The worksheet on Matrix Multiplication consists of all problems related to multiplication properties, Multiplication of Two Matrices, Multiplication of more than two matrices, etc. Solve all the Matrix Multiplication Problems with Solutions on your own to know how to multiply matrices.
The multiplication of two matrices A and B is possible when the number of columns in the A matrix equals the number of rows in the B matrix. In other words, if the order of the matrix A is m x n and the order of the matrix B is n x p, then AB exists and the order of the resultant matrix is m x p. Different tricks and tips are also included in this multiplying matrices notes PDF for easy solving of matrix multiplication questions.
So, first, quickly go through the entire article and gain a high knowledge of matrix problems. The complete matrices information is given in the Matrices Worksheet for Grade 10 PDF articles. This worksheet will help students to solve the problems based on the properties of matrices.
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Multiplying Matrices Worksheet with Answers PDF
Check out the below problems to learn the Multiplication of matrices concepts deeply. The different matrix problems along with their answers are given below.
Problem 1: The matrix is,
\( A = \left[
\begin{matrix}
2& 3\cr
1 & 4\cr
\end{matrix}
\right]
\)Â and \( B = \left[
\begin{matrix}
6 & 8\cr
4 & 3\cr
\end{matrix}
\right]
\). What is the final product value of two matrices?
Solution:
As given in the question,
The matrix A is \( A = \left[
\begin{matrix}
2 & 3\cr
1 & 4\cr
\end{matrix}
\right]
\) and \( B = \left[
\begin{matrix}
6 & 8\cr
4 & 3\cr
\end{matrix}
\right]
\)
Now, we will find the final product value of two matrices.
After, multiplication of two matrix is \( \left[
\begin{matrix}
(2*6) + (3*4) & (2*8) + (3*3) \cr
(1*6) + (4*4) & (1*8) + (4*3) \cr
\end{matrix}\right] \)
\( \left[
\begin{matrix}
12+12 & 16+9\cr
6+16 & 8+12\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
24 & 25\cr
22 & 20\cr
\end{matrix}
\right] \)
Therefore, the final product value of the two matrix is \( \left[
\begin{matrix}
24 & 25\cr
22 & 20\cr
\end{matrix}
\right] \)
Problem 2: Show that the equation ZX = Z and XZ = Z holds if the matrix is,
\( X = \left[
\begin{matrix}
1 & 2\cr
3 & 4\cr
\end{matrix}
\right]
\)Â and \( Z = \left[
\begin{matrix}
0 & 0\cr
0 & 0\cr
\end{matrix}
\right]
\)?
Solution:
As given in the question,
The matrix X is \( X = \left[
\begin{matrix}
1 & 2\cr
3 & 4\cr
\end{matrix}
\right]
\) and \( Z = \left[
\begin{matrix}
0 & 0\cr
0 & 0\cr
\end{matrix}
\right]
\)
Now, we will equate the ZX =Z and XZ=Z
First, look at the equation ZX =Z
Calculating ZX is,
\( \left[
\begin{matrix}
1 & 2\cr
3 & 4\cr
\end{matrix}
\right]
\). \(Â \left[
\begin{matrix}
0 & 0\cr
0 & 0\cr
\end{matrix}
\right]
\)
= \( \left[
\begin{matrix}
0(1)+0(3) & 0(2)+0(4)\cr
0(1)+0(3) & 0(2)+0(4)\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
0 & 0\cr
0 & 0\cr
\end{matrix}
\right] \) = Z
Now, equate XZ=Z, that is
\( \left[
\begin{matrix}
1 & 2\cr
3 & 4\cr
\end{matrix}
\right]
\). \(Â \left[
\begin{matrix}
0 & 0\cr
0 & 0\cr
\end{matrix}
\right]
\) =Â \( \left[
\begin{matrix}
0(1)+0(3) & 0(2)+0(4)\cr
0(1)+0(3) & 0(2)+0(4)\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
0 & 0\cr
0 & 0\cr
\end{matrix}
\right] \) = Z
Hence, XZ is equate to Z and ZX is equate to Z.
Problem 3: Let \( A =\left[
\begin{matrix}
2 & 5\cr
-1 & 3\cr
\end{matrix}
\right]
\) and \( B= \left[
\begin{matrix}
2 & 5\cr
-1 & 3\cr
\end{matrix}
\right]
\). Find the values of AB and BA. Is AB = BA?
Solution:
Given that, the A and B matrices.
The matrix A is the order of 2 × 2 and the matrix B is the order of 2 × 2.
So, the number of columns in the A matrix equals the number of rows in the B matrix. Hence, AB can be found. Also, the number of columns in B equals the number of rows in A. Hence, BA can also be found.
Now, find the value of AB,
AB = \( \left[
\begin{matrix}
2 & 5\cr
-1 & 3\cr
\end{matrix}
\right]
\) \( \left[
\begin{matrix}
2 & 5\cr
-1 & 3\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
2*1 + 5*4Â & 2*1 + 5*(-2)\cr
-1*1 +3 x 4 &(-1)*1 +3*(-2)\cr
\end{matrix}
\right]
\)Â = \( \left[
\begin{matrix}
22 & -8\cr
11 & -7\cr
\end{matrix}
\right]
\)
Next, find the value of the BA.
\( BA= \left[
\begin{matrix}
2 & 5\cr
-1 & 3\cr
\end{matrix}
\right]
\) \( \left[
\begin{matrix}
2 & 5\cr
-1 & 3\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
1*2 + 1*(-1)& 1*5 + 1*3\cr
4*2+(-2)*(-1) & 4*5 + (-2)*3\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
11 & 8\cr
0 & 14\cr
\end{matrix}
\right]
\).
Clearly, \( \left[
\begin{matrix}
21 & -8\cr
11 & -7\cr
\end{matrix}
\right]
\) ≠\( \left[
\begin{matrix}
11& 8\cr
0 & 14\cr
\end{matrix}
\right]
\)
Therefore, the value of AB and the value of BA is not equal. Hence it is AB ≠BA.
Problem 4: Let \( A =\left[
\begin{matrix}
11 & 4\cr
-5 & 2\cr
\end{matrix}
\right]
\) and \( B= \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right]
\). Prove that AB = BA = A?
Solution:
Given that, the A and B matrices values.
The matrix A is the order of 2 × 2 and the matrix B is the order of 2 × 2.
So, the number of columns in matrix A equals the number of rows in matrix B. Hence, AB can be found. Also, the number of columns in B equals the number of rows in A. So, BA can also be found.
Now, find the value of AB,
AB = \( \left[
\begin{matrix}
11 & 4\cr
-5 & 2\cr
\end{matrix}
\right]
\) \( \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
11*1 + 4*0 & 11*0+ 4*1\cr
-5*1 + 2*0 & -5*0+ 2*1\cr
\end{matrix}
\right]
\)Â = \( \left[
\begin{matrix}
11 & 4\cr
-5 & 2\cr
\end{matrix}
\right]
\) = A
Next, find the value of the BA.
\( BA= \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right]
\) \( \left[
\begin{matrix}
11 & 4\cr
-5 & 2\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
1*11 + 0*(-5) & 1*4 + 0*2\cr
0*11 + 1*(-5) & 0*4 + 1*2\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
11 & 2\cr
-5 & 4\cr
\end{matrix}
\right]
\) = A
Clearly, \( \left[
\begin{matrix}
11 & 4\cr
-5 & 2\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
11& 4\cr
-5 & 2\cr
\end{matrix}
\right]
\)
Therefore, the value of AB and BA is equal. Hence its proved AB = BA = A.
Problems 5: Multiply the given matrix by 6. The matrix is,
\( A = \left[
\begin{matrix}
2 & 4\cr
5 & 3\cr
\end{matrix}
\right]
\) What is the final value of Matrix?
Solution:
As given in the question,
The matrix A is \( A = \left[
\begin{matrix}
2 & 4\cr
5 & 3\cr
\end{matrix}
\right]
\)
Now, multiply given matrix by 6.
After, multiplication the matrix is \( A = \left[
\begin{matrix}
2*6 & 4*6\cr
5*6 & 3*6\cr
\end{matrix}
\right]
\)
Therefore, the final matrix value of A is \( A = \left[
\begin{matrix}
12 & 24\cr
30 & 18\cr
\end{matrix}
\right]
\)
Problem 6: The matrix is,
\( A = \left[
\begin{matrix}
1& 0\cr
2 & 3\cr
\end{matrix}
\right]
\)Â and \( B = \left[
\begin{matrix}
7 & 9\cr
4 & 2\cr
\end{matrix}
\right]
\). What is the final product value of two matrices?
Solution:
As given in the question,
The matrix A is \( A = \left[
\begin{matrix}
1 & 0\cr
2 & 3\cr
\end{matrix}
\right]
\) and \( B = \left[
\begin{matrix}
7 & 9\cr
4 & 2\cr
\end{matrix}
\right]
\)
Now, we need to find out the final product value of two matrices.
After the multiplication of two matrices is \( \left[
\begin{matrix}
(1*7) + (0*4) & (1*9) + (0*2) \cr
(2*7) + (3*4) & (2*9) + (3*2) \cr
\end{matrix}\right] \)
\( \left[
\begin{matrix}
7+0 & 9+0\cr
14+12 & 18+6\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
7 & 9\cr
26 & 24\cr
\end{matrix}
\right] \)
Therefore, the final product value of the two matrices is \( \left[
\begin{matrix}
7 & 9\cr
26 & 24\cr
\end{matrix}
\right] \)
Problem 7: For the matrix \( C = \left[
\begin{matrix}
12 & 15\cr
-3 & -8\cr
5 &Â 10\cr
\end{matrix}
\right] \), Answer the following.
(i) What is the order of matrix C?
(ii) Find the (2, 1), (1, 2) and (3, 2) elements?
(iii) Is it a square matrix or a rectangular matrix?
Solution:
As in the given question, matrix C is given.
Now, we will find the following.
(i) The order of matrix A is 3 × 2.
(ii) In the given matrix, the (2, 1)th element is -3.
Next, the (1, 2)th element is 15.
The (3, 2)th element is 10.
(iii) The matrix A will have 3 rows and 2 columns.
The number of rows is not equal to the number of columns.
Therefore matrix A is a rectangular matrix.
Problem 8: If \( A = \left[
\begin{matrix}
cosθ & 0\cr
0 & cosθ\cr
\end{matrix}
\right] \) and \( B = \left[
\begin{matrix}
sinθ & 0\cr
0 & sinθ\cr
\end{matrix}
\right] \) then show that A2 +B2 = I.
Solution:
In the given question, the matrices are \( A = \left[
\begin{matrix}
cosθ & 0\cr
0 & cosθ\cr
\end{matrix}
\right] \) and \( B = \left[
\begin{matrix}
sinθ & 0\cr
0 & sinθ\cr
\end{matrix}
\right] \)
Now, we will prove that the value of A2 +B2 = I.
First, find the value of A2
So, A2 = A x A = \( \left[
\begin{matrix}
cosθ & 0\cr
0 & cosθ\cr
\end{matrix}
\right] \) x \( \left[
\begin{matrix}
cosθ & 0\cr
0 & cosθ\cr
\end{matrix}
\right] \)
=Â \( \left[
\begin{matrix}
cos^2θ+0 &(0+0)\cr
(0+0) & cos^2θ+0\cr
\end{matrix}
\right] \)
The value of \( A^2 = \left[
\begin{matrix}
cos^2θ & 0\cr
0 & cos^2θ\cr
\end{matrix}
\right] \)
Now, find the value of B^2
So, the value is B x B = \( \left[
\begin{matrix}
Sinθ & 0\cr
0 & Sinθ\cr
\end{matrix}
\right] \) x \( \left[
\begin{matrix}
Sinθ & 0\cr
0 & Sinθ\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
sin^2θ+0 &(0+0)\cr
(0+0) & sin^2θ+0\cr
\end{matrix}
\right]\)
The value of \( B^2= \left[
\begin{matrix}
sin^2θ & 0\cr
0 & sin^2θ\cr
\end{matrix}
\right] \)
Now, adding the corresponding terms. We get,
A^2+B^2 = \( \left[
\begin{matrix}
cos^2θ+sin^2θ & 0 \cr
0 & cos^2θ+sin^2θ \cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right] \) = I.
Problem 9: Find the value of 4Z where
\( Z= \left[
\begin{matrix}
7 & 8\cr
9 & 10\cr
\end{matrix}
\right] \)
Solution:
The given matrix is \( Z = \left[
\begin{matrix}
7 & 8\cr
9 & 10\cr
\end{matrix}
\right] \)
To find the value of 4Z, we need to multiply the value of 4 by all the elements of matrix Z.
i.e., 4Z = \( \left[
\begin{matrix}
4*7 & 4*8\cr
4*9 & 4*10\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
28 & 32\cr
36 & 40\cr
\end{matrix}
\right] \)
Therefore, the required matrix is 4Z \( =\left[
\begin{matrix}
28 & 32\cr
36 & 40\cr
\end{matrix}
\right] \).
Problem 10: The matrix is,
\( A = \left[
\begin{matrix}
1 & 0\cr
2 & 4\cr
\end{matrix}
\right]
\)Â and \( B = \left[
\begin{matrix}
6 & 8\cr
4 & 3\cr
\end{matrix}
\right]
\). What is the final product value of two matrices?
Solution:
As given in the question,
The matrix A is \( A = \left[
\begin{matrix}
1 & 0\cr
2 & 4\cr
\end{matrix}
\right]
\) and \( B = \left[
\begin{matrix}
6 & 8\cr
4 & 3\cr
\end{matrix}
\right]
\)
Now, we will find the final product value of two matrices.
After, multiplication of two matrix is \( \left[
\begin{matrix}
(1*6) + (0*4) & (1*8) + (0*3) \cr
(2*6) + (4*4) & (2*8) + (4*3) \cr
\end{matrix}\right] \)
\( \left[
\begin{matrix}
6+0 & 8+0\cr
12+16 & 16+12\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
6 & 8\cr
28 & 28\cr
\end{matrix}
\right] \)
Therefore, the final product value of the two matrix is \( \left[
\begin{matrix}
6 & 8\cr
28 & 28\cr
\end{matrix}
\right] \)
Problem 11: Verify that A2 = I when \( A = \left[
\begin{matrix}
5 & -4\cr
6 & -5\cr
\end{matrix}
\right] \) ?
Solution:
As given in the question, the value of A is \( \left[
\begin{matrix}
5 & -4\cr
6 & -5\cr
\end{matrix}
\right] \)
Now, we will verify that A2 = I.
First, we will find the value of the A2.
The value of A2 is A x A = \( \left[
\begin{matrix}
5 & -4\cr
6 & -5\cr
\end{matrix}
\right] \) x \( \left[
\begin{matrix}
5 & -4\cr
6 & -5\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(25-24) & (20-20)\cr
(30-30) & (-24+25)\cr
\end{matrix}
\right] \)
So, the final value is \( \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right] \)
Hence, proved the value of A2 is I.
Problem 12: Prove that A(B + C) = AB + AC. The matrix of \( A=\left[
\begin{matrix}
1 & 3\cr
5 & -1\cr
\end{matrix}
\right] \), \( B= \left[
\begin{matrix}
1 & -1 & 2\cr
3 & 5 & 2\cr
\end{matrix}
\right] \), and \( C = \left[
\begin{matrix}
1 & 3 & 2\cr
-4 & 1 & 3\cr
\end{matrix}
\right] \) ?
Solution:
In the given questions are, the matrices are \( A=\left[
\begin{matrix}
1 & 3\cr
5 & -1\cr
\end{matrix}
\right] \), \( B= \left[
\begin{matrix}
1 & -1 & 2\cr
3 & 5 & 2\cr
\end{matrix}
\right] \), and \( C = \left[
\begin{matrix}
1 & 3 & 2\cr
-4 & 1 & 3\cr
\end{matrix}
\right] \)
Now, we will prove that A(B + C) = AB + AC.
First, take the L.H.S part that is A(B + C).
(B+C) = \( \left[
\begin{matrix}
1 & -1 & 2\cr
3 & 5 & 2\cr
\end{matrix}
\right] \) + \( \left[
\begin{matrix}
1 & 3 & 2\cr
-4 & 1 & 3\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(1+1) & (-1+3) & (2+2)\cr
(3-4) & (5+1) & (2+3)\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
2 & 2 & 4\cr
-1 & 6 & 5\cr
\end{matrix}
\right] \)
Now, multiply the B+C value with A, We get,
A(B+C) = \( \left[
\begin{matrix}
1 & 3 \cr
5 & -1 \cr
\end{matrix}
\right] \). \( \left[
\begin{matrix}
2 & 2 & 4\cr
-1 & 6 & 5\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(2-3)Â & (2+18) & (4+15)\cr
(10+1) & (10-6) & (20-5)\cr
\end{matrix}
\right] \)
A(B+C) = \( \left[
\begin{matrix}
-1 & 20 & 19\cr
11 & 4 & 15\cr
\end{matrix}
\right] \)
Next, find the value of the R.H.S that is AB+AC
First take AB= \( \left[
\begin{matrix}
1 & 3\cr
5 & -1\cr
\end{matrix}
\right] \) x \(\left[
\begin{matrix}
1 & -1 & 2\cr
3 & 5 & 2\cr
\end{matrix}
\right] \)
= \(\left[
\begin{matrix}
(1+9) & (-1+15) & (2+6)\cr
(5-3) & (-5-5) & (10-2)\cr
\end{matrix}
\right] \)
\( AB=\left[
\begin{matrix}
10 & 14 & 8\cr
2 & -10 & 8\cr
\end{matrix}
\right] \)
Now, find the value of AC,
AC = \( \left[
\begin{matrix}
1 & 3\cr
5 & -1\cr
\end{matrix}
\right] \) x \( \left[
\begin{matrix}
1 & 3 & 2\cr
-4 & 1 & 3\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(1-12)Â & (3+3) & (2+19)\cr
(5+4) & (15-1) & (10-3)\cr
\end{matrix}
\right] \)
So, the value of AC is \( \left[
\begin{matrix}
-11Â & 6 & 11\cr
9 & 14 & 7\cr
\end{matrix}
\right] \)
Now, let us add AB+ AC.
AB+AC = \( \left[
\begin{matrix}
10Â & 14 & 8\cr
2 & -10 & 8\cr
\end{matrix}
\right] \) + \( \left[
\begin{matrix}
-11Â & 6 & 11\cr
9 & 14 & 7\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(10-11)Â & (14+6) & (8+11)\cr
(2+9) & (-10+14) & (8+7)\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
-1 & 20 & 19\cr
11 & 4 & 15\cr
\end{matrix}
\right] \)
The value of A(B+C) and AB+AC both are Equal.
Hence, prove that L.H.S = R.H.S.