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Chapter 2: Q14Q (page 93)

Suppose \(\left( {B - C} \right)D = 0\), where Band Care \(m \times n\) matrices and \(D\) is invertible. Show that B = C.

Short Answer

Expert verified

It is proved that \(B = C\).

Step by step solution

01

Condition for an invertible matrix

If Ais an invertible \(n \times n\) matrix, for each b in \({\mathbb{R}^n}\), the equation \(Ax = b\) has the unique solution \(x = {A^{ - 1}}b\).

02

Show that B = C

Multiply both sides of the equation \(\left( {B - C} \right)D = 0\) by \({D^{ - 1}}\).

\(\begin{aligned}{c}\left( {B - C} \right)D{D^{ - 1}} = 0{D^{ - 1}}\\\left( {B - C} \right)I = 0\\B - C = 0\\B = C\end{aligned}\)

Hence, it is proved that \(B = C\).

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Most popular questions from this chapter

Suppose Ais an \(n \times n\) matrix with the property that the equation \[A{\mathop{\rm x}\nolimits} = 0\] has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.

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 \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

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

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

Explain why the columns of an \(n \times n\) matrix Aspan \({\mathbb{R}^{\bf{n}}}\) when

Ais invertible. (Hint:Review Theorem 4 in Section 1.4.)

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)
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