/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Solve for \(X\) in the equation,... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve for \(X\) in the equation, where \(A=\left[ \begin{array}{rr}{-2} & {-1} \\ {1} & {0} \\ {3} & {-4}\end{array}\right]\) and \(B=\left[ \begin{array}{rr}{0} & {3} \\ {2} & {0} \\\ {-4} & {-1}\end{array}\right].\) \(2 X+3 A=B\)

Short Answer

Expert verified
\(X = \left[ \begin{array}{cc}{3} & {3} \ {1} & {0} \ {-7} & {6} \end{array} \right]\)

Step by step solution

01

Isolate the term with X on one side

To do that, subtract matrix \(3A\) from both sides of the equation. It will look like \(2X = B - 3A\).
02

Perform the subtraction operation

Compute \(B - 3A\) which is \[\left[ \begin{array}{cc}{0} & {3} \ {2} & {0} \ {-4} & {-1}\end{array}\right] - 3 \cdot \left[ \begin{array}{cc}{-2} & {-1} \ {1} & {0} \ {3} & {-4}\end{array}\right]. \] After performing the matrix subtraction, you get \[\left[ \begin{array}{cc}{6} & {6} \ {2} & {0} \ {-13} & {11}\end{array}\right]. \]
03

Solve for X by dividing the scalar

To solve for \(X\), divide both sides by the scalar 2, hence \(X\) is equal to \[\frac{1}{2} \cdot \left[ \begin{array}{cc}{6} & {6} \ {2} & {0} \ {-13} & {11} \end{array} \right] \] which then turns out to be \[\left[ \begin{array}{cc}{3} & {3} \ {1} & {0} \ {-7} & {6} \end{array} \right]. \] So, \(X = \left[ \begin{array}{cc}{3} & {3} \ {1} & {0} \ {-7} & {6} \end{array} \right]\).

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Key Concepts

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

Solving Matrix Equations
When tackling matrix equations, it's essential to be familiar with both algebra rules and special matrix operations. A matrix equation like 2X + 3A = B might seem daunting at first, but if we tackle it step by step, it's not so different from the equations you've encountered before. The first thing you need to do is isolate the variable matrix, which, in this case, is X. This involves performing matrix addition or subtraction to move all the other matrices to the other side. Once X is isolated, we can then 'divide' by any scalar coefficients by using scalar multiplication. Although we cannot divide matrices per se, scalar multiplication allows us to simplify the matrix equation just like you would when you divide both sides of an algebraic equation by a number. Thus, solving for X involves simplifying the matrix operation until you isolate X on one side of the equation.
Matrix Subtraction
Matrix subtraction is analogous to arithmetic subtraction but with one crucial distinction: it's not a single number operation but involves multiple elements. To perform matrix subtraction, both matrices must have the same dimensions. Then, subtract corresponding elements in the matrices from one another. It's as if you're subtracting two sets of numbers at the same time, element by element. A helpful tip is to line up the matrices so your work is organized, ensuring you subtract the correct elements. This plays an important part in solving our matrix equation because we had to subtract 3A from B to isolate X.
Scalar Multiplication
To further decode matrix equations, scalar multiplication is used to ‘scale’ a matrix, which means to multiply each element within a matrix by the same number, called a scalar. This process is also key in solving for X in our original equation. After isolating X, and ending up with an expression such as 2X = C, where C is a resultant matrix after matrix subtraction, we can divide both sides of the equation by 2. However, in matrix terms, this division translates to multiplying by the reciprocal of the scalar, which is 1/2 in this case. So, using scalar multiplication, we turn 2X into X by multiplying every element in matrix C by 1/2.
Precalculus
The field of precalculus is a preparatory stage for calculus, where you encounter concepts like matrix equations. It's a crucial part of mathematics education, serving as the bridge between algebra, geometry, and the advanced concepts of calculus. By understanding the structured ways of manipulating numbers and variables through precalculus, students gain the ability to solve complex problems like the matrix equation we discussed.
Remember that patience and practice are keys. Matrix operations aren't always intuitive, but with repetition, they begin to feel as manageable as the algebra you are used to. Mastering matrix equations is not only satisfying, but will also boost your mathematical confidence as you step into the world of calculus.

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Most popular questions from this chapter

True or False? In Exercises 95 and 96, determine whether the statement is true or false. Justify your answer If two columns of a square matrix are the same, then the determinant of the matrix will be zero.

Curve Fitting, use a system of equations to find the quadratic function \(f(x)=a x^{2}+b x+c\) that satisfies the given conditions. Solve the system using matrices. $$f(1)=2, f(2)=9, f(3)=20$$

A corporation has three factories, each of which manufactures acoustic guitars and electric guitars. The production levels are represented by \(A\) . \(A=\left[ \begin{array}{ccc}{70} & {50} & {25} \\ {35} & {100} & {70}\end{array}\right]\) (a) Interpret the value of \(a_{22}\) (b) How could you find the production levels when production is increased by 20\(\% ?\) (c) Each acoustic guitar sells for \(\$ 80\) and each electric guitar sells for \(\$ 120 .\) How could you use matrices to find the total sales value of the guitars produced at each factory?

Comparing Solutions of Two Systems In, determine whether the two systems of linear equations yield the same solution. If so, find the solution using matrices. $$(a)\left\\{\begin{array}{rrr}{x+3 y-z=} & {19} \\ {y+6 z=} & {-18} \\ {z=} & {-4}\end{array}\right.$$ $$(b)\left\\{\begin{array}{rrr}{x-y+3 z=} & {-15} \\ {y-2 z=} & {14} \\ {z=} & {-4}\end{array}\right.$$

Museum A natural history museum borrowed \(\$ 2,000,000\) at simple annual interest to purchase new exhibits. Some of the money was borrowed at \(7 \%,\) some at \(8.5 \%,\) and some at 9.5\(\% .\) Use a system of linear equations to determine how much was borrowed at each rate given that the total annual interest was \(\$ 169,750\) and the amount borrowed at 8.5\(\%\) was four times the amount borrowed at 9.5\(\% .\) Solve the system of linear equations using matrices.

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