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  • Find the matrix A satisfying the matrix equation: {2 1 3 2} A {-3 -2 5 -3}

Find the matrix A satisfying the matrix equation:

\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]

Answers (1)

The given matrix equation is,

\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]

We need to find matrix A.

Let matrix A be of order 2 × 2, and can be represented as

A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]

Then, we have

\begin{aligned} &\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\text { Take L.H.S: }\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]\\ &\text { So, first let us calculate }\\ &\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \end{array}\right]=\mathrm{X} . \mathrm{Y}(\text { say }) \end{aligned}

If A and B are two given matrices and we have to multiply them, then the number of columns in matrix A should be equal to the number of rows in matrix B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Multiply 1st row of matrix X by matching members of 1st column of matrix Y, then finally end by summing them up.

\\(2, 1).(a, c) = (2 $ \times $ a) + (1 $ \times $ c) \\$ \Rightarrow $ (2, 1).(a, c) = 2a + c

\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \end{array}\right]=\left[\begin{array}{l} 2 \mathrm{a}+\mathrm{c} \end{array}\right]

Multiply 1st row of matrix X by matching members of 2nd column of matrix Y, then finally end by summing them up.

\\(2, 1).(b, d) = (2 $ \times $ b) + (1 $ \times $ d) \\$ \Rightarrow $ (2, 1).(b, d) = 2b + d

\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll} 2 a+c & 2 b+d \end{array}\right]

Multiply 2nd row of matrix X by matching members of 1st column of matrix Y, then finally end by summing them up.

\\(3, 2).(a, c) = (3 $ \times $ a) + (2 $ \times $ c) \\$ \Rightarrow $ (3, 2).(a, c) = 3a + 2c

\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & \end{array}\right]

Multiply 2nd row of matrix X by matching members of 2nd column of matrix Y, then finally end by summing them up.

\\(3, 2).(b, d) = (3 $ \times $ b) + (2 $ \times $ d) \\$ \Rightarrow $ (3, 2).(b, d) = 3b + 2d

\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]

Let X.Y = Z

Now, we need to find  \left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix}

Z.Q=\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix}

Where, let Q=\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix}

Multiply 1st row of matrix Z by matching members of 1st column of matrix Q, then finally end by summing them up.

\\(2a + c, 2b + d).(-3, 5) = ((2a + c) $ \times $ -3) + ((2b + d) $ \times $ 5) \\$ \Rightarrow $ (2a + c, 2b + d).(-3, 5) = -6a - 3c + 10b + 5d \\$ \Rightarrow $ (2a + c, 2b + d).(-3, 5) = -6a + 10b - 3c + 5d

\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix} = \begin{bmatrix} -6a+10b-3c+5d & \\ & \end{bmatrix}

Multiply 1st row of matrix Z by matching members of 2nd column of matrix Q, then finally end by summing them up.

\\(2a + c, 2b + d).(2, -3) = ((2a + c) $ \times $ 2) + ((2b + d) $ \times $ -3) \\$ \Rightarrow $ (2a + c, 2b + d).(2, -3) = 4a + 2c - 6b - 3d \\$ \Rightarrow $ (2a + c, 2b + d).(2, -3) = 4a - 6b + 2c - 3d

\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix} = \begin{bmatrix} -6a+10b-3c+5d &4a-6b+2c-3d \\ & \end{bmatrix}

Multiply 2nd row of matrix Z by matching members of 1st column of matrix Q, then finally end by summing them up.

\\(3a + 2c, 3b + 2d).(-3, 5) = ((3a + 2c) $ \times $ -3) + ((3b + 2d) $ \times $ 5) \\$ \Rightarrow $ (3a + 2c, 3b + 2d).(-3, 5) = -9a - 6c + 15b + 10d \\$ \Rightarrow $ (3a + 2c, 3b + 2d).(-3, 5) = -9a + 15b - 6c + 10d

\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix} = \begin{bmatrix} -6a+10b-3c+5d &4a-6b+2c-3d \\ -9a + 15b - 6c + 10d& \end{bmatrix}

Multiply 2nd row of matrix Z by matching members of 2nd column of matrix Q, then finally end by summing them up.

\\(3a + 2c, 3b + 2d).(2, -3) = ((3a + 2c) $ \times $ 2) + ((3b + 2d) $ \times $ -3) \\$ \Rightarrow $ (3a + 2c, 3b + 2d).(2, -3) = 6a + 4c - 9b - 6d \\$ \Rightarrow $ (3a + 2c, 3b + 2d).(2, -3) = 6a - 9b + 4c - 6d

\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix} = \begin{bmatrix} -6a+10b-3c+5d &4a-6b+2c-3d \\ -9a + 15b - 6c + 10d& 6a - 9b + 4c - 6d\end{bmatrix}
So, we have
{\left[\begin{array}{lll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} -6 a+10 b-3 c+5 d & 4 a-6 b+2 c-3 d \\ -9 a+15 b-6 c+10 d & 6 a-9 b+4 c-6 d \end{array}\right]} \\
Now, for  L . H . S=R . H . S 
{\left[\begin{array}{ll} -6 a+10 b-3 c+5 d & 4 a-6 b+2 c-3 d \\ -9 a+15 b-6 c+10 d & 6 a-9 b+4 c-6 d \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]}

If the matrices have the same order then we can write them as,

 \\-6a + 10b - 3c + 5d = 1 $ \ldots $ (i) \\4a - 6b + 2c - 3d = 0 $ \ldots $ (ii) \\-9a + 15b - 6c + 10d = 0 $ \ldots $ (iii) \\6a - 9b + 4c - 6d = 1 $ \ldots $ (iv)

We have to find four variables: a, b, c, d and four equations

So, on adding equations (i) and (iv), we get

 \\(-6a + 10b - 3c + 5d) + (6a - 9b + 4c - 6d) = 1 + 1 \\$ \Rightarrow $ -6a + 6a + 10b - 9b - 3c + 4c + 5d - 6d = 2 \\$ \Rightarrow $ 0 + b + c - d = 2 \\$ \Rightarrow $ d = b + c - 2 $ \ldots $ (v)

Now, adding equations (ii) and (iii), we get

 \\(4a - 6b + 2c - 3d) + (-9a + 15b - 6c + 10d) = 0 + 0 \\$ \Rightarrow $ 4a - 9a - 6b + 15b + 2c - 6c - 3d + 10d = 0 \\$ \Rightarrow $ -5a + 9b - 4c + 7d = 0 $ \ldots $ (vi)

By adding equations (iv) and (vi), we get

 \\(6a - 9b + 4c - 6d) + (-5a + 9b - 4c + 7d) = 1 + 0 \\$ \Rightarrow $ 6a - 5a - 9b + 9b + 4c - 4c - 6d + 7d = 1 \\$ \Rightarrow $ a + 0 + 0 + d = 1 \\$ \Rightarrow $ d = 1 - a $ \ldots $ (vii)

Substituting the value of d from equation (vii) in (v), we get

 \\(1 - a) = b + c - 2 \\$ \Rightarrow $ b + c - 2 - 1 = -a \\$ \Rightarrow $ b + c - 3 = -a \\$ \Rightarrow $ a = 3 - b - c $ \ldots $ (viii)

Now, by substituting values of a and d from equations (vii) and (viii) in equation (iii), we get

 \\-9(3 - b - c) + 15b - 6c + 10(1 - a) = 0 \\$ \Rightarrow $ -9(3 - b - c) + 15b - 6c + 10(1 - (3 - b - c)) = 0 [$\because$ a = 3 - b - c] \\$ \Rightarrow $ -27 + 9b + 9c + 15b - 6c + 10(1 - 3 + b + c) = 0 \\$ \Rightarrow $ -27 + 9b + 9c + 15b - 6c + 10(-2 + b + c) = 0 \\$ \Rightarrow $ -27 + 9b + 9c + 15b - 6c - 20 + 10b + 10c = 0 \\$ \Rightarrow $ 9b + 15b + 10b + 9c - 6c + 10c - 27 - 20 = 0 \\$ \Rightarrow $ 34b + 13c - 47 = 0 \\$ \Rightarrow $ 34b + 13c = 47 $ \ldots $ (ix)

Also, substituting values of a and d from equations (vii) and (viii) in equation (ii), we get

 \\4(3 - b - c) - 6b + 2c - 3(1 - a) = 0 \\$ \Rightarrow $ 12 - 4b - 4c - 6b + 2c - 3(1 - (3 - b - c)) = 0 \\$ \Rightarrow $ 12 - 4b - 4c - 6b + 2c - 3(1 - 3 + b + c) = 0 \\$ \Rightarrow $ 12 - 4b - 4c - 6b + 2c - 3(-2 + b + c) = 0 \\$ \Rightarrow $ 12 - 4b - 4c - 6b + 2c + 6 - 3b - 3c = 0 \\$ \Rightarrow $ -4b - 6b - 3b - 4c + 2c - 3c + 12 + 6 = 0 \\$ \Rightarrow $ -13b - 5c + 18 = 0 \\$ \Rightarrow $ 13b + 5c = 18 $ \ldots $ (x)

On multiplication of equation (ix) by 5 and equation (x) by 13, we get

 \\(ix) $ \Rightarrow $ 5(34b + 13c) = 5 $ \times $ 47 \\$ \Rightarrow $ 170b + 65c = 235 $ \ldots $ (xi) \\(x) $ \Rightarrow $ 13(13b + 5c) = 13 $ \times $ 18 \\$ \Rightarrow $ 169b + 65c = 234 $ \ldots $ (xii)

By subtracting equations (xi) and (xii), we get

 \\(170b + 65c) - (169b + 65c) = 235 - 234 \\$ \Rightarrow $ 170b - 169b + 65c - 65c = 1 \\$ \Rightarrow $ b = 1

By substituting b = 1 in equation (x), we get

 \\13(1) + 5c = 18 \\$ \Rightarrow $ 13 + 5c = 18 \\$ \Rightarrow $ 5c = 18 - 13 \\$ \Rightarrow $ 5c = 5

\\$ \Rightarrow $ c = 1

By substituting b = 1 and c = 1 in equation (viii), we get

\\a = 3 - b - c \\$ \Rightarrow $ a = 3 - 1 - 1 \\$ \Rightarrow $ a = 3 - 2 \\$ \Rightarrow $ a = 1

By substituting a = 1 in equation (vii), we get

 \\d = 1 - a \\$ \Rightarrow $ d = 1 - 1 \\$ \Rightarrow $ d = 0

Thus, the matrix A is

A= \begin{bmatrix} 1 & 1\\1 & 0 \end{bmatrix}

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