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Chapter 7: Problem 4

Let \(\mathbf{A}=\left[\begin{array}{rr}1 & -2 \\ 3 & 5 \\ 0 & 4\end{array}\right]\) and \(\quad \mathbf{B}=\left[\begin{array}{rr}0 & -1 \\ 2 & 7 \\ 1 & 6\end{array}\right]\) (1) Find (a) \(2 \mathrm{~A}\) (b) \(2 \mathrm{~B}\) (c) \(\mathbf{A}+\mathbf{B}\) (d) \(2(\mathbf{A}+\mathbf{B})\) Hence verify that $$ 2(\mathbf{A}+\mathbf{B})=2 \mathbf{A}+2 \mathbf{B} $$ (2) Find (a) \(3 \mathrm{~A}\) (b) \(-6 \mathbf{A}\) Hence verify that $$ -2(3 \mathbf{A})=-6 \mathbf{A} $$

Short Answer

Expert verified
2(A + B) = 2A + 2B and -2(3A) = -6A

Step by step solution

01

Finding 2A

Multiply each element of matrix A by 2.\[ \mathbf{A} = \left[ \begin{array}{rr} 1 & -2 \ 3 & 5 \ 0 & 4 \end{array} \right] \] So,\[ 2\mathbf{A} = 2 \times \left[ \begin{array}{rr} 1 & -2 \ 3 & 5 \ 0 & 4 \end{array} \right] = \left[ \begin{array}{rr} 2 & -4 \ 6 & 10 \ 0 & 8 \end{array} \right] \]
02

Finding 2B

Multiply each element of matrix B by 2.\[ \mathbf{B} = \left[ \begin{array}{rr} 0 & -1 \ 2 & 7 \ 1 & 6 \end{array} \right] \] So,\[ 2\mathbf{B} = 2 \times \left[ \begin{array}{rr} 0 & -1 \ 2 & 7 \ 1 & 6 \end{array} \right] = \left[ \begin{array}{rr} 0 & -2 \ 4 & 14 \ 2 & 12 \end{array} \right] \]
03

Addition of A and B

Add corresponding elements of matrices A and B.\[ \mathbf{A} + \mathbf{B} = \left[ \begin{array}{rr} 1 & -2 \ 3 & 5 \ 0 & 4 \end{array} \right] + \left[ \begin{array}{rr} 0 & -1 \ 2 & 7 \ 1 & 6 \end{array} \right] = \left[ \begin{array}{rr} 1+0 & -2-1 \ 3+2 & 5+7 \ 0+1 & 4+6 \end{array} \right] = \left[ \begin{array}{rr} 1 & -3 \ 5 & 12 \ 1 & 10 \end{array} \right] \]
04

Finding 2(A + B)

Take the result from the previous addition of A and B, and multiply each element by 2.\[ 2 (\mathbf{A} + \mathbf{B}) = 2 \times \left[ \begin{array}{rr} 1 & -3 \ 5 & 12 \ 1 & 10 \end{array} \right] = \left[ \begin{array}{rr} 2 & -6 \ 10 & 24 \ 2 & 20 \end{array} \right] \]
05

Verification of 2(A + B) = 2A + 2B

From the steps above,\[ 2 (\mathbf{A} + \mathbf{B}) = \left[ \begin{array}{rr} 2 & -6 \ 10 & 24 \ 2 & 20 \end{array} \right] \] and\[ 2\mathbf{A} + 2\mathbf{B} = \left[ \begin{array}{rr} 2 & -4 \ 6 & 10 \ 0 & 8 \end{array} \right] + \left[ \begin{array}{rr} 0 & -2 \ 4 & 14 \ 2 & 12 \end{array} \right] = \left[ \begin{array}{rr} 2 & -6 \ 10 & 24 \ 2 & 20 \end{array} \right] \] Hence,\[ 2 (\mathbf{A} + \mathbf{B}) = 2 \mathbf{A} + 2 \mathbf{B} \]
06

Finding 3A

Multiply each element of matrix A by 3.\[ \mathbf{A} = \left[ \begin{array}{rr} 1 & -2 \ 3 & 5 \ 0 & 4 \end{array} \right] \] So,\[ 3\mathbf{A} = 3 \times \left[ \begin{array}{rr} 1 & -2 \ 3 & 5 \ 0 & 4 \end{array} \right] = \left[ \begin{array}{rr} 3 & -6 \ 9 & 15 \ 0 & 12 \end{array} \right] \]
07

Finding -6A

Multiply each element of matrix A by -6.\[ \mathbf{A} = \left[ \begin{array}{rr} 1 & -2 \ 3 & 5 \ 0 & 4 \end{array} \right] \] So,\[ -6\mathbf{A} = -6 \times \left[ \begin{array}{rr} 1 & -2 \ 3 & 5 \ 0 & 4 \end{array} \right] = \left[ \begin{array}{rr} -6 & 12 \ -18 & -30 \ 0 & -24 \end{array} \right] \]
08

Verification of -2(3A) = -6A

From the steps above,\[ -2 (3\mathbf{A}) = -2 \times \left[ \begin{array}{rr} 3 & -6 \ 9 & 15 \ 0 & 12 \end{array} \right] = \left[ \begin{array}{rr} -6 & 12 \ -18 & -30 \ 0 & -24 \end{array} \right] \] and\[ -6\mathbf{A} = \left[ \begin{array}{rr} -6 & 12 \ -18 & -30 \ 0 & -24 \end{array} \right] \] Hence,\[ -2 (3 \mathbf{A}) = -6 \mathbf{A} \]

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Key Concepts

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

Matrix Multiplication
Matrix multiplication is a crucial operation in both mathematics and economics. It involves combining two matrices to produce a new matrix. This process is not as intuitive as addition and requires a specific formula. To multiply two matrices \( \textbf{A} \) and \( \textbf{B} \), the number of columns in \( \textbf{A} \) must be equal to the number of rows in \( \textbf{B} \). For each element in the product matrix, take the dot product of the corresponding row from \( \textbf{A} \) and column from \( \textbf{B} \). Mathematically, if \( \textbf{C} \) is the product of \( \textbf{A} \) and \( \textbf{B} \), then \( \textbf{C}[i][j] = \textbf{A}[i][k] \times \textbf{B}[k][j] \) summed over \( k \). This operation is useful in economic modeling where multiple variables are analyzed together.
Matrix Addition
Matrix addition is one of the simplest matrix operations. To add two matrices, \( \textbf{A} \) and \( \textbf{B} \), they must have the same dimensions. Each corresponding element from \( \textbf{A} \) and \( \textbf{B} \) is added together to form a new matrix. For example, if \( \textbf{A} \) and \( \textbf{B} \) both have dimensions \( m \times n \), then \( \textbf{C}[i][j] = \textbf{A}[i][j] + \textbf{B}[i][j] \) for all \( i \) and \( j \). This operation is particularly useful in economics when you want to combine different sets of data, like revenue and expenses, to get a comprehensive view.
Scalar Multiplication
Scalar multiplication involves multiplying each element of a matrix by a single number, called a scalar. For example, if you have matrix \( \textbf{A} \) and a scalar \( c \), then the result of the scalar multiplication is a new matrix where each element \( \textbf{A}[ij] \) is multiplied by \( c \). Mathematically, \( (c \times \textbf{A})[i][j] = c \times \textbf{A}[i][j] \). This is useful in economics for scaling data up or down. For instance, adjusting prices based on inflation rates can be done through scalar multiplication.
Verification of Matrix Equations
Verification of matrix equations involves proving that both sides of an equation are equal by performing matrix operations. For example, to verify \( 2(\textbf{A} + \textbf{B}) = 2 \textbf{A} + 2 \textbf{B} \), you first calculate \( \textbf{A} + \textbf{B} \), then multiply the result by 2. Separately, calculate 2 \( \textbf{A} \) and 2 \( \textbf{B} \), and then add them up. If both results match, the equation holds true. This process helps to confirm the correctness of matrix operations and is particularly useful in solving complex economic models.

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Most popular questions from this chapter

Let $$ \mathbf{A}=\left[\begin{array}{ll} 2 & 1 \\ 5 & 1 \end{array}\right] \text { and } \quad \mathbf{B}=\left[\begin{array}{ll} 1 & 0 \\ 2 & 4 \end{array}\right] $$ (1) Find (a) \(|\mathbf{A}|\) (b) \(|\mathbf{B}|\) (c) \(\mid \mathbf{A B} \|\) Do you notice any connection between \(|\mathbf{A}|, \mid \mathrm{B} \|\) and \(|\mathrm{AB}| ?\) (2) Find (a) \(\mathbf{A}^{-1}\) (b) \(\mathrm{B}^{-1}\) (c) \((\mathrm{AB})^{-1}\) Do you notice any connection between \(A^{-1}, B^{-1}\) and \((A B)^{-1} ?\)

(1) Let $$ \mathbf{A}=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array}\right] \text { and } \quad \mathbf{B}=\left[\begin{array}{rr} 1 & -1 \\ 2 & 1 \\ -3 & 4 \end{array}\right] $$ Find (a) \(\mathbf{A}^{\mathrm{T}}\) (b) \(\mathbf{B}^{\mathrm{T}}\) (c) \(\mathbf{A}+\mathbf{B}\) (d) \((\mathbf{A}+\mathbf{B})^{\mathrm{T}}\) Do you notice any connection between \((A+B)^{\top}, A^{\top}\) and \(B^{\top} ?\) (2) Let $$ \mathbf{C}=\left[\begin{array}{ll} 1 & 4 \\ 5 & 9 \end{array}\right] \text { and } \quad \mathbf{D}=\left[\begin{array}{rrr} 2 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] $$ Find (a) \(\mathrm{C}^{\mathrm{T}}\) (b) \(\mathrm{D}^{\mathrm{T}}\) (c) \(\mathrm{CD}\) (d) \((\mathbf{C D})^{\mathrm{T}}\) Do you notice any connection between \((\mathrm{CD})^{\top}, C^{\top}\) and \(D^{T} ?\)

Consider the macroeconomic model defined by $$ \begin{array}{ll} Y=C+I^{*}+G^{*}+X^{*}-M & \\ C=a Y+b & (00) \\ M=m Y+M^{*} & \left(00\right) \end{array} $$ Show that this system can be written as \(\mathrm{Ax}=\mathrm{b}\), where $$ \mathbf{A}=\left[\begin{array}{ccc} 1 & -1 & 1 \\ -a & 1 & 0 \\ -m & 0 & 1 \end{array}\right] \mathbf{x}=\left[\begin{array}{c} Y \\ C \\ M \end{array}\right] \quad \mathbf{b}=\left[\begin{array}{c} I^{*}+G^{*}+X^{*} \\ b \\ M^{*} \end{array}\right] $$ Use Cramer's rule to show that $$ Y=\frac{b+I^{*}+G^{*}+X^{*}-M^{*}}{1-a+m} $$ Write down the autonomous investment multiplier for \(Y\) and deduce that \(Y\) increases as as \(/^{*}\) increases.

(Maple) (a) Find the determinant of the general \(3 \times 3\) matrix, \(A\) $$ \left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] $$ Deduce that this matrix is singular whenever the second row is a multiple of the first row. (b) Find the inverse, \(A^{-1}\), of the general \(3 \times 3\) matrix given in part (a) and verify that \(A A^{-1}=I\).

(Maple) Use the Maple instruction det \((A)\) to find the determinant of the \(4 \times 4\) matrix $$ \left[\begin{array}{rrrr} a & 2 & 1 & -1 \\ 2 & 3 & 4 & 1 \\ 0 & -1 & 5 & 6 \\ 1 & 2 & 4 & -3 \end{array}\right] $$ For what value of \(a\) does this matrix fail to possess an inverse? [Do not forget to load the linear algebra package before you begin. You do this by typing with (linalg) :]
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