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Chapter 2: Q15Q (page 93)

Can a square matrix with two identical columns be invertible? Why or why not?

Short Answer

Expert verified

It is not invertible as the identical column vectors are linearly dependent.

Step by step solution

01

State the condition of an invertible matrix

If a matrix contains two identical columns, then the columns are linearly dependent on each other.

02

Consider the matrix

For the matrix \(\left[ {\begin{aligned}{*{20}{c}}{ - 2}&3&3\\1&{ - 1}&{ - 1}\\{ - 1}&{ - 2}&{ - 2}\end{aligned}} \right]\),column 2 and column 3 are identical.

The equation \({x_1}\left[ {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 2}\end{aligned}} \right] + {x_2}\left[ {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 2}\end{aligned}} \right] = 0\) has one solution. Therefore, the vectors are linearly dependent.

So, the square matrix with two identical columns cannot be inverted as the columns are linearly dependent.

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Most popular questions from this chapter

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 \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).

Suppose AB = AC, where Band Care \(n \times p\) matrices and A is invertible. Show that B = C. Is this true, in general, when A is not invertible.

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 \({\mathbb{R}^n}\).

Suppose A, B, and Care \(n \times n\) matrices with A, X, and \(A - AX\) invertible, and suppose

\({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) 鈥�(3)

  1. Explain why B is invertible.
  2. Solve (3) for X. If you need to invert a matrix, explain why that matrix is invertible.
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