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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.Let $A=\left[\begin{array}{ll}{4} & {-1} \\ {5} & {-2}\end{array}\right] .$ Compute $3 I_{2}-A$ and $\left(3 I_{2}\right) A$

$3 I_{2}-A=3\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]-\left[\begin{array}{cc}{4} & {-1} \\ {5} & {-2}\end{array}\right]$$=\left[\begin{array}{ll}{3} & {0} \\ {0} & {3}\end{array}\right]-\left[\begin{array}{ll}{4} & {-1} \\ {5} & {-2}\end{array}\right]$$=\left[\begin{array}{cc}{3-4} & {0-(-1)} \\ {0-5} & {3-(-2)}\end{array}\right]=\left[\begin{array}{cc}{-1} & {1} \\ {-5} & {5}\end{array}\right]$

$\left(3 I_{2}\right) A=3\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right] \times\left[\begin{array}{ll}{4} & {-1} \\ {5} & {-2}\end{array}\right]$$=\left[\begin{array}{ll}{3} & {0} \\ {0} & {3}\end{array}\right] \times\left[\begin{array}{ll}{4} & {-1} \\ {5} & {-2}\end{array}\right]$$=\left[\begin{array}{lll}{3 \times 4+0 \times 5} & {3 \times(-1)+0 \times(-2)} \\ {0 \times 4+3 \times 5} & {0 \times(-1)+3 \times(-2)}\end{array}\right]$$=\left[\begin{array}{ll}{12} & {-3} \\ {15} & {-6}\end{array}\right]$

Algebra

Chapter 2

Matrix Algebra

Section 1

Matrix Operations

Introduction to Matrices

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for this problem were given a matrix A which is four or five negative one negative too. We want calculate three times the identity minus a. And we also want to calculate three times the identity kinds of a so three times our identity matrix. It specified as I too, because we want our two by two identity matrix. Oh, I said to And so I thought Thio wrote the wrong number. There we go. We have 3003 minus. So I'm going to write this, actually, as plus negative a So plus negative four and negative five. Positive one Positive too, which is going to equal. So we have three minus four. Gives us negative 10 minus five. Yeah, zero minus five is going to give us negative. Five zero plus one is one and three plus two is positive. Five. Next doing three times our identity. Identity Identity matrix times A. You have 3003 times. Uh, not negative. A regular A. So we have four or five. Negative one Negative too. It's not negative. 12 Negative too. So we'll have three times four plus zero times five, then down below. We will have zero times four plus three times five. We'll have three times negative one plus zero times, too, and we'll end off with zero. A time is negative one zero times negative, one plus three times negative, too. So that's going to you? No, these are all pretty easy to evaluate. Here. We'll end up with 12 15 a few times. Negative one is negative three and then negative six. And I'll point out that that is actually the exact same thing as three times a.

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