Hit  for B (without the ENTER), or scroll up or down to “rref(“ and hit ENTER. In this chapter, we will typically assume that our matrices contain only numbers. -5 & 3 & -9 \\ (It doesn’t matter which side; just watch for negatives). 1 & -1 & 9 When you try to these types of systems in your calculator (using matrices), you’ll get an error since the determinant of the coefficient matrix will be 0. (b)  When we square P, we just multiply it by itself. Her supplier has provided the following nutrition information: Her first mixture, Mixture 1, consists of 6 cups of almonds, 3 cups of cashews, and 1 cup of pecans. Just remember when you put matrices together with matrix multiplication, the columns (what you see across) on the first matrix have to correspond to the rows down on the second matrix. One row of the coefficient matrix (and the corresponding constant matrix) is a multiple of another row. But, like we learned in the Systems of Linear Equations and Word Problems Section here, sometimes we have systems where we either have no solutions or an infinite number of solutions. Now we know that $$x=5$$, $$y=1$$, and $$z=-2$$. (These equations are called independent or consistent). The following matrix consists of a shoe store’s inventory of flip flops, clogs, and Mary Janes in sizes small, medium, and large: The store wants to know how much their inventory is worth for all the shoes. \end{bmatrix} $(A^T)_{ij} = (A)_{ji}$ A nut distributor wants to know the nutritional content of various mixtures of almonds, cashews, and pecans. Here is the information we have in table/matrix form: Then we can multiply the matrices (we can use a graphing calculator) since we want to end up with the amount of Protein, Carbs, and Fat in each of the mixtures. It turns out that we have extraneous information in this matrix; we only need the information where the girls’ names line up. $11 million of energy is consumed internally and$5.75 million of manufacturing is consumed internally. F = \begin{bmatrix} 23 & 4 & 9 \\ Then hit ENTER once more and you’ll get the determinant! Think of turning the first matrix to the right and sideways, multiplying each number by the numbers in the second matrix, and then adding them together. $\newcommand{\bfr}{\mathbf{r}}$ \displaystyle \begin{align}\pm \frac{1}{2}\left| {\begin{array}{*{20}{c}} {{{a}_{1}}} & {{{b}_{1}}} & 1 \\ {{{a}_{2}}} & {{{b}_{2}}} & 1 \\ {{{a}_{3}}} & {{{b}_{3}}} & 1 \end{array}} \right|&=\pm \frac{1}{2}\left| {\begin{array}{*{20}{c}} {-1} & 3 & 1 \\ 0 & {-5} & 1 \\ 2 & 8 & 1 \end{array}} \right|=\pm \frac{1}{2}\left[ {\left( {-1} \right)\left( {-5\cdot 1-1\cdot 8} \right)-3\left( {0\cdot 1-1\cdot 2} \right)+1\left( {0\cdot 8–5\cdot 2} \right)} \right]\\&=\pm \frac{1}{2}\left( {29} \right)=\frac{1}{2}\left( {29} \right)=14.5\end{align}. Let’s say we want to find the final grades for 3 girls, and we know what their averages are for tests, projects, homework, and quizzes. $\newcommand{\bfx}{\mathbf{x}}$ ), $$\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,AX=B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5x=10\\{{A}^{{-1}}}AX={{A}^{{-1}}}B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{5}\cdot 5x=\frac{1}{5}\cdot 10\\\,\,\,\,\,\,\,\,\,\,IX={{A}^{{-1}}}B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1x=2\\\,\,\,\,\,\,\,\,\,\,\,X={{A}^{{-1}}}B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\end{array}$$. Let’s take the system of equations that we worked with earlier and show that it can be solved using matrices: $$\displaystyle \begin{array}{l}(1)x+(1)y=\text{ }6\\25x+50y=200\end{array}$$, $$\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,A\,\,\,\,\,\,\,\times \,\,\,\,\,X\,\,\,=\,\,\,\,B\\\left[ {\begin{array}{*{20}{c}} 1 & 1 \\ {25} & {50} \end{array}} \right]\,\,\times \,\,\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]\,\,=\,\,\left[ {\begin{array}{*{20}{c}} 6 \\ {200} \end{array}} \right]\end{array}$$. -7 & 3 & 2 \\ For example, to find out how many healthy males we would have, we’d set up the following equation and do the calculation: $$.15(100)+.25(80)=35$$. Without going too much into Geometry, let’s look at what it looks like when three systems (each system looks like a “plane” or a piece of paper) have an infinite number of solutions, no solutions, and one solution, respectively: eval(ez_write_tag([[300,250],'shelovesmath_com-large-mobile-banner-2','ezslot_10',134,'0','0']));eval(ez_write_tag([[300,250],'shelovesmath_com-large-mobile-banner-2','ezslot_11',134,'0','1']));eval(ez_write_tag([[300,250],'shelovesmath_com-large-mobile-banner-2','ezslot_12',134,'0','2']));Systems that have an infinite number of solutions (called dependent or coincident) will have two equations that are basically the same. Again, matrices are great for storing numbers and variables – and also great for solving systems of equations, which we’ll see later. (a)  When we multiply a matrix by a scalar (number), we just multiply all elements in the matrix by that number. The order (or dimensions or size) of a matrix indicates the number of rows and the number of columns of the matrix. A = \begin{bmatrix} of a matrix A (capital letter A) is denoted by the symbol $$(A)_{ij}$$ or $$a_{ij}$$ (small letter a). Let’s add the second matrix to both sides, to get $$X$$ and its coefficient matrix alone by themselves. From counting through calculus, making math make sense! $$\begin{pmatrix}1&1\\2&c\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$$, For what values of $\lambda$ are there nontrivial solutions to $$\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \lambda \begin{pmatrix}x\\y\\z\end{pmatrix}$$, Are there any real values of $c$ for which there is a nontrivial (nonzero) solution to $$\begin{pmatrix}1&c\\-c&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}?$$, How many solutions are there to $$\begin{pmatrix}1&1&1\\1&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix}?$$. -4 & -3 & 4 \\ eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-3','ezslot_6',320,'0','0'])); Example 1The following matrix has 3 rows and 6 columns. $\newcommand{\bfb}{\mathbf{b}}$ \end{bmatrix} We can store a collection of values in an array. Multiply each of the top numbers by the determinant of the 2 by 2 matrix that you get by crossing out the other numbers in that top number’s row and column.