Question

application problems on matrix​

Answer 1

Answer:

They are used for plotting graphs, statistics and also to do scientific studies and research in almost different fields. Matrices can also be used to represent real world data like the population of people, infant mortality rate, etc.

Step-by-step explanation:

llustration 3: If [latex]A=\begin{bmatrix} 1 &-2 &3 \\ -4 & 2 & 5 \end{bmatrix} and B=\begin{bmatrix} 1 &3 \\ -1&0 \\ 2&4 \end{bmatrix}[/latex] . then prove that (AB)T = BTAT.

Solution:

By obtaining the transpose of AB i.e. (AB)T and multiplying BT and AT we can easily get the result.

Here, AB =[latex]\begin{bmatrix} 1 & -2 &3 \\ -4& 2& 5 \end{bmatrix}  \begin{bmatrix} 1 &3 \\ -1&0 \\ 2& 4 \end{bmatrix}=\begin{bmatrix} 1(1)-2(-1)+3(2) &1(3)-2(0)+3(4) \\ -4(1)+2(-1)+5(2) &-4(3)+2(0)+5(4) \end{bmatrix} = \begin{bmatrix} 9 &15 \\ 4& 8 \end{bmatrix}[/latex].

∴ [latex](AB)^{T}=\begin{bmatrix} 9 & 4\\ 15& 8 \end{bmatrix};B^{T}A^{T} ==\begin{bmatrix} 1 & -1 &2 \\ 3& 0& 4 \end{bmatrix} \begin{bmatrix} 1 &-4 \\ -2&2 \\ 3& 5 \end{bmatrix}=\begin{bmatrix} 1(1)-1(-2)+2(3) &1(-4)-1(2)+2(5) \\ 3(1)+0(-2)+4(3) &3(-4)+0(2)+4(5) \end{bmatrix} = \begin{bmatrix} 9 &4 \\ 15 & 8 \end{bmatrix}=(AB)^{T}[/latex]

Illustration 4: If [latex]A=\begin{bmatrix} 5 &-1 &3 \\ 0& 1& 2 \end{bmatrix} and B=\begin{bmatrix} 0 &2 &3 \\ 1& -1 &4 \end{bmatrix}[/latex]. then what is (B’)’A’ equal to?

Solution:

In this problem, we use the properties of the transpose of a matrix to get the required result.

We have =[latex]{({B}’)}'{A}’ =B{A}’=\begin{bmatrix} 0 &2 &3 \\ 1& -1 &4 \end{bmatrix}\begin{bmatrix} 5 &0 \\ -1&1 \\ 3& 2 \end{bmatrix}=\begin{bmatrix} 7 & 8\\ 18& 7 \end{bmatrix}[/latex].

Illustration 5: If the matrix [latex]A=\left[ \begin{matrix} 3-x & 2 & 2 \\ 2 & 4-x & 1 \\ -2 & -4 & -1-x \\ \end{matrix} \right][/latex]

is a singular matrix then find x. Verify whether AAT = I for that value of x.

Solution:

Using the condition of a singular matrix, i.e. |A| = 0, we get the value of x and then substituting the value of x in matrix A and multiplying it to its transpose we will obtain the required result.

Here, A is a singular matrix if |A| = 0, i.e., [latex]\left| \begin{matrix} 3-x & 2 & 2 \\ 2 & 4-x & 1 \\ -2 & -4 & -1-x \\ \end{matrix} \right|=0[/latex]

R3 –> R3 + R2

[latex]\left| \begin{matrix} 3-x & 2 & 2 \\ 2 & 4-x & 1 \\ 0 & -x & -x \\ \end{matrix} \right|=0[/latex]

C2 → C2-C3

[latex]\left| \begin{matrix} 3-x & 0 & 2 \\ 2 & 3-x & 1 \\ 0 & 0 & -x \\ \end{matrix} \right|=0[/latex]

Here, x = 0, 3.

When x = 0, A = [latex]\left[ \begin{matrix} 3 & 2 & 2 \\ 2 & 4 & 1 \\ -2 & -4 & -1 \\ \end{matrix} \right][/latex]

∴ [latex]A{{A}^{T}}=\left[ \begin{matrix} 3 & 2 & 2 \\ 2 & 4 & 1 \\ -2 & -4 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 2 & -2 \\ 2 & 4 & -4 \\ 2 & 1 & -1 \\ \end{matrix} \right][/latex]

= [latex]\left[ \begin{matrix} 17 & 16 & -16 \\ 16 & 21 & -21 \\ -16 & -21 & 21 \\ \end{matrix} \right]\ne I[/latex]

When x = 3, A = [latex]\left[ \begin{matrix} 0 & 2 & 2 \\ 2 & 1 & 1 \\ -2 & -4 & -4 \\ \end{matrix} \right];\,\,\,[/latex]

∴ [latex]A{{A}^{T}}=\left[ \begin{matrix} 0 & 2 & 2 \\ 2 & 1 & 1 \\ -2 & -4 & -4 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 2 & -2 \\ 2 & 1 & -4 \\ 2 & 1 & -4 \\ \end{matrix} \right]=\left[ \begin{matrix} 8 & 4 & -16 \\ 4 & 6 & -12 \\ -16 & -12 & 36 \\ \end{matrix} \right]\ne I[/latex]

Note: simple way to solve is that if A is a singular matrix then |A| = 0 and |AT| = 0. But |I| is 1.

Hence, AAT ≠ I if |A| = 0.

Illustration 6: If the matrix A = [latex]\left[ \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right][/latex]

where a, b, c, are positive real numbers such that abc = 1 and ATA = I then find the value of a3 + b3 + c3.

Solution:

Given: abc = 1 and ATA = I

Here, A= [latex]\left[ \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right].So,{{A}^{T}}=\left[ \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right],[/latex]

Interchanging rows and columns.

⇒ [latex]But{{A}^{T}}A=I(given).[/latex] [latex]\begin{bmatrix}a & b &c \\ b&c & a\\ c & a & b\end{bmatrix}\begin{bmatrix}a & b &c \\ b&c & a\\ c & a & b\end{bmatrix}=\begin{bmatrix}1 & 0 &0 \\ 0&1 & 0\\ 0 & 0 & 1\end{bmatrix}[/latex]

Solving above equation, we have

(a2 + b2 + c2) = 1 and ab + bc + ca = 0  . …..(i)

We know, (a + b + c)2 = (a2 + b2 + c2) + 2(ab + bc + ca)

= 1

or (a + b + c) = 1  ….(ii)

Again ,we have (a3 + b3 + c3 -3abc) = (a + b + c) (a2 + b2 + c2 -ab – bc – ca)

Since abc = 1

Using  (i) and (ii), we have

(a3 + b3 + c3 -3) = 1

or a3 + b3 + c3 =4