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Chapter 2: Problem 21

a. Verify that \(A^{2}=I\) when \(A=\left[\begin{array}{rr}{1} & {0} \\ {3} & {-1}\end{array}\right]\) b. Use partitioned matrices to show that \(M^{2}=I\) when \(M=\left[\begin{array}{rrrr}{1} & {0} & {0} & {0} \\ {3} & {-1} & {0} & {0} \\\ {1} & {0} & {-1} & {0} \\ {0} & {1} & {-3} & {1}\end{array}\right]\)

Short Answer

Expert verified
Both matrices satisfy the condition: \( A^2 = I \) and \( M^2 = I \).

Step by step solution

01

Find Matrix A Squared

To verify that \( A^2 = I \), we need to find \( A^2 \) where \( A = \left[\begin{array}{rr}1 & 0 \ 3 & -1\end{array}\right] \). Multiply matrix \( A \) by itself: \[A^2 = \left[\begin{array}{rr}1 & 0 \ 3 & -1\end{array}\right] \times \left[\begin{array}{rr}1 & 0 \ 3 & -1\end{array}\right] = \left[\begin{array}{rr}1\times1 + 0\times3 & 1\times0 + 0\times(-1) \ 3\times1 + (-1)\times3 & 3\times0 + (-1)\times(-1) \end{array}\right]\]Calculating the elements:1. First row, first column: \(1\times1 + 0\times3 = 1\)2. First row, second column: \(1\times0 + 0\times(-1) = 0\)3. Second row, first column: \(3\times1 + (-1)\times3 = 3 - 3 = 0\)4. Second row, second column: \(3\times0 + (-1)\times(-1) = 1\)So, \( A^2 = \left[\begin{array}{rr}1 & 0 \ 0 & 1\end{array}\right] = I \).
02

Partition Matrix M

To verify that \( M^2 = I \) for the matrix \( M = \left[\begin{array}{rrrr}1 & 0 & 0 & 0 \ 3 & -1 & 0 & 0 \ 1 & 0 & -1 & 0 \ 0 & 1 & -3 & 1\end{array}\right] \), we divide \( M \) into partitions. Consider \( M \) as a block matrix:\[ M = \left[\begin{array}{cc}A & B \ C & D\end{array}\right] = \left[\begin{array}{cc}\left[\begin{array}{rr}1 & 0 \ 3 & -1 \end{array}\right] & \left[\begin{array}{rr}0 & 0 \ 0 & 0 \end{array}\right] \ \left[\begin{array}{rr}1 & 0 \ 0 & 1 \end{array}\right] & \left[\begin{array}{rr}-1 & 0 \ -3 & 1 \end{array}\right] \end{array}\right]. \]
03

Verify M Squared Using Partitioned Matrices

Calculate the product \( M^2 \). Using the fact that the product of block matrices:\[ M^2 = \left[\begin{array}{cc}A & B \ C & D\end{array}\right] \cdot \left[\begin{array}{cc}A & B \ C & D\end{array}\right] = \left[\begin{array}{cc}AA + BC & AB + BD \ CA + DC & CB + DD \end{array}\right] \].Substitute the blocks from step 2:1. \( AA + BC = \left[\begin{array}{rr}1 & 0 \ 3 & -1 \end{array}\right] \cdot \left[\begin{array}{rr}1 & 0 \ 3 & -1 \end{array}\right] + \left[\begin{array}{rr}0 & 0 \ 0 & 0 \end{array}\right] \cdot \left[\begin{array}{rr}1 & 0 \ 0 & 1 \end{array}\right] = I \) - already calculated.2. \( AB + BD = O + BD = BD = \left[\begin{array}{rr}0 & 0 \ 0 & 0 \end{array}\right] \cdot \left[\begin{array}{rr}-1 & 0 \ -3 & 1 \end{array}\right] = O \).3. \( CA + DC = \left[\begin{array}{rr}1 & 0 \ 0 & 1 \end{array}\right] \cdot \left[\begin{array}{rr}1 & 0 \ 3 & -1 \end{array}\right] + \left[\begin{array}{rr}-1 & 0 \ -3 & 1 \end{array}\right] \cdot \left[\begin{array}{rr}1 & 0 \ 0 & 1 \end{array}\right] = O \).4. \( CB + DD = O + DD = \left[\begin{array}{rr}-1 & 0 \ -3 & 1 \end{array}\right] \cdot \left[\begin{array}{rr}-1 & 0 \ -3 & 1 \end{array}\right] = I \) - calculated similarly.The resulting matrix \( M^2 = I \).
04

Final Verification

The calculations confirm that both matrices \( A^2 \) and \( M^2 \) indeed equal the identity matrix \( I \). This verifies that for both cases \( A \) and \( M \), the squares of the matrices are identity matrices, thus \( A^2 = I \) and \( M^2 = I \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Identity Matrix
An identity matrix is a special kind of matrix that plays a crucial role in matrix operations, especially in determining matrix inverses. It acts like the number 1 in multiplication for real numbers. When you multiply any matrix by the identity matrix, you get the original matrix back. This is why it's called "identity." For an identity matrix, all the diagonal elements are 1, and all other elements are 0. For example, a 2x2 identity matrix looks like this:
  • \[ I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]
No matter how you multiply any compatible matrix with the identity matrix, the result remains the same as the original matrix. This property is what helps us verify if an inverse matrix is correct. If multiplying a matrix by another matrix results in the identity matrix, the latter is the inverse of the former.
Block Matrices
Block matrices are a neat way to organize larger matrices by grouping elements into smaller matrix 'blocks' within the original matrix. This makes operations such as addition, multiplication, and especially solving matrix equations more manageable. In the exercise, a matrix is partitioned into blocks to verify it forms an identity matrix when squared. Think of a block matrix like a big chocolate bar, with each section filled with smaller blocks of chocolate. This structure allows the leveraging of smaller, simpler calculations when dealing with large matrices. In other words, a complex 4x4 matrix can be divided into four 2x2 sub-matrices:
  • The top-left corner could be named 'A'.
  • The top-right could be 'B'.
  • The bottom-left 'C'.
  • The bottom-right, 'D'.
For matrices that are invertible, utilizing block matrices simplifies the computation needed for operations such as finding an inverse or performing multiplication.
Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra and is used to solve systems of equations, transform geometric vectors, and model various physical phenomena. Unlike regular multiplication, matrix multiplication is not commutative, meaning the order matters. To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix.The resulting matrix has dimensions that are specified by the rows of the first and columns of the second matrix. To compute a single element of the resulting matrix, you perform a dot product of the corresponding row from the first matrix and the column of the second matrix. For example:\[ A = \begin{bmatrix}1 & 2 \ 3 & 4 \end{bmatrix}, B = \begin{bmatrix}5 & 6 \ 7 & 8 \end{bmatrix} \]The product \(C = A \times B\) is computed as follows:
  • First element: \(1 \times 5 + 2 \times 7 = 19\)
  • Second element: \(1 \times 6 + 2 \times 8 = 22\)
  • Third element: \(3 \times 5 + 4 \times 7 = 43\)
  • Fourth element: \(3 \times 6 + 4 \times 8 = 50\)
Thus, the product matrix C is:\[ C = \begin{bmatrix} 19 & 22 \ 43 & 50 \end{bmatrix} \]This principle is applied extensively in solving problems about inverses and transformations in linear algebra.

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Most popular questions from this chapter

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations. [M] Use at least three pairs of random \(4 \times 4\) matrices \(A\) and \(B\) to test the equalities \((A+B)^{T}=A^{T}+B^{T}\) and \((A B)^{T}=\) \(A^{T} B^{T}\) . (See Exercise \(37 . )\) Report your conclusions. [Note: Most matrix programs use \(A^{\prime}\) for \(A^{T} . ]\)

\([\mathbf{M}]\) Let \(H=\operatorname{Span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) and \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\} .\) Show that \(\mathbf{x}\) is in \(H,\) and find the \(\mathcal{B}\) -coordinate vector of \(\mathbf{x},\) when $$ \mathbf{v}_{1}=\left[\begin{array}{c}{11} \\ {-5} \\ {10} \\\ {7}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{c}{14} \\ {-8} \\\ {13} \\ {10}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{19} \\\ {-13} \\ {18} \\ {15}\end{array}\right] $$

In Exercises 19鈥�24, justify each answer or construction. If the subspace of all solutions of \(A \mathbf{x}=\mathbf{0}\) has a basis consisting of three vectors and if \(A\) is a \(5 \times 7\) matrix, what is the rank of \(A ?\)

Give the \(4 \times 4\) matrix that rotates points in \(\mathbb{R}^{3}\) about the \(z\) -axis through an angle of \(-30^{\circ},\) and then translates by \(\mathbf{p}=(5,-2,1)\)

In Exercises 13 and 14, find a basis for the subspace spanned by the given vectors. What is the dimension of the subspace? $$ \left[\begin{array}{r}{1} \\ {-1} \\ {-2} \\\ {5}\end{array}\right],\left[\begin{array}{r}{2} \\ {-3} \\ {-1} \\\ {6}\end{array}\right],\left[\begin{array}{r}{0} \\ {2} \\ {-6} \\\ {8}\end{array}\right],\left[\begin{array}{r}{-1} \\ {4} \\ {-7} \\\ {7}\end{array}\right],\left[\begin{array}{r}{3} \\ {-8} \\ {9} \\\ {-5}\end{array}\right] $$
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