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Discovery guide 4.1 Discovery guide

Discovery 4.1.

Consider matrices
\begin{align*} A \amp = \begin{abmatrix}{rcc} 1 \amp 2 \amp 3 \\ -1 \amp 3 \amp 2 \end{abmatrix}, \amp B \amp = \begin{abmatrix}{rc} 0 \amp 1 \\ -1 \amp 4 \\ 1 \amp 0 \end{abmatrix}, \amp C \amp = \begin{abmatrix}{rc} -6 \amp 1 \\ 1 \amp 2 \end{abmatrix}. \end{align*}
For each matrix, how would you describe its size (or dimensions)?

Discovery 4.2.

Consider matrices
\begin{align*} A \amp = \begin{abmatrix}{rcc} 1 \amp 1 \amp 3 \\ -1 \amp 3 \amp 2 \end{abmatrix}, \amp B \amp = \begin{abmatrix}{rcc} x^2 \amp 2 x + 3 \amp 3 \\ -1 \amp 3 \amp 2 \end{abmatrix},\\ C \amp = \begin{bmatrix} 1 \amp 1 \amp 3\\ x^2 \amp 3 \amp 2 \end{bmatrix}, \amp D \amp = \begin{bmatrix} x^2 \amp 2 x + 3 \\ -1 \amp 3 \end{bmatrix}. \end{align*}

(a)

For what value(s) of \(x\) is \(B\) equal to \(A\text{?}\) \(C\) equal to \(A\text{?}\) \(D\) equal to \(A\text{?}\)

Discovery 4.3.

Consider matrices
\begin{align*} A \amp = \begin{abmatrix}{rcc} 1 \amp 2 \amp 3\\ -1 \amp 3 \amp 2 \end{abmatrix}, \amp B \amp = \begin{abmatrix}{rcc} 0 \amp 2 \amp 1\\ -1 \amp 0 \amp 4 \end{abmatrix}, \amp C \amp = \begin{abmatrix}{rc} -6 \amp 1\\1 \amp 2 \end{abmatrix}. \end{align*}

(c)

Now letโ€™s consider the sum \(A + C\text{.}\)
(i)
Compute \(A + C\text{.}\) Call this result matrix \(D\text{.}\) What are the dimensions of \(D\text{?}\)
(ii)
Now compute \(D - A\text{.}\) Do this numerically, not algebraically; that is, forget where your result matrix \(D\) came from and actually compute \(D - A\) using the same procedure that you used to subtract matrices in Taskย a. What are the dimensions of this result?
(iii)
Now letโ€™s remember that \(D = A + C\text{.}\) Algebraically, what result would you expect from computing \((A + C) - A\text{?}\) Does your numerical computation in the previous step agree with your algebraic expectation? (Keep in mind your answer to what it means for two matrices to be equal from Taskย b of Discoveryย 4.2.)
(iv)
Given how things worked out, how do you feel about performing \(A + C\) in the first place?

Discovery 4.4.

The number zero is important in algebra, it lets us do things like the following.
\begin{align*} a+5 \amp = 7\\ a+5-5 \amp = 7-5\\ a+0 \amp = 2\\ a \amp = 2. \end{align*}
The critical step for us right now is the last simplification of the left-hand side:
\begin{equation*} a + 0 = a \text{.} \end{equation*}

(a)

What matrix do you think will act like zero in matrix addition? Is the answer different for different dimensions?

Discovery 4.5.

(a)

Use your idea from Taskย b of Discoveryย 4.2 to turn the following single matrix equation into a system of two equations in the unknowns \(c\) and \(d\text{.}\) (Donโ€™t bother to actually solve for the values of \(c\) and \(d\text{.}\))
\begin{equation*} \begin{bmatrix} c + 2 d \\ 3 d \end{bmatrix} = \begin{abmatrix}{r} 5 \\ -3 \end{abmatrix} \end{equation*}
Careful. What sizes are the two matrices above?

(b)

Now do the reverse of Taskย a: write the following system of equations as a single matrix equation using a column matrix on each side of the equation:
\begin{equation*} \begin{sysofeqns}{rcrcrcr} x_1 \amp - \amp 3 x_2 \amp - \amp x_3 \amp = \amp -4 \text{,} \\ -2 x_1 \amp + \amp 7 x_2 \amp + \amp 2 x_3 \amp = \amp 9 \text{.} \end{sysofeqns} \end{equation*}
Again, be careful about the sizes of your matrices! If you have an equals sign between two matrices, they must adhere to your principle from Taskย b of Discoveryย 4.2.

(c)

The simplest system of equations is one equation in one unknown, i.e.
\begin{equation*} a x = b\text{.} \end{equation*}
But we donโ€™t usually just think of this as left-hand side and right-hand side, we think of it in the pattern
\begin{equation*} \text{coefficient} \times \text{unknown} = \text{constant} \text{.} \end{equation*}
Can we represent the system from Taskย b in a similar pattern using a matrix equation
\begin{equation*} A \uvec{x} = \uvec{b} \text{?} \end{equation*}

(d)

On the left-hand side of the matrix equation \(A \uvec{x} = \uvec{b}\text{,}\) the operation matrix-times-matrix should compute to a single matrix. What size of matrix should this multiplication result be?
Hint.
The result of computing \(A \uvec{x}\) must make sense in the matrix equality \(A \uvec{x} = \uvec{b}\text{,}\) per the pattern of matrix equality you described in Taskย b of Discoveryย 4.2.

(e)

Finally, we want \(A \uvec{x} = \uvec{b}\) to represent in one matrix equation the full system of two number equations from Taskย b. We already came up with a matrix equation to represent that system in Taskย b. Looking at your matrices \(A\) and \(\uvec{x}\) from Taskย c, and comparing with the left-hand side of your matrix equation from Taskย b, what procedure should be used to carry out the operation matrix \(A\) times column \(\uvec{x}\)?

(f)

The values \(x_1 = 2\text{,}\) \(x_2 = 1\text{,}\) \(x_3 = 3\text{,}\) represent a solution to the system in Taskย b. Verify this by carrying out the multiplication \(A \uvec{x}\text{,}\) using your calculation procedure from Taskย e, and with the unknowns \(x_1,x_2,x_3\) in the column matrix \(\uvec{x}\) replaced by these solution values. Then compare your calculation result with \(\uvec{b}\text{.}\)

Discovery 4.6.

Consider
\begin{align*} A \amp = \begin{abmatrix}{rrr} 1 \amp -3 \amp -1 \\ -2 \amp 7 \amp 2 \end{abmatrix}, \amp X \amp = \begin{abmatrix}{rrr} 2 \amp 0 \amp 2 \\ 1 \amp 3 \amp 0 \\ 3 \amp -1 \amp -2 \end{abmatrix}. \end{align*}
Compute the product \(A X\) by considering \(X\) as a collection of three columns
\begin{equation*} X = \begin{bmatrix} \mtrxvbar \amp \mtrxvbar \amp \mtrxvbar \\ \uvec{x}_1 \amp \uvec{x}_2 \amp \uvec{x}_3 \\ \mtrxvbar \amp \mtrxvbar \amp \mtrxvbar \end{bmatrix} \end{equation*}
and using the procedure for โ€œmatrix times columnโ€ that you developed in Discoveryย 4.5.

Discovery 4.7.

We all know that \(3\) times \(5\) and \(5\) times \(3\) have the same result. Algebraically, we write that \(a b = b a \) is true for all numbers \(a,b\text{.}\) What about matrices?

(a)

Try it with matrices
\begin{align*} A \amp = \begin{abmatrix}{rr} 1 \amp 0 \\ 1 \amp -1 \end{abmatrix}, \amp B \amp = \begin{abmatrix}{rr} 3 \amp 2 \\ 1 \amp -1 \end{abmatrix}. \end{align*}

(b)

Look back at matrices \(A\) and \(X\) from Discoveryย 4.6, where you computed the matrix product \(A X\text{.}\) Does multiplying \(X A\) in the opposite order even make sense?

Discovery 4.8.

Considering the previous three activities about matrix multiplication, what patterns have you observed about the required sizes of the two matrices involved for things to work out?
In particular, if \(A\) has \(m\) rows and \(n\) columns, and \(B\) has \(k\) rows and \(\ell\) columns, what relationship must there be between these numbers for the matrix-times-columns calculation method to make sense when computing \(A B\text{?}\) And in that case, what size will the resulting product matrix \(A B\) be?