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In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved “match” appropriately.Compute $A-5 I_{3}$ and $\left(5 I_{3}\right) A,$ when$$A=\left[\begin{array}{rrr}{9} & {-1} & {3} \\ {-8} & {7} & {-6} \\ {-4} & {1} & {8}\end{array}\right]$$

$A-5 I_{3}=\left[\begin{array}{ccc}{9} & {-1} & {3} \\ {-8} & {7} & {-6} \\ {-4} & {1} & {8}\end{array}\right]-5\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right]$$=\left[\begin{array}{ccc}{9} & {-1} & {3} \\ {-8} & {7} & {-6} \\ {-4} & {1} & {8}\end{array}\right]-\left[\begin{array}{ccc}{5} & {0} & {0} \\ {0} & {5} & {0} \\ {0} & {0} & {5}\end{array}\right]$$=\left[\begin{array}{ccc}{9-5} & {-1-0} & {3-0} \\ {-8-0} & {7-5} & {-6-0} \\ {-4-0} & {1-0} & {8-5}\end{array}\right]=\left[\begin{array}{ccc}{4} & {-1} & {3} \\ {-8} & {2} & {-6} \\ {-4} & {1} & {3}\end{array}\right]$

$\left(5 I_{3}\right) A=5\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right] \times\left[\begin{array}{ccc}{9} & {-1} & {3} \\ {-8} & {7} & {-6} \\ {-4} & {1} & {8}\end{array}\right]$$=\left[\begin{array}{lll}{5} & {0} & {0} \\ {0} & {5} & {0} \\ {0} & {0} & {5}\end{array}\right] \times\left[\begin{array}{ccc}{9} & {-1} & {3} \\ {-8} & {7} & {-6} \\ {-4} & {1} & {8}\end{array}\right]$$=\left[\begin{array}{ccc}{5 \times 9+0+0} & {5 \times(-1)+0+0} & {5 \times 3+0+0} \\ {0+5 \times(-8)+0} & {0+5 \times 7+0} & {0+5 \times(-6)+0} \\ {0+0+5 \times(-4)} & {0+0+5 \times 1} & {0+0+5 \times 8}\end{array}\right]=\left[\begin{array}{ccc}{45} & {-5} & {15} \\ {-40} & {35} & {-30} \\ {-20} & {5} & {40}\end{array}\right]$

Algebra

Chapter 2

Matrix Algebra

Section 1

Matrix Operations

Introduction to Matrices

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University of Michigan - Ann Arbor

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Okay. And for this question, we are asked to compute Uh, thes two piece of a riff. Is it here? Uh, and using this major is a So let's begin. So a minus five times, the three by three identity will be the 1st 1 So that will be nine minus 13 minus 87 minus six minus 418 On this, five times three by three Identity, which is 111 000000 And we'll start by multiplying in the scaler. Okay? And I'll continue over here, right? And so this will result in nine minus five, which is for the other entries won't change. Only the diagonal entries will change. Here is the only A things in this matrix are diagonal entries. So this will still be minus eight. This will still be minus 57 Minus five is, too that this was simply minus six. This will still be 18 miles. Five is three. And so we're done. And for the next one, we will do, uh, five times three by three. Identity multiplied by a so five times the, uh, three by three identities. This matrix multiplied by a again minus eight seven. My six, Nice for one and eight, and the result will be five times nine plus zero zero. So that will be five times minus 45 and then zero times, or five times minus one plus zero times seven plus zero times one. So that will be minus five and again five times three and the rest of the zeroes. So that will be five times 3 15 and then zero times nine plus five times minus eight, which is minus 40 and zeros for the others, dear. It's as minus 15 times 7 35 And so the pattern here is that we only have to multiply correctly, corresponding element by five. And this will be five times minus six. Leaved Maya's 30. This will be five times minus four, minus 20. This will be five times one. This will be five times eight. Okay, we're done.

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