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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 43
Chapter 7, Problem 43

In Exercises 37–44, use Cramer's Rule to solve each system. {x+2z=42y−z=52x+3y=13\(\begin{cases}\)x + 2z = 4 \\2y - z = 5 \\2x + 3y = 13\(\end{cases}\)

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Write the system of equations in standard form, aligning variables x, y, and z: \(\begin{cases} x + 0y + 2z = 4 \\ 0x + 2y - z = 5 \\ 2x + 3y + 0z = 13 \end{cases}\)
Form the coefficient matrix \(A\) from the coefficients of \(x\), \(y\), and \(z\): \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \end{bmatrix}\)
Calculate the determinant of matrix \(A\), denoted as \(\det(A)\), to ensure it is not zero (which would mean the system has a unique solution).
Form matrices \(A_x\), \(A_y\), and \(A_z\) by replacing the respective columns of \(A\) with the constants vector \(\begin{bmatrix} 4 \\ 5 \\ 13 \end{bmatrix}\): - \(A_x\) replaces the first column, - \(A_y\) replaces the second column, - \(A_z\) replaces the third column.
Calculate the determinants \(\det(A_x)\), \(\det(A_y)\), and \(\det(A_z)\), then use Cramer's Rule to find the variables: \( x = \frac{\det(A_x)}{\det(A)}\), \( y = \frac{\det(A_y)}{\det(A)}\), \( z = \frac{\det(A_z)}{\det(A)}\).

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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 when the system has the same number of equations as unknowns and the coefficient matrix has a non-zero determinant. Each variable is found by replacing the corresponding column in the coefficient matrix with the constants vector and calculating the determinant ratio.
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Determinants of Matrices

The determinant is a scalar value that can be computed from a square matrix and provides important properties about the matrix, such as invertibility. For a 3x3 matrix, the determinant is calculated using a specific formula involving minors and cofactors. A non-zero determinant indicates the system has a unique solution.
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Setting up the Coefficient Matrix and Constants Vector

To apply Cramer's Rule, the system of equations must be expressed in matrix form: Ax = b, where A is the coefficient matrix, x is the vector of variables, and b is the constants vector. Correctly identifying coefficients and constants from the equations is essential for forming these matrices accurately.
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Related Practice
Textbook Question

In Exercises 37 - 44, perform the indicated matrix operations given that A, B and C are defined as follows. If an operation is not defined, state the reason.

A=[40−3501],B=[51−2−2],C=[1−1−11]A=\(\begin{bmatrix}\)4 & 0\\ -3 & 5\\ 0 & 1\(\end{bmatrix}\),B=\(\begin{bmatrix}\)5 & 1\\ -2 & -2\(\end{bmatrix}\),C=\(\begin{bmatrix}\)1 & -1\\ -1 & 1\(\end{bmatrix}\)

A(BC)

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Textbook Question

Solve the system: (Hint: Let A = ln w, B = ln x, C = ln y, and D = ln z. Solve the system for A, B, C, and D. Then use the logarithmic equations to find w, x, y, and z.)

{2lnw+lnx+3lny−2lnz=−64lnw+3lnx+lny−lnz=−2lnw+lnx+lny+lnz=−5lnw+lnx−lny−lnz=5\(\begin{cases}\)2 \(\ln\) w + \(\ln\) x + 3 \(\ln\) y - 2 \(\ln\) z = -6 \\4 \(\ln\) w + 3 \(\ln\) x + \(\ln\) y - \(\ln\) z = -2 \(\ln\) w + \(\ln\) x + \(\ln\) y + \(\ln\) z = -5 \(\ln\) w + \(\ln\) x - \(\ln\) y - \(\ln\) z = 5\(\end{cases}\)

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Textbook Question

In Exercises 46–51, evaluate each determinant.

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Textbook Question

In Exercises 45–48, explain why the system of equations cannot be solved using Cramer's Rule. Then use Gaussian elimination to solve the system.

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Textbook Question

In Exercises 43–44, (a) Write each linear system as a matrix equation in the form AX = B (b) Solve the system using the inverse that is given for the coefficient matrix.

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Textbook Question

In Exercises 37–44, use Cramer's Rule to solve each system. {x+y+z=4x−2y+z=7x+3y+2z=4\(\begin{cases}\)x + y + z = 4 \(\x\) - 2y + z = 7 \(\x\) + 3y + 2z = 4\(\end{cases}\)

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