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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 6 - Matrices and DeterminantsProblem 17
Chapter 7, Problem 17

For Exercises 11–22, use Cramer's Rule to solve each system. {x+2y=33x−4y=4\(\begin{cases}\)x + 2y = 3 \\3x - 4y = 4\(\end{cases}\)

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Write the system of equations in matrix form: \( A\mathbf{x} = \mathbf{b} \), where \( A = \begin{bmatrix} 1 & 2 \\ 3 & -4 \end{bmatrix} \), \( \mathbf{x} = \begin{bmatrix} x \\ y \end{bmatrix} \), and \( \mathbf{b} = \begin{bmatrix} 3 \\ 4 \end{bmatrix} \).
Calculate the determinant of matrix \( A \), denoted as \( \det(A) \), using the formula for a 2x2 matrix: \( \det(A) = a_{11}a_{22} - a_{12}a_{21} \). For this matrix, \( \det(A) = (1)(-4) - (2)(3) \).
Form matrix \( A_x \) by replacing the first column of \( A \) with vector \( \mathbf{b} \), so \( A_x = \begin{bmatrix} 3 & 2 \\ 4 & -4 \end{bmatrix} \). Then calculate \( \det(A_x) \) using the same determinant formula.
Form matrix \( A_y \) by replacing the second column of \( A \) with vector \( \mathbf{b} \), so \( A_y = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} \). Then calculate \( \det(A_y) \) using the determinant formula.
Use Cramer's Rule to find the solutions: \( x = \frac{\det(A_x)}{\det(A)} \) and \( y = \frac{\det(A_y)}{\det(A)} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Cramer's Rule

Cramer's Rule is a method for solving systems of linear equations using determinants. It applies to square systems where the number of equations equals the number of variables. The solution for each variable is found by replacing the corresponding column of the coefficient matrix with the constants vector and dividing the determinant of this new matrix by the determinant of the coefficient matrix.
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Cramer's Rule - 2 Equations with 2 Unknowns

Determinants of 2x2 Matrices

The determinant of a 2x2 matrix [[a, b], [c, d]] is calculated as ad - bc. This scalar value helps determine if the system has a unique solution (non-zero determinant) or not. In Cramer's Rule, determinants are used to find the values of variables by comparing the determinant of the coefficient matrix and modified matrices.
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Solving Systems of Linear Equations

Solving systems of linear equations involves finding values for variables that satisfy all equations simultaneously. Methods include substitution, elimination, and matrix approaches like Cramer's Rule. Understanding how to manipulate and interpret equations is essential for applying these methods effectively.
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Related Practice
Textbook Question

Use the fact that if A=[abcd]A=\(\begin{bmatrix}\)a & b\\ c & d\(\end{bmatrix}\), then A−1=1ad−bc[d−b−ca]A^{-1}=\(\frac{1}{ad-bc}\]\begin{bmatrix}\)d & -b\\ -c & a\(\end{bmatrix}\) to find the inverse of each matrix, if possible. Check that AA−1=I2AA^{-1} = I_2 and A−1A=I2A^{-1}A = I_2.

A=[10−2−51]A = \(\begin{bmatrix}\)10 & -2 \\-5 & 1\(\end{bmatrix}\)

Textbook Question

In Exercises 14–27, perform the indicated matrix operations given that and D are defined as follows. If an operation is not defined, state the reason. D-A

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Textbook Question

Let A=[−3−72−950]A = \(\begin{bmatrix}\)-3 & -7 \\2 & -9 \\5 & 0\(\end{bmatrix}\) and B=[−5−1003−4]B = \(\begin{bmatrix}\)-5 & -1 \\0 & 0 \\3 & -4\(\end{bmatrix}\) Solve each matrix equation for X. X - A = B

Textbook Question

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. {x+2y+3z=5y−5z=0\(\begin{cases}\)x + 2y + 3z = 5 \(\y\) - 5z = 0\(\end{cases}\)

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Textbook Question

In Exercises 14–27, perform the indicated matrix operations given that and D are defined as follows. If an operation is not defined, state the reason. 3A+2D

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Textbook Question

Perform each matrix row operation and write the new matrix.

[1−111∣301−2−1∣02034∣115124∣6]−2R1+R3−5R1+R4\(\begin{bmatrix}\)1 & -1 & 1 & 1 & \(\vert\) & 3 \\0 & 1 & -2 & -1 & \(\vert\) & 0 \\2 & 0 & 3 & 4 & \(\vert\) & 11 \\5 & 1 & 2 & 4 & \(\vert\) & 6\(\end{bmatrix}\[\quad\]\begin{array}{l}\)-2R_1 + R_3 \\-5R_1 + R_4\(\end{array}\)

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