/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 The matrices \(A, B, C,\) and \(... [FREE SOLUTION] | 91Ó°ÊÓ

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The matrices \(A, B, C,\) and \(D\) are defined as follows. $$\begin{aligned}&A=\left(\begin{array}{rr}4 & -1 \\\\-5 & 2\end{array}\right) \quad B=\left(\begin{array}{ll}\frac{1}{2} & 5 \\\3 & 1\end{array}\right)\\\&C=\left(\begin{array}{rrr}3 & 0 & -2 \\\0 & 5 & 6 \\\1 & 4 & -7 \end{array}\right) \quad D=\left(\begin{array}{lll}1 & 2 & 3 \\\4 & 5 & 6 \\\7 & 8 & 9\end{array}\right) \end{aligned}$$ Compute \(A I_{2}\) and \(I_{2} A\) to verify that \(A I_{2}=I_{2} A=A\)

Short Answer

Expert verified
Multiplying matrix \(A\) by the identity matrix \(I_2\) results in \(A\), confirming \(A I_2 = I_2 A = A\).

Step by step solution

01

Identify Identity Matrix

The identity matrix of order 2, denoted by \(I_2\), is a 2x2 matrix given by \(I_2 = \left( \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right)\). This matrix acts as the multiplicative identity for 2x2 matrices.
02

Compute \( A I_2 \)

To compute \(A I_2\), multiply matrix \(A\) by the identity matrix \(I_2\).\[A = \begin{pmatrix} 4 & -1 \ -5 & 2 \end{pmatrix}, I_2 = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \A I_2 = \begin{pmatrix} 4 & -1 \ -5 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} (4)(1) + (-1)(0) & (4)(0) + (-1)(1) \ (-5)(1) + (2)(0) & (-5)(0) + (2)(1) \end{pmatrix}\]\[A I_2 = \begin{pmatrix} 4 & -1 \ -5 & 2 \end{pmatrix}\]
03

Compute \( I_2 A \)

To compute \(I_2 A\), multiply the identity matrix \(I_2\) by matrix \(A\).\[I_2 = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}, A = \begin{pmatrix} 4 & -1 \ -5 & 2 \end{pmatrix} \I_2 A = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 4 & -1 \ -5 & 2 \end{pmatrix} = \begin{pmatrix} (1)(4) + (0)(-5) & (1)(-1) + (0)(2) \ (0)(4) + (1)(-5) & (0)(-1) + (1)(2) \end{pmatrix} \]\[I_2 A = \begin{pmatrix} 4 & -1 \ -5 & 2 \end{pmatrix}\]
04

Verify the Equality

Both \(A I_2\) and \(I_2 A\) result in \(A\). Thus, \(A I_2 = I_2 A = A\). This verifies that multiplying any matrix by the identity matrix of the same order results in the original matrix.

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Key Concepts

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

Identity Matrix
The identity matrix is like the number 1 for matrix multiplication. When you multiply any matrix by an identity matrix, the original matrix remains unchanged, just like how any number multiplied by 1 remains the same. This fascinating characteristic makes the identity matrix fundamental in matrix operations. For a 2x2 matrix, the identity matrix is given by:
  • \[ I_2 = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \]
You can think of the identity matrix as a mirror reflecting the original matrix's properties. It's called the multiplicative identity because it keeps the identity of the matrix it's multiplying. In our exercise, we saw that multiplying matrix \(A\) with \(I_2\) resulted in matrix \(A\) itself. This is a key property used in many applications such as solving linear equations or transforming coordinate systems.
2x2 Matrices
In the world of linear algebra, 2x2 matrices are the simplest form of square matrices, which have both rows and columns equal in number. A matrix like:
  • \[ \begin{pmatrix} a & b \ c & d \end{pmatrix} \]
is a 2x2 matrix with elements \(a, b, c,\) and \(d\). These matrices are widely used for various calculations because they are easy to handle yet capable of representing complex operations. When learning about matrix algebra, starting with 2x2 matrices is beneficial. They introduce you to key concepts without the complication of larger dimensions. Their simplicity allows for easier visualization and comprehension of matrix operations, such as addition, scalar multiplication, and especially matrix multiplication as demonstrated in our exercise.
Matrix Operations
Matrix operations are essential in performing calculations within linear algebra. They include operations like addition, subtraction, scalar multiplication, and matrix multiplication. Among these, matrix multiplication is crucial because it combines information from two matrices to produce a new matrix, often representing a transformation or a combined set of data.

To multiply two matrices, remember to follow the row-by-column rule. For a matrix \(A\) and matrix \(B\), each element in the resulting matrix is the sum of products from the corresponding row in \(A\) and column in \(B\). The operation might seem challenging due to the series of computations involved, but breaking it down step-by-step, as with our example, aids in understanding. Matrix operations, and especially multiplication, form the backbone of more advanced topics like eigenvectors and determinants, making them indispensable in fields such as physics, engineering, and computer science. By mastering these basic operations, you're building a foundation for exploring these more complex areas.

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Most popular questions from this chapter

So the input (3,5) yields an output of \(\sqrt{2} .\) We define the do- main for this function just as we did in Chapter 3: The domain is the set of all inputs that yield real-number outputs. For instance, the ordered pair (1,4) is not in the domain of the function we have been discussing, because (as you should check for yourself\() f(1,4)=\sqrt{-1},\) which is not a real number. We can determine the domain of the function in equation ( 1 ) by requiring that the quantity under the radical sign be non negative. Thus we require that \(2 x-y+1 \geq 0\) and, consequently, \(y \leq 2 x+1\) (Check this.) The following figure shows the graph of this inequality; the domain of our function is the set of ordered pairs making up the graph. In Exercises follow a similar procedure and sketch the domain of the given function. (Graph cant copy) $$h(x, y)=\sqrt{x}+\sqrt{y}$$

In this exercise, let's agree to write the coordinates \((x, y)\) of a point in the plane as the \(2 \times 1\) matrix \(\left(\begin{array}{l}x \\\ y\end{array}\right) .\) (a) Let \(A=\left(\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right)\) and \(Z=\left(\begin{array}{l}x \\ y\end{array}\right) .\) Compute the ma- trix \(A Z .\) After computing \(A Z,\) observe that it represents the point obtained by reflecting \(\left(\begin{array}{l}x \\ y\end{array}\right)\) about the \(x\) -axis. (b) Let \(B=\left(\begin{array}{rr}-1 & 0 \\ 0 & 1\end{array}\right)\) and \(Z=\left(\begin{array}{l}x \\ y\end{array}\right) .\) Compute the ma- trix \(B Z\). After computing \(B Z\), observe that it represents the point obtained by reflecting \(\left(\begin{array}{l}x \\ y\end{array}\right)\) about the \(y\) -axis. (c) Let \(A, B,\) and \(Z\) represent the matrices defined in parts \((a)\) and \((b) .\) Compute the matrix \((A B) Z\) and then interpret it in terms of reflection about the axes.

(a) Compute the inverse of the coefficient matrix for the system. (b) Use the inverse matrix to solve the system. In cases in which the final answer involves decimals, round to three decimal places. $$\left\\{\begin{array}{c} 2 x+3 y+z+w=3 \\ 6 x+6 y-5 z-2 w=15 \\ x-y+z+\frac{1}{6} w=-3 \\ 4 x+9 y+3 z+2 w=-3 \end{array}\right.$$

Consider the following system: $$ \left\\{\begin{array}{l} x+y+3 z=0 \\ x+2 y+5 z=0 \\ x-4 y-8 z=0 \end{array}\right. $$ (a) Without doing any calculations, find one obvious solution of this system. (b) Calculate the determinant \(D\) (c) List all solutions of this system.

Solve the following system for \(x, y,\) and \(z:\) $$\left\\{\begin{array}{l} a x+b y+c z=k \\ a^{2} x+b^{2} y+c^{2} z=k^{2} \\ a^{3} x+b^{3} y+c^{3} z=k^{3} \end{array}\right.$$

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