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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 13
Chapter 7, Problem 13

For Exercises 11–22, use Cramer's Rule to solve each system. {12x+3y=152x−3y=13\(\begin{cases}\)12x + 3y = 15 \\2x - 3y = 13\(\end{cases}\)

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Write the system of equations in standard form: \[\begin{cases} 12x + 3y = 15 \\ 2x - 3y = 13 \end{cases}\]
Identify the coefficient matrix, the variable matrix, and the constant matrix: Coefficient matrix \[A = \begin{bmatrix} 12 & 3 \\ 2 & -3 \end{bmatrix}\] Constants matrix \[\mathbf{b} = \begin{bmatrix} 15 \\ 13 \end{bmatrix}\]
Calculate the determinant of the coefficient matrix \[D = \det(A) = \begin{vmatrix} 12 & 3 \\ 2 & -3 \end{vmatrix}\] using the formula \[D = (12)(-3) - (3)(2)\]
Form matrices \[D_x\] and \[D_y\] by replacing the respective columns of \[A\] with the constants matrix \[\mathbf{b}\]: - Replace the first column of \[A\] with \[\mathbf{b}\] to get \[D_x = \begin{bmatrix} 15 & 3 \\ 13 & -3 \end{bmatrix}\] - Replace the second column of \[A\] with \[\mathbf{b}\] to get \[D_y = \begin{bmatrix} 12 & 15 \\ 2 & 13 \end{bmatrix}\]
Calculate the determinants \[D_x = \det(D_x)\] and \[D_y = \det(D_y)\] using the determinant formula for 2x2 matrices, then find the solutions for \[x\] and \[y\] using Cramer's Rule: \[x = \frac{D_x}{D}\] \[y = \frac{D_y}{D}\]

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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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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

A system of linear equations consists of multiple linear equations with the same variables. The goal is to find values for the variables that satisfy all equations simultaneously. Methods include substitution, elimination, and matrix-based approaches like Cramer's Rule, which is efficient for small systems.
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Related Practice
Textbook Question

Find the following matrices: A+BA + B

A=[2−41],B=[−53−1]A = \(\begin{bmatrix}\)2 \\-4 \\1\(\end{bmatrix}\), B = \(\begin{bmatrix}\)-5 \\3 \\-1\(\end{bmatrix}\)

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=[23−12]A = \(\begin{bmatrix}\) 2 & 3 \\ -1 & 2 \(\end{bmatrix}\)

Textbook Question

In Exercises 9 - 16, find the following matrices: b. A - B

Textbook Question

Find the following matrices: - 3A + 2B

A=[311−125],B=[2−36−31−4]A = \(\begin{bmatrix}\)3 & 1 & 1 \\-1 & 2 & 5\(\end{bmatrix}\), B = \(\begin{bmatrix}\)2 & -3 & 6 \\-3 & 1 & -4\(\end{bmatrix}\)

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Textbook Question

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. {w−3x+y−4z=4−2w+x+2y=−23w−2x+y−6z=2−w+3x+2y−z=−6\(\begin{cases}\)w - 3x + y - 4z = 4 \\-2w + x + 2y = -2 \\3w - 2x + y - 6z = 2 \\-w + 3x + 2y - z = -6\(\end{cases}\)

Textbook Question

Perform each matrix row operation and write the new matrix.

[2−64∣1015−5∣0304∣7]12R1\(\begin{bmatrix}\)2 & -6 & 4 & \(\vert\) & 10 \\1 & 5 & -5 & \(\vert\) & 0 \\3 & 0 & 4 & \(\vert\) & 7\(\end{bmatrix}\]\quad\) \(\frac{1}{2}\)R_1

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