Show that if ABis invertible, so is B.

Short Answer

Expert verified

Both AB and B are invertible.

Step by step solution

01

Write the algorithm for obtaining \({A^{ - 1}}\)

The inverse of an\(m \times m\)matrix A can be computed by using the augmented matrix\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\), where\(I\)is the identity matrix. Matrix Ahas an inverse only if \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\) is row equivalent to \(\left( {\begin{aligned}{*{20}{c}}I&{{A^{ - 1}}}\end{aligned}} \right)\).

02

Show that A is invertible

Product AB is invertible, so there should be an inverse of matrix AB. Let D be the inverse matrix of AB.

Then, it can be represented as shown below:

\(\begin{aligned}{c}D\left( {AB} \right) = I\\\left( {DA} \right)B = I\end{aligned}\)

The equation\(\left( {DA} \right)B = I\) shows that matrix\(DA\)is the inverse of matrix B.

Therefore, B is invertible.

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Most popular questions from this chapter

In Exercises 13 and 14, determine if \({\mathop{\rm b}\nolimits} \) is a linear combination of the vectors formed from the columns of the matrix \(A\).

14. \(A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 6}\\0&3&7\\1&{ - 2}&5\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}{11}\\{ - 5}\\9\end{array}} \right]\)

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

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. (Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)).

Question: Determine whether the statements that follow are true or false, and justify your answer.

14: rank.|111123136|=3

In Exercise 23 and 24, make each statement True or False. Justify each answer.

23.

a. Another notation for the vector \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\3\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}{ - 4}&3\end{array}} \right]\).

b. The points in the plane corresponding to \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\5\end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}}{ - 5}\\2\end{array}} \right]\) lie on a line through the origin.

c. An example of a linear combination of vectors \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) is the vector \(\frac{1}{2}{{\mathop{\rm v}\nolimits} _1}\).

d. The solution set of the linear system whose augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}&b\end{array}} \right]\) is the same as the solution set of the equation\({{\mathop{\rm x}\nolimits} _1}{a_1} + {x_2}{a_2} + {x_3}{a_3} = b\).

e. The set Span \(\left\{ {u,v} \right\}\) is always visualized as a plane through the origin.

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