Suppose Ais \(n \times n\) and the equation \(A{\bf{x}} = {\bf{b}}\) has a solution for each b in \({\mathbb{R}^{\bf{n}}}\). Explain why Amust be invertible. (Hint:Is Arow equivalent to \({I_n}\)?)

Short Answer

Expert verified

Matrix A is invertible because each row contains pivots.

Step by step solution

01

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

The inverse of an\(m \times m\)matrix A can be computed using theaugmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\), where\(I\)is theidentity 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

Explain the invertible and matrix span \({\mathbb{R}^{\bf{n}}}\)

It is given that each value of vector bin the equation\(A{\bf{x}} = {\bf{b}}\)has aunique solution. It means the columns of the\(n \times n\)matrix Amust span\({\mathbb{R}^{\bf{n}}}\). Also, there should be pivots in each row.

The diagonal entries in the matrix are non-zero. So, matrix A isrow equivalent to the identity matrix.

Thus, matrix Amust beinvertible.

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Most popular questions from this chapter

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

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 \(U\left( {\mathop{\rm v}\nolimits} \right)\)].

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{9}}&{ - {\bf{1}}}&{\bf{3}}\\{ - {\bf{8}}}&{\bf{7}}&{ - {\bf{6}}}\\{ - {\bf{4}}}&{\bf{1}}&{\bf{8}}\end{aligned}} \right)\)

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.

1. \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\)

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

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