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Use matrices \(A, B, C,\) and \(D\) to find the indicated matrices. \(A=\left[\begin{array}{cc}6 & -3 \\ 4 & -5\end{array}\right] \quad B=\left[\begin{array}{rr}3 & 12 \\ -9 & -6\end{array}\right]\) \(C=\left[\begin{array}{rrr}-1 & 4 & -7 \\ 2 & -6 & 11\end{array}\right] \quad D=\left[\begin{array}{rrr}7 & 9 & -6 \\ -4 & 0 & 8\end{array}\right]\) $$-C-2 D$$

Short Answer

Expert verified
The matrix \(-C - 2D\) is \(\left[\begin{array}{rrr}15 & 14 & -5 \\ -10 & 6 & 5 \end{array}\right]\).

Step by step solution

01

Multiply Matrix D by 2

First, we need to calculate \(2D\). To do this, we multiply each element of matrix \(D\) by 2. \[2D = \left[\begin{array}{rrr}2 \cdot 7 & 2 \cdot 9 & 2 \cdot (-6) \ 2 \cdot (-4) & 2 \cdot 0 & 2 \cdot 8 \end{array}\right] = \left[\begin{array}{rrr}14 & 18 & -12 \ -8 & 0 & 16 \end{array}\right]\]
02

Negate Matrix C

Next, we need to find \(-C\). This means multiplying each element of matrix \(C\) by -1. \[-C = \left[\begin{array}{rrr}-1 \cdot (-1) & -1 \cdot 4 & -1 \cdot (-7) \ -1 \cdot 2 & -1 \cdot (-6) & -1 \cdot 11 \end{array}\right] = \left[\begin{array}{rrr}1 & -4 & 7 \ -2 & 6 & -11 \end{array}\right]\]
03

Add Matrices -C and 2D

Finally, we need to add \(-C\) and \(2D\). We do this by adding corresponding elements of the matrices. \[-C - 2D = \left[\begin{array}{rrr}1 & -4 & 7 \ -2 & 6 & -11 \end{array}\right] + \left[\begin{array}{rrr}14 & 18 & -12 \ -8 & 0 & 16 \end{array}\right] = \left[\begin{array}{rrr}1+14 & -4+18 & 7+(-12) \ -2+(-8) & 6+0 & -11+16 \end{array}\right]\]
04

Simplify the Result

Compute the arithmetic of each element from the addition in step 3. \[\left[\begin{array}{rrr}15 & 14 & -5 \ -10 & 6 & 5 \end{array}\right]\] This is the matrix result for \(-C - 2D\).

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Key Concepts

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

Matrix Addition
Matrix Addition is a fundamental operation where two matrices of the same dimensions are combined by adding their corresponding elements. For example, if we have two 2x2 matrices:
  • Matrix A: \[\begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}\]
  • Matrix B: \[\begin{bmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{bmatrix}\]
Then, the sum of these matrices, \(A + B\), is computed as:\[\begin{bmatrix} a_{11} + b_{11} & a_{12} + b_{12} \ a_{21} + b_{21} & a_{22} + b_{22} \end{bmatrix}\]
In the given problem, after negating Matrix C and multiplying Matrix D by 2, the matrices were added:
  • Negated C: \[\begin{bmatrix} 1 & -4 & 7 \ -2 & 6 & -11 \end{bmatrix}\]
  • Matrix 2D: \[\begin{bmatrix} 14 & 18 & -12 \ -8 & 0 & 16 \end{bmatrix}\]
Each corresponding element was added to form the resulting matrix. This operation underlines the importance of ensuring matrices have the same dimensions, which allows each element to align perfectly and be summed correctly.
Scalar Multiplication
Scalar multiplication in matrices involves multiplying every element of a matrix by a constant, known as the scalar. This operation maintains the original layout of the matrix, only changing the values inside based on the scalar applied.
Let's consider a matrix \(D\):\[\begin{bmatrix} d_{11} & d_{12} & d_{13} \ d_{21} & d_{22} & d_{23} \end{bmatrix}\]If we need to find \(2D\), we multiply each element by 2:\[\begin{bmatrix} 2 \cdot d_{11} & 2 \cdot d_{12} & 2 \cdot d_{13} \ 2 \cdot d_{21} & 2 \cdot d_{22} & 2 \cdot d_{23} \end{bmatrix}\]
In this exercise, multiplying every element of Matrix D by the scalar 2 transformed it into:
  • \[\begin{bmatrix} 14 & 18 & -12 \ -8 & 0 & 16 \end{bmatrix}\]
Scalar multiplication is useful when you need to scale a matrix up or down while retaining the positional integrity of the elements.
Matrix Negation
Matrix Negation is akin to multiplying all elements of a matrix by -1. This operation flips the sign of each element within the matrix, providing a complementary perspective for addition or subtraction tasks.
To negate a matrix like \(C\):\[\begin{bmatrix} c_{11} & c_{12} & c_{13} \ c_{21} & c_{22} & c_{23} \end{bmatrix}\]The negated matrix, \(-C\), is represented as:\[\begin{bmatrix} -c_{11} & -c_{12} & -c_{13} \ -c_{21} & -c_{22} & -c_{23} \end{bmatrix}\]
In practice, negating Matrix C turned it into:
  • \[\begin{bmatrix} 1 & -4 & 7 \ -2 & 6 & -11 \end{bmatrix}\]
This process is foundational in combining matrices with addition, as it allows for the easy cancellation of values or the creation of "opposite" matrices for further computations.

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Most popular questions from this chapter

Solve the indicated systems of equations using the inverse of the coefficient matrix. It is necessary to set up the appropriate equations. A river tour boat takes \(5.0 \mathrm{h}\) to cruise downstream and \(7.0 \mathrm{h}\) for the return upstream. If the river flows at \(4.0 \mathrm{mi} / \mathrm{h}\), how fast does the boat travel in still water, and how far downstream does the boat go before starting the return trip?

Evaluate the given determinants by expansion by minors. $$\left|\begin{array}{llll} 4 & 3 & 6 & 0 \\ 3 & 0 & 0 & 4 \\ 5 & 0 & 1 & 2 \\ 2 & 1 & 1 & 7 \end{array}\right|$$

Solve the given systems of equations by determinants. Evaluate by using the properties of determinants. $$\begin{aligned} &3 x+y+t=0\\\ &3 z+2 t=8\\\ &6 x+2 y+2 z+t=3\\\ &3 x-y-z-t=0 \end{aligned}$$

Perform the indicated matrix operations. One serving of brand \(\mathrm{K}\) of breakfast cereal provides the given percentages of the given vitamins and minerals: vitamin A, \(15 \% ;\) vita\(\min \mathrm{C}, 25 \% ;\) calcium, \(10 \% ;\) iron, \(25 \% .\) One serving of brand \(\mathrm{G}\) provides: vitamin A, 10\%; vitamin C, 10\%; calcium, 10\%; iron, \(45 \% .\) One serving of tomato juice provides: vitamin A, \(15 \%\) vitamin \(\mathrm{C}, 30 \% ;\) calcium, \(3 \% ;\) iron, \(3 \% .\) One serving of orangepineapple juice provides vitamin A, \(0 \% ;\) vitamin \(\mathrm{C}, 100 \% ;\) calcium, \(2 \% ;\) iron, \(2 \% .\) Set up a two-row, four-column matrix \(B\) to represent the data for the cereals and a similar matrix \(J\) for the juices.

Perform the indicated matrix multiplications. Show that \(A^{2}-I=(A+I)(A-I)\) for \(A=\left[\begin{array}{ll}2 & 4 \\ 3 & 5\end{array}\right].\)

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