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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

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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

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

16. a. If A and B are \({\bf{3}} \times {\bf{3}}\) and \(B = \left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}\end{aligned}} \right)\), then \(AB = \left( {A{{\bf{b}}_1} + A{{\bf{b}}_2} + A{{\bf{b}}_3}} \right)\).

b. The second row of ABis the second row of Amultiplied on the right by B.

c. \(\left( {AB} \right)C = \left( {AC} \right)B\)

d. \({\left( {AB} \right)^T} = {A^T}{B^T}\)

e. The transpose of a sum of matrices equals the sum of their transposes.

Suppose P is invertible and \(A = PB{P^{ - 1}}\). Solve for Bin terms of A.

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

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