Note again that only square matrices have inverses, but there are square matrices that don’t have one (when the determinant is 0): $$\displaystyle \text{Inverse }\left[ {\begin{array}{*{20}{c}} {{{a}_{{11}}}} & {{{a}_{{12}}}} \\ {{{a}_{{21}}}} & {{{a}_{{22}}}} \end{array}} \right]=\frac{1}{{\det A}}\left[ {\begin{array}{*{20}{c}} {{{a}_{{22}}}} & {-{{a}_{{12}}}} \\ {-{{a}_{{21}}}} & {{{a}_{{11}}}} \end{array}} \right]$$, $$\displaystyle \color{#800000}{{\text{Inverse }\left[ {\begin{array}{*{20}{c}} 3 & 1 \\ 4 & 8 \end{array}} \right]}}=\frac{1}{{20}}\left[ {\begin{array}{*{20}{c}} 8 & {-1} \\ {-4} & 3 \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} {\frac{2}{5}} & {-\frac{1}{{20}}} \\ {-\frac{1}{5}} & {\frac{3}{{20}}} \end{array}} \right]$$, $$\displaystyle \color{#800000}{{\text{Inverse }\left[ {\begin{array}{*{20}{c}} 3 & 6 \\ 2 & 4 \end{array}} \right]}}=\frac{1}{0}\left[ {\begin{array}{*{20}{c}} 4 & {-6} \\ {-2} & 3 \end{array}} \right]=\text{No Inverse}$$. Note that you don’t need a “times” sign between [A]-1  and [B]. -5 & 3 & -9 \\ F = \begin{bmatrix} -6 & -4 & 23 \\ A nut distributor wants to know the nutritional content of various mixtures of almonds, cashews, and pecans. 3 \\ Then hit ENTER once more and you’ll get the determinant! Then hit (without the 2nd before it), and ENTER to get the inverse matrix. B = \begin{bmatrix} (You always go down first, and then over to get the dimensions of the matrix). eval(ez_write_tag([[250,250],'shelovesmath_com-mobile-leaderboard-1','ezslot_17',148,'0','0']));We can tell that this looks like matrix multiplication. (It doesn’t matter which side; just watch for negatives). Determine the amount of protein, carbs, and fats in a 1 cup serving of each of the mixtures. Multiplying matrices is a little trickier. eval(ez_write_tag([[300,250],'shelovesmath_com-large-mobile-banner-1','ezslot_3',127,'0','0']));eval(ez_write_tag([[300,250],'shelovesmath_com-large-mobile-banner-1','ezslot_4',127,'0','1']));eval(ez_write_tag([[300,250],'shelovesmath_com-large-mobile-banner-1','ezslot_5',127,'0','2']));Oh, one more thing! Donate or volunteer today! $\newcommand{\bfI}{\mathbf{I}}$ Let’s multiply the following matrix using the calculator: By definition, the inverse of a matrix is the reciprocal of the determinant, multiplied by a “, \displaystyle \begin{align}\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]&={{\left[ {\begin{array}{*{20}{c}} 1 & 1 \\ {25} & {50} \end{array}} \right]}^{{-1}}}\times \,\,\,\,\left[ {\begin{array}{*{20}{c}} 6 \\ {200} \end{array}} \right]\\\,\,&=\,\frac{1}{{25\,}}\left[ {\begin{array}{*{20}{c}} {50} & {-1} \\ {-25} & 1 \end{array}} \right]\times \,\left[ {\begin{array}{*{20}{c}} 6 \\ {200} \end{array}} \right]\\\,\,\,&=\left[ {\begin{array}{*{20}{c}} 2 & {-\frac{1}{{25}}} \\ {-1} & {\frac{1}{{25}}} \end{array}} \right]\times \,\left[ {\begin{array}{*{20}{c}} 6 \\ {200} \end{array}} \right]\\\,\,\,&=\left[ {\begin{array}{*{20}{c}} {(2\times 6)+(-\frac{1}{{25}}\times 200)} \\ {(-1\times 6)+(\frac{1}{{25}}\times 200)} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right]\end{align}. $\newcommand{\bfC}{\mathbf{C}}$ To get the answers, we have to divide each answer by 10 to get grams per cup. An upper triangular matrix is a square matrix with all its elements below the main diagonal equal to zero. G = \begin{bmatrix} ,\quad Note that, like the other systems, we can do this for any system where we have the same numbers of equations as unknowns. For example, for the matrix above, “Ashley’s number of pairs of shoes (5)” would be identified as $${{a}_{{2,1}}}$$, since it’s on the 2nd row and it’s the 1st entry. -6 & 0 & 0 \\ (b)   How much energy and manufacturing must be produced to have $8 million worth of energy and$5 million worth of manufacturing available for consumer use? Compute the matrix multiplications $$\begin{pmatrix} 1 & 2 & 3 \end{pmatrix}\begin{pmatrix} 1 \\2\\3\end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 1 \\2\\3\end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}.$$, Compute the matrix multiplication $$\begin{pmatrix}1 & 0 & 2 \\ -1 & 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}$$, Find the $3 \times 3$ matrix $\bfA$ satisfying \begin{align}, For what value of $c$ is there a nonzero solution to the following equation? $\newcommand{\bfb}{\mathbf{b}}$ For example, to find out how many healthy males we would have, we’d set up the following equation and do the calculation: $$.15(100)+.25(80)=35$$. It’s really not too difficult; it can just be a lot of work, so again, I’ll take the liberty of using the calculator to do most of the work , Let’s just show an example; let’s solve the following system using Cramer’s rule:  $$\displaystyle \begin{array}{l}\,2x+3y-\,\,z\,=\,15\\4x-3y-\,\,z\,=\,19\\\,\,x\,-\,3y+\,3z\,=\,-4\end{array}$$. Here are a couple more types of matrices problems you might see: Let $$P=\left[ {\begin{array}{*{20}{c}} 4 & {-6} \\ {-2} & 8 \end{array}} \right]$$. Each number or variable inside the matrix is called an entry or element, and can be identified by subscripts. Note that when the determinant is 0, the reciprocal is undefined; therefore, there is no inverse matrix. That is, show that $(AB)C = A(BC)$ for any matrices $A$, $B$, and $C$ that are of the appropriate dimensions for matrix multiplication. Here are some basic steps for storing, multiplying, adding, and subtracting matrices: $$\color{#800000}{{\left[ {\begin{array}{*{20}{c}} 2 & {-1} \\ 3 & 2 \\ 7 & 5 \end{array}} \right]\,\times \,\left[ {\begin{array}{*{20}{c}} 0 & {-4} & 3 & 1 & 4 \\ 6 & 7 & 2 & 9 & {-3} \end{array}} \right]\,\,}}\,=\,\,\left[ {\begin{array}{*{20}{c}} {-6} & {-15} & 4 & {-7} & {11} \\ {12} & 2 & {13} & {21} & 6 \\ {30} & 7 & {31} & {52} & {13} \end{array}} \right]$$, (Note that you can also enter matrices using ALPHA ZOOM and the arrow keys in the newer graphing calculators.).

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