Theory Questions – Matrices & Determinants

Question 1

Question: Given that

M = \(\scriptsize \begin{pmatrix}1& 2\\4& 3\end{pmatrix}\)

N = \(\scriptsize \begin{pmatrix}m& x\\n& y\end{pmatrix}\)

MN = \(\scriptsize \begin{pmatrix}2& 1\\3& 4\end{pmatrix}\)

Find the matrix N.

Solution

Given that

MN = \(\scriptsize \begin{pmatrix}2& 1\\3& 4\end{pmatrix}\)

∴ M × N = MN

\(\scriptsize \begin{pmatrix}m+2n & x + 2y\\4m + 3n & 4x + 3y\end{pmatrix} \scriptsize  = \scriptsize \begin{pmatrix}2& 1\\3& 4\end{pmatrix}\)

Step 2: Simplifying the above we have

Step 3: using equality of matrix, then we have

m + 2n = 2 ————– (1)

x + 2y = 1 ————— (2)

4m + 3n = 3 ————- (3)

4x + 3y = 4 ————– (4)

Solving equations (1) and (3)

m + 2n = 2

m = 2 – 2n ————- (*)

Substitute m in equation (3)

4(2 – 2n) + 3n = 3

8 – 8n + 3n = 3

8 – 5n = 3

-5n = 3 – 8

-5n = -5

n = \(\frac{-5}{-5}\)

n = 1

Step 4: substitute n = 1 into (*)

m = 2 – 2 (1)

m = 2 – 2

m = 0

  n = 1 and m = 0

Step 5: Solving equations (2) and (4) from (2)

x = 1 – 2y ———– (**)

Step 6: substitute x = 1 – 2y in equation (4)

  4 (1 -2y) + 3y = 4

4 – 8y + 3y = 4

4 – 5y = 4

-5y = 4 – 4

-5y = 0

y = \(\frac{0}{-5}\)

y = 0

Step 7: substitute y = 0 in equation (**)

x = 1 – 2(0)

x = 1 – 0

x = 1

Step 8: hence the matrix

N = \(\scriptsize \begin{pmatrix}m& x\\n& y\end{pmatrix}\)

= \(\scriptsize \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}\)