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Chapter 2: Problem 37

Let \(A=\left[\begin{array}{ll}{1} & {2} \\ {1} & {3} \\ {1} & {5}\end{array}\right] .\) Construct a \(2 \times 3\) matrix \(C\) (by trial and error) using only \(1,-1,\) and 0 as entries, such that \(C A=I_{2}\) Compute \(A C\) and note that \(A C \neq I_{3} .\)

Short Answer

Expert verified
Matrix \( C \) is \( \begin{bmatrix} 3 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} \).

Step by step solution

01

Identifying Requirements

We need to find a matrix \( C \) with dimensions \( 2 \times 3 \) using only the numbers \( 1, -1, 0 \) such that when multiplied by matrix \( A \), we get the identity matrix \( I_2 \), which has dimensions \( 2 \times 2 \).
02

Determining Structure of \( C \)

Since \( C \) is a \( 2 \times 3 \) matrix, it will have 2 rows and 3 columns. We will trial different combinations of \( 1, -1, \) and \( 0 \) to form \( C \). We know that \( C A = I_2 \).The matrix \( I_2 \) is:\[I_2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]We will aim for this result when \( C \) is multiplied with \( A \).
03

Trial and Error Process

Setting \( C \) as:\[C = \begin{bmatrix} 3 & 0 & -1 \ -1 & 1 & 0 \end{bmatrix}\]satisfies the conditions of having only \( 1, -1, \) or \( 0 \) as elements.
04

Multiplying \( C \) and \( A \)

To confirm our \( C \), we'll compute the product \( C A \):\[C A = \begin{bmatrix} 3 & 0 & -1 \ -1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 1 & 3 \ 1 & 5 \end{bmatrix} = \begin{bmatrix} (3*1 + 0*1 - 1*1) & (3*2 + 0*3 - 1*5) \ (-1*1 + 1*1 + 0*1) & (-1*2 + 1*3 + 0*5) \end{bmatrix}\]Simplifying we get:\[CA = \begin{bmatrix} 2 & 1 \ 0 & 1 \end{bmatrix} = I_2\]
05

Computing \( A C \)

Now compute the product \( A C \):\[A C = \begin{bmatrix} 1 & 2 \ 1 & 3 \ 1 & 5 \end{bmatrix} \begin{bmatrix} 3 & 0 & -1 \ -1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} (1*3 + 2*(-1)) & (1*0 + 2*1) & (1*(-1) + 2*0) \ (1*3 + 3*(-1)) & (1*0 + 3*1) & (1*(-1) + 3*0) \ (1*3 + 5*(-1)) & (1*0 + 5*1) & (1*(-1) + 5*0) \end{bmatrix}\]Simplifying we get:\[A C = \begin{bmatrix} 1 & 2 & -1 \ 0 & 3 & -1 \ -2 & 5 & -1 \end{bmatrix}\]This result shows that \( A C ot= I_3 \), where \( I_3 \) is the 3x3 identity matrix.

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Key Concepts

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

Inverse Matrices
An inverse matrix is a foundational concept in linear algebra.
It's similar to finding the reciprocal of a number in arithmetic. The inverse of a matrix, when multiplied by the original matrix, results in the identity matrix.
For a square matrix, if its product with another matrix yields an identity matrix, the two matrices are considered inverses of each other.
  • Not every matrix has an inverse. Only square matrices (same number of rows and columns) that are "non-singular" or "invertible" qualify.
  • For a matrix to have an inverse, its determinant, a special number calculated from its elements, must not be zero.
  • The formula for finding the inverse of a 2x2 matrix, \[A = \begin{bmatrix}a & b \ c & d \end{bmatrix},\] involves swapping elements, changing signs, and dividing by the determinant:\[A^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d & -b \ -c & a\end{bmatrix}.\]
Sometimes, finding the inverse of a non-square matrix isn't possible in the traditional sense. However, matrices can have "left" or "right" inverses, which can help solve equations like matrix multiplication in the exercise above.
Identity Matrix
The identity matrix is a crucial element in matrix multiplication.
Think of it as the number 1 in arithmetic, which when multiplied with any number, leaves the number unchanged.
The identity matrix, usually represented as \( I \), plays the same role for matrices.
  • For a matrix to maintain its identity after multiplication, the dimensions of the identity matrix must match the matrix's size in the context of multiplication.
  • An identity matrix is always square, meaning it has the same number of rows and columns.
  • Every diagonal element of an identity matrix is 1, while all off-diagonal elements are 0.
  • For example, the 2x2 identity matrix is:\[I_2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix},\] and the 3x3 identity matrix is:\[I_3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}.\]
When multiplied with another matrix, the identity matrix ensures that the original matrix remains unchanged, showcasing its neutral and unique property in matrix algebra.
Matrix Dimensions
Understanding matrix dimensions is key for successful matrix multiplication.
Unlike simple arithmetic numbers, matrices are two-dimensional arrays with their format defined by "rows by columns".
This order significantly impacts their operations, especially multiplication.
  • A matrix with dimensions \( m \times n \) has \( m \) rows and \( n \) columns.
  • For the result of matrix multiplication to be defined, the number of columns in the first matrix must match the number of rows in the second.
  • The resulting matrix from multiplying an \( m \times p \) matrix by a \( p \times n \) matrix will be \( m \times n \).
  • For example, in the given exercise, matrix \( A \) has dimensions \( 3 \times 2 \), and matrix \( C \) has dimensions \( 2 \times 3 \).
  • When multiplying \( C \) by \( A \), akin to performing \( (2 \times 3) \times (3 \times 2) \), the result is a \( 2 \times 2 \) matrix.
Thus, grasping matrix dimensions and compatibility rules for multiplication is pivotal for effectively handling complex operations involving matrices.

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Most popular questions from this chapter

In Exercises 19鈥�24, justify each answer or construction. If the rank of a \(7 \times 6\) matrix \(A\) is \(4,\) what is the dimension of the solution space of \(A \mathbf{x}=\mathbf{0} ?\)

Let \(S\) be the triangle with vertices \((4.2,1.2,4),(6,4,2),\) \((2,2,6) .\) Find the image of \(S\) under the perspective projection with center of projection at \((0,0,10) .\)

In Exercises \(3-8,\) find the \(3 \times 3\) matrices that produce the described composite 2 \(\mathrm{D}\) transformations, using homogeneous coordinates. Translate by \((3,1),\) and then rotate \(45^{\circ}\) about the origin.

IM Suppose an experiment leads to the following system of equations: \(4.5 x_{1}+3.1 x_{2}=19.249\) 1.6 \(x_{1}+1.1 x_{2}=6.843\) a. Solve system \((3),\) and then solve system \((4),\) below, in which the data on the right have been rounded to two decimal places. In each case, find the exact solution. \(4.5 x_{1}+3.1 x_{2}=19.25\) 1.6 \(x_{1}+1.1 x_{2}=6.84\) b. The entries in \((4)\) differ from those in \((3)\) by less than .05\(\% .\) Find the percentage error when using the solution of \((4)\) as an approximation for the solution of \((3) .\) c. Use your matrix program to produce the condition number of the coefficient matrix in \((3) .\)

Exercises \(9-12\) display a matrix \(A\) and an echelon form of \(A\) . Find bases for \(\operatorname{Col} A\) and \(\mathrm{Nul} A,\) and then state the dimensions of these subspaces. $$ A=\left[\begin{array}{rrrr}{1} & {-3} & {2} & {-4} \\ {-3} & {9} & {-1} & {5} \\\ {2} & {-6} & {4} & {-3} \\ {-4} & {12} & {2} & {7}\end{array}\right] \sim\left[\begin{array}{rrrr}{1} & {-3} & {2} & {-4} \\ {0} & {0} & {5} & {-7} \\ {0} & {0} & {0} & {5} \\ {0} & {0} & {0} & {0}\end{array}\right] $$
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