right)\)

At row four, subtract row one from row four, i.e., \({R_4} \to {R_4} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\1&1&0&0\\1&1&1&0\\0&1&1&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\{ - 1}&0&0&1\end{aligned}} \right)\)

At row three, subtract row one from row three, i.e., \({R_3} \to {R_3} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\1&1&0&0\\0&1&1&0\\0&1&1&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\{ - 1}&0&1&0\\{ - 1}&0&0&1\end{aligned}} \right)\)

At row two, subtract row one from row two, i.e., \({R_2} \to {R_2} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&1&1&0\\0&1&1&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\{ - 1}&0&1&0\\{ - 1}&0&0&1\end{aligned}} \right)\)

04

Apply row operations to \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

At row four, subtract row three from row four, i.e., \({R_4} \to {R_4} - {R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&1&1&0\\0&0&0&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\{ - 1}&0&1&0\\0&0&{ - 1}&1\end{aligned}} \right)\)

At row three, subtract row two from row three, i.e., \({R_3} \to {R_3} - {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\0&{ - 1}&1&0\\0&0&{ - 1}&1\end{aligned}} \right)\)

So, the inverse of the matrix is \(\left( {\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\0&{ - 1}&1&0\\0&0&{ - 1}&1\end{aligned}} \right)\).

05

Guess the matrix \(B\)

The inverse matrix of \(A\) is \(B\). Matrix \(B\) of the order \(n \times n\) can be expressed as

\(B = \left( {\begin{aligned}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&{}\\ \vdots &{}& \ddots & \ddots & \vdots \\0&0& \cdots &{ - 1}&1\end{aligned}} \right)\).

For \(j = 1,...,n\), let \({{\bf{a}}_j}\), \({{\bf{b}}_j}\) and \({{\bf{e}}_j}\) denote the \(j\)th columns of \(A\), \(B\) and \(I\), respectively.

For \(j = 1,...,n - 1\),

\({{\bf{a}}_j} - {{\bf{a}}_{j + 1}} = {{\bf{e}}_j}\),

\({{\bf{b}}_j} = {{\bf{e}}_j} - {{\bf{e}}_{j + 1}}\), and

\({{\bf{a}}_n} = {{\bf{b}}_n} = {{\bf{e}}_n}\).

06

Find the value of \(A{{\bf{b}}_j}\)

\(\begin{aligned}{c}A{b_j} = A\left( {{{\bf{e}}_j} - {{\bf{e}}_{j + 1}}} \right)\\ = A{{\bf{e}}_j} - A{{\bf{e}}_{j + 1}}\\ = {{\bf{a}}_j} - {{\bf{a}}_{j + 1}}\\ = {{\bf{e}}_j}\end{aligned}\)

So, \(AB = I\).

07

Find the value of \(B{{\bf{a}}_j}\)

As \({a_n} = {b_n} = {e_n}\),

\(\begin{aligned}{c}B{{\bf{a}}_j} = B\left( {{{\bf{e}}_j} + ..... + {{\bf{e}}_n}} \right)\\ = {{\bf{b}}_j} + ..... + {{\bf{b}}_n}\\ = \left( {{{\bf{e}}_j} - {{\bf{e}}_{j + 1}}} \right) + \left( {{{\bf{e}}_{j + 1}} - {{\bf{e}}_{j + 2}}} \right) + ..... + \left( {{{\bf{e}}_{n - 1}} - {{\bf{e}}_n}} \right) + {{\bf{e}}_n}\\ = {{\bf{e}}_j}.\end{aligned}\)

So, \(BA = I\).

The inverse of the given matrices are \(\left( {\begin{aligned}{*{20}{c}}1&0&0\\{ - 1}&1&0\\0&{ - 1}&1\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\0&{ - 1}&1&0\\0&0&{ - 1}&1\end{aligned}} \right)\), and \(AB = BA = I\).

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Most popular questions from this chapter

Suppose Ais an \(n \times n\) matrix with the property that the equation \(Ax = 0\)has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation \(Ax = b\) must have a solution for each b in \({\mathbb{R}^n}\).

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5鈥�8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

5. \[\left[ {\begin{array}{*{20}{c}}A&B\\C&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&Y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\Z&{\bf{0}}\end{array}} \right]\]

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

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.

36. Write the command(s) that will create a \(6 \times 4\) matrix with random entries. In what range of numbers do the entries lie? Tell how to create a \(3 \times 3\) matrix with random integer entries between \( - {\bf{9}}\) and 9. (Hint:If xis a random number such that 0 < x < 1, then \( - 9.5 < 19\left( {x - .5} \right) < 9.5\).
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