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Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 2: Q19E (page 108)

TRUE OR FALSE?
There exists an upper triangular $2 脳 2$ matrixAsuch that $A^{2} = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ .

Short Answer

Expert verified

The given statement is true.

Step by step solution

01

Compute the square of the matrix

An upper triangular matrix is a matrix with zero lower diagonal elements.

Consider the matrix.

$A = (\begin{matrix} 1 & a \\ 0 & 1 \end{matrix})$

Let鈥檚 check for the square of the matrix.

$A^{2} = (\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}) (\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}) = (\begin{matrix} 1.1 + a . 0 & 1 . a + a . 1 \\ 0.1 + 1.0 & 0 . a + 1.1 \end{matrix}) = (\begin{matrix} 1 & 2 a \\ 0 & 1 \end{matrix})$

02

Compare the matrices

Compare the above matrix with $A^{2} = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ .

$(\begin{matrix} 1 & 2 a \\ 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) 鈬� 2 a = 1 鈭� a = \frac{1}{2}$

Therefore, the matrix $(\begin{matrix} 1 & \frac{1}{2} \\ 0 & 1 \end{matrix})$ has a square, $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ .

Hence, the given statement is true.

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